Maharashtra Board Class 12 Chemistry Chapter 6 Chemical Kinetics Solutions

Chemistry Class 12 Chapter 6 Exercise Solutions

1. Choose The Most Correct Option.

Question 1. The rate law for the reaction \(aA + bB \to P\) is rate = \(k[A][B]\). The rate of reaction doubles if
(a) concentrations of A and B are both doubled.
(b) [A] is doubled and [B] is kept constant
(c) [B] is doubled and [A] is halved
(d) [A] is kept constant and [B] is halved.
Answer: (b) [A] is doubled and [B] is kept constant
In simple words: If the rate law is \(rate = k[A][B]\), doubling [A] while keeping [B] constant will double the rate of reaction because the rate is directly proportional to [A].

🎯 Exam Tip: Understanding the rate law is crucial. Pay attention to how changes in reactant concentrations affect the overall reaction rate, as this indicates the order of the reaction with respect to each reactant.

 

Question 2. The order of the reaction for which the units of rate constant are \(mol\;dm^{-3}\;s^{-1}\) is
(a) 1
(b) 3
(c) 0
(d) 2
Answer: (c) 0
In simple words: For a zero-order reaction, the rate constant's units are concentration per unit time, which is \(mol\;L^{-1}\;s^{-1}\) or \(mol\;dm^{-3}\;s^{-1}\).

🎯 Exam Tip: Memorizing the units of rate constants for different reaction orders (0, 1, 2) is a quick way to determine reaction order from given unit data in exams.

 

Question 3. The rate constant for the reaction \(2N_2O_5(g) \to 2N_2O_4(g) + O_2(g)\) is \(4.98 \times 10^{-4}\;s^{-1}\). The order of reaction is
(a) 2
(b) 1
(c) 0
(d) 3
Answer: (b) 1
In simple words: The unit of the rate constant (\(s^{-1}\)) directly indicates that this is a first-order reaction.

🎯 Exam Tip: The units of the rate constant are a direct indicator of the overall order of the reaction. For first-order reactions, the unit is \(s^{-1}\).

 

Question 4. Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9% completion will be
(a) t
(b) 2t
(c) t/2
(d) 3t
Answer: (d) 3t
In simple words: For a first-order reaction, the time required for 99.9% completion is approximately three times the time required for 90% completion. This is a characteristic property derived from the integrated rate law.

🎯 Exam Tip: Understand the integrated rate law for first-order reactions. Problems involving completion percentages (like 90%, 99%, 99.9%) are common and can often be solved by recognizing proportional relationships.

 

Question 5. Slope of the graph \(ln[A]_t\) versus t for first order reaction is
(a) -k
(b) k
(c) k/2.303
(d) -k/2.303
Answer: (a) -k
In simple words: The integrated rate law for a first-order reaction is \(\ln[A]_t = -kt + \ln[A]_0\). Comparing this to \(y = mx + c\), the slope (m) is -k.

🎯 Exam Tip: Graphical representations of integrated rate laws are important. Know the expected plots and their slopes/intercepts for zero, first, and second-order reactions.

 

Question 6. What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
(a) 12 h
(b) 3 h
(c) 1.5 h
(d) 6 h
Answer: (d) 6 h
In simple words: The concentration decreases from 0.8 M to 0.4 M (one half-life), then from 0.4 M to 0.2 M (another half-life). Since two half-lives occur in 12 hours, one half-life is 6 hours.

🎯 Exam Tip: For first-order reactions, half-life is constant. You can determine the number of half-lives that have passed by observing the concentration reduction (e.g., 0.8 M to 0.2 M is two half-lives: 0.8 -> 0.4 -> 0.2). Then divide the total time by the number of half-lives.

 

Question 7. The reaction, \(3ClO_3^- \to ClO_3^- + 2Cl^-\) occurs in two steps,
(i) \(2ClO^- \to ClO_2^-\)
(ii) \(ClO_2^- + ClO^- \to ClO_3^- + Cl^-\)
The reaction intermediate is
(a) Cl-
(b) ClO2-
(c) ClO3-
(d) ClO-
Answer: (b) ClO2-
In simple words: A reaction intermediate is a species that is formed in one step of a reaction mechanism and consumed in a subsequent step. In this case, \(ClO_2^-\) is formed in step (i) and consumed in step (ii).

🎯 Exam Tip: To identify reaction intermediates, look for species that appear as products in an earlier elementary step and as reactants in a later elementary step, and do not appear in the overall balanced equation.

 

Question 8. The elementary reaction \(O_2(g) + O(g) \to 2O_2(g)\) is
(a) unimolecular and second order
(b) bimolecular and first order
(c) bimolecular and second order
(d) unimolecular and first order
Answer: (c) bimolecular and second order
In simple words: An elementary reaction's molecularity is the number of reacting species, which is two (\(O_2\) and O), making it bimolecular. Its order is also two, corresponding to its molecularity.

🎯 Exam Tip: For elementary reactions, molecularity (number of colliding species) is equal to the order of the reaction. This is not necessarily true for complex reactions.

 

Question 9. Rate law for the reaction, \(2NO + Cl_2 \to 2NOCl\) is rate = \(k[NO_2]^2[Cl_2]\). Thus k would increase with
(a) increase of temperature
(b) increase of concentration of NO
(c) increase of concentration of \(Cl_2\)
(d) increase of concentrations of both \(Cl_2\) and NO
Answer: (a) increase of temperature
In simple words: The rate constant (k) is dependent on temperature (Arrhenius equation), but it is independent of reactant concentrations.

🎯 Exam Tip: Distinguish between rate and rate constant. The rate of reaction depends on both concentration and temperature, while the rate constant depends only on temperature and the nature of the reactants.

 

Question 10. For an endothermic reaction, \(X \rightleftharpoons Y\). If \(E_f\) is activation energy of the forward reaction and \(E_r\) that for reverse reaction, which of the following is correct?
(a) \(E_f = E_r\)
(b) \(E_f < E_r\)
(c) \(E_f > E_r\)
(d) \(\Delta H = E_f - E_r\) is negative
Answer: (c) \(E_f > E_r\)
In simple words: For an endothermic reaction, the products have higher energy than the reactants, meaning the activation energy for the forward reaction (\(E_f\)) must be greater than that for the reverse reaction (\(E_r\)). \(\Delta H = E_f - E_r\) would be positive.

🎯 Exam Tip: Relate activation energies to the enthalpy change of a reaction. For endothermic reactions, the forward activation energy is larger than the reverse, leading to a positive \(\Delta H\).

 

2. Answer The Following In One Or Two Sentences.

Question 1. For the reaction, \(N_2(g) + 3H_2(g) \to 2NH_3(g)\), what is the relationship among \(\frac{d[N_2]}{dt}\), \(\frac{d[H_2]}{dt}\) and \(\frac{d[NH_3]}{dt}\)?
Answer:
\(N_2(g) + 3H_2(g) \to 2NH_3(g)\)
From the above reaction, when 1 mole of \(N_2\) reacts, 3 moles of \(H_2\) are consumed and 2 moles of \(NH_3\) are formed.
If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of \(N_2\) then, \(R = -\frac{d[N_2]}{dt}\)

\(\therefore\) Rate of consumption of \(H_2 = -\frac{d[H_2]}{dt}\)

\(= 3\left(-\frac{d[N_2]}{dt}\right) = 3R\)

\(\therefore R = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt}\)
Rate of formation of \(NH_3 = +\frac{d[NH_3]}{dt}\)

\(= 2\left(-\frac{d[N_2]}{dt}\right) = 2R\)

\(\therefore R = -\frac{d[N_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}\)
Hence the rate of reaction in terms of concentration changes in \(N_2\), \(H_2\) and \(NH_3\) may be represented as,
Rate (R) \( = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}\)
In simple words: The rate of reaction is expressed by dividing the change in concentration of a reactant or product by the stoichiometric coefficient and time, with a negative sign for reactants and a positive sign for products.

🎯 Exam Tip: Remember that the rate of reaction is always positive. The negative sign for reactants indicates their decreasing concentration, while the positive sign for products indicates their increasing concentration.

 

Question 2. For the reaction,
\(CH_3Br(aq) + OH^-(aq) \to CH_3OH^-(aq) + Br^-(aq)\), rate law is rate = \(k[CH_3Br][OH^-]\)
a. How does reaction rate changes if \([OH^-]\) is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Answer:
Solution:
Given:
(a) Rate = \(R = k [CH_3Br] \times [OH^-]\)
If \(R_1\) and \(R_2\) are initial and final rates of reaction then,
\(R_1 = k [CH_3Br]_1 \times [OH^-]_1\)
\(R_2 = k [CH_3Br]_1 \times [OH^-]_2\)
If \([OH^-]\) is decreased by a factor of 5, then \([OH^-]_2 = \frac{1}{5}[OH^-]_1\)

\(= k [CH_3Br]_1 \times \frac{1}{5} \times [OH^-]_1\)

\(\frac{R_2}{R_1} = \frac{k [CH_3Br]_1 \times \frac{1}{5} \times [OH^-]_1}{k [CH_3Br]_1 \times [OH^-]_1} = \frac{1}{5}\)

\(\therefore R_2 = \frac{1}{5} \times R_1\)
The rate will be decreased by a factor of 5.
(b) \(R_1 = k \times [CH_3Br] \times [OH^-]\)
If concentrations of both reactants are doubled, then \([CH_3Br]_2 = 2[CH_3Br]\) and \([OH^-]_2 = 2[OH^-]\)
\(R_2 = k \times 2 \times [CH_3Br] \times 2 \times [OH^-]\)

\(\therefore \frac{R_2}{R_1} = \frac{k \times 2 \times [CH_3Br] \times 2 \times [OH^-]}{k \times [CH_3Br] \times [OH^-]} = 4\)

\(\therefore R_2 = 4R_1\)
Rate will be increased 4 time.
In simple words: If the rate law is first order with respect to \([OH^-]\), decreasing its concentration by a factor of 5 will decrease the rate by a factor of 5. If both reactant concentrations are doubled, and the reaction is first order with respect to each, the rate will increase by a factor of \(2 \times 2 = 4\).

🎯 Exam Tip: To solve such problems, always write down the given rate law. Then, substitute the new concentrations into the rate law and compare the new rate with the initial rate to find the factor of change.

 

Question 3. What is the relationship between coefficients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coefficients are the exponents?
Answer:
Explanation: Consider the following reaction, \(aA + bB \to products\)
If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
\(R \alpha [A]^a [B]^b\)

\(\therefore R = k [A]^a [B]^b\)
where [A] = concentration of A and
[B] = concentration of B
The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.
a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.
The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
\(R = k [A]^x [B]^y\)
For example, if x = 1 and y = 2, then,
\(R = k [A] [B]^2\)
The coefficients of reactants in a balanced chemical equation (a and b) are not necessarily equal to the exponents (x and y) in the rate law (order of reaction). The exponents (x and y) are determined experimentally and reflect how the rate depends on reactant concentrations.
The coefficients are equal to the exponents *only* when the reaction is an elementary reaction (occurs in a single step). For complex reactions, the rate law is determined by the slowest step (rate-determining step) of the mechanism, and its exponents may or may not match the stoichiometric coefficients of the overall reaction.
In simple words: The exponents in a rate law (reaction order) show how concentration affects rate and are found experimentally. They are only equal to the stoichiometric coefficients in the balanced equation if the reaction occurs in a single step (elementary reaction).

🎯 Exam Tip: Always remember that reaction order is an experimentally determined value, not derived from the stoichiometric coefficients of the balanced chemical equation, unless explicitly stated that the reaction is elementary.

 

Question 4. Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules: The basic requirements of a reaction are that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.
However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.
(ii) Energy requirement (Activation energy): The colliding molecules must possess a certain minimum energy called activation energy required for breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.
(iii) Orientation of reactant molecules: The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.
This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, \(A - B + C \to A + B - C\). For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रासायनिक अभिक्रिया में "फायदेमंद टक्कर" और "कोई अभिक्रिया नहीं" के लिए आणविक अभिविन्यास के महत्व को दर्शाता है। शीर्ष आरेख में, एक अणु (A-B) दूसरे अणु (C) के साथ इस तरह से टकराता है कि C, B से जुड़ सकता है, जिससे A + B-C बनता है - इसे एक फायदेमंद टक्कर माना जाता है। निचले आरेख में, अणु (C) अणु (A-B) के एक अलग हिस्से (A) से टकराता है; क्योंकि C को B से जुड़ना है, यह टक्कर कोई अभिक्रिया नहीं करती है। यह दर्शाता है कि एक सफल अभिक्रिया के लिए अणुओं का सही अभिविन्यास आवश्यक है।

\(A-B + C \to A + B-C\) (fruitful collision)
\(C + A-B \to \text{no reaction}\)
In simple words: Not all collisions lead to a reaction because molecules must collide with sufficient energy (activation energy) to break and form bonds, and they must also have the correct orientation for the reacting parts to interact effectively.

🎯 Exam Tip: The collision theory explains reaction rates. Remember the two key factors for effective collisions: sufficient kinetic energy (greater than or equal to activation energy) and proper orientation of molecules.

 

Question 5. What is the activation energy of a reaction?
Answer:
Activation energy: The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.
In simple words: Activation energy is the minimum energy that reacting molecules must possess for a reaction to occur, representing an energy barrier that must be overcome.

🎯 Exam Tip: Activation energy is a critical concept in kinetics. It determines how fast a reaction proceeds; a higher activation energy typically means a slower reaction rate.

 

Question 6. What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, \(mol\;L^{-1}\;s^{-1}\).
(b) For second order reaction,
Rate = \(k \times [Reactant]^2\)

\(\therefore k = \frac{\text{Rate}}{[Reactant]^2} = \frac{mol\;L^{-1}\;s^{-1}}{(mol\;L^{-1})^2}\)
Units of k are \(mol^{-1}\;L\;s^{-1}\)
In simple words: For zero order, the rate constant unit is concentration/time (\(mol\;L^{-1}\;s^{-1}\)). For second order, it's \(L\;mol^{-1}\;s^{-1}\).

🎯 Exam Tip: Derive the units of the rate constant from the rate law: \(Rate = k[A]^n\). The unit of rate is always concentration/time, and the unit of concentration is usually \(mol\;L^{-1}\).

 

Question 7. Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as \(k = A \times e^{-E_a/RT}\)
where
k = Rate constant at absolute temperature T
\(E_a\) = Energy of activation
R = Gas constant
A = Frequency factor or pre-exponential factor.
In simple words: The Arrhenius equation relates the rate constant of a reaction to temperature and activation energy, with 'A' being the frequency factor, 'Ea' the activation energy, 'R' the gas constant, and 'T' the absolute temperature.

🎯 Exam Tip: Understand how each term in the Arrhenius equation (k, A, \(E_a\), R, T) influences the rate of reaction. This equation is fundamental for explaining temperature dependence of reaction rates.

 

Question 8. What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.
Rate determining step: The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.
In simple words: The rate-determining step is the slowest elementary step in a multi-step reaction mechanism, which dictates the overall rate of the chemical reaction.

🎯 Exam Tip: Identifying the rate-determining step is key to predicting the rate law for complex reactions. The rate law of the overall reaction is determined by the stoichiometry of the reactants involved in the slowest step.

 

Question 9. Write the relationships between rate constant and half life of first order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period \(t_{1/2}\) is, \(t_{1/2} = \frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (\(t_{1/2}\)) is, \(t_{1/2} = \frac{[A]_0}{2k}\) where k is the rate constant and \([A]_0\) is initial concentration of the reactant.
In simple words: For a first-order reaction, half-life is constant and independent of initial concentration (\(t_{1/2} = 0.693/k\)). For a zero-order reaction, half-life is directly proportional to initial concentration (\(t_{1/2} = [A]_0/2k\)).

🎯 Exam Tip: Memorize these relationships for zero and first-order reactions. They are often used in problems to calculate rate constants or half-lives, or to determine the order of a reaction from half-life data.

 

Question 10. How do half lives of the first order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, \(t_{1/2}\) is given by, \(t_{1/2} = \frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.
(B) Zero order reaction,
\(t_{1/2} = \frac{[A]_0}{2k}\) where \([A]_0\) is initial concentration of the reactant.
Hence, half life period increases with the increase in concentration of the reactant.
In simple words: A first-order reaction's half-life doesn't change with initial concentration, while a zero-order reaction's half-life increases as the initial concentration increases.

🎯 Exam Tip: This distinction is a key characteristic to differentiate between first and zero-order reactions. Experiments showing half-life dependence on initial concentration can help determine reaction order.

 

3. Answer The Following In Brief.

Question 1. How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinitesimal change in concentration (-dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, \(A \to B\), let an infinitesimal change in A be –dc in time dt, then Rate \( = -\frac{d[A]}{dt}\).
Hence, it is represented as,

\(\therefore \text{Instantaneous rate} = -\frac{d[A]}{dt}\)
The negative sign indicates a decrease in the concentration of A.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अभिक्रिया में अभिकारक की सांद्रता [A] को समय (t) के विरुद्ध प्लॉट करता है। जैसे-जैसे समय बढ़ता है, अभिकारक की सांद्रता घटती जाती है। वक्र पर किसी भी बिंदु पर खींची गई स्पर्श रेखा का ढलान -\(\frac{d[A]}{dt}\) द्वारा दर्शाया गया है, जो उस विशिष्ट क्षण पर तात्कालिक अभिक्रिया दर को इंगित करता है।
(For the reactant)
It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.
(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).
Let dB be an infinitesimal change in the concentration of product B in time dt, then
Rate \( = \frac{d[B]}{dt} = \frac{dx}{dt}\)
Hence,
Instantaneous rate \( = \frac{dx}{dt}\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अभिक्रिया में उत्पाद की सांद्रता (x) को समय (t) के विरुद्ध प्लॉट करता है। जैसे-जैसे समय बढ़ता है, उत्पाद की सांद्रता बढ़ती जाती है। वक्र पर किसी भी बिंदु पर खींची गई स्पर्श रेखा का ढलान \(\frac{dx}{dt}\) द्वारा दर्शाया गया है, जो उस विशिष्ट क्षण पर तात्कालिक अभिक्रिया दर को इंगित करता है।
(For the product)
It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.
The instantaneous rate is more useful in obtaining the rate law integrated equations.
In simple words: The instantaneous rate of reaction is determined by finding the slope of a tangent drawn to the concentration-time curve at a specific point, representing the rate at that exact moment.

🎯 Exam Tip: Remember to differentiate between average rate and instantaneous rate. Instantaneous rate provides a more precise measure of reaction speed at a particular time and is obtained graphically.

 

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero order reaction is given by, \( t_{1/2} = \frac{[\text{A}]_0}{2k} \) where [A]0 is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of \( t_{1/2} \) is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \( \frac{1}{2k} \), where k is the rate constant.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख शून्य-कोटि अभिक्रिया के लिए दर (Rate) को अभिकर्मक की सांद्रता (Concentration) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जिसका ढलान शून्य है, जो यह बताता है कि अभिक्रिया की दर अभिकर्मक की सांद्रता पर निर्भर नहीं करती है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख शून्य-कोटि अभिक्रिया के लिए अर्ध-आयु काल (t½) को प्रारंभिक सांद्रता ([A]₀) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जो मूलबिंदु से गुजरती है, जिसका ढलान \( \frac{1}{2k} \) है, जो दर्शाता है कि अर्ध-आयु काल प्रारंभिक सांद्रता के सीधे आनुपातिक है।

Question vi. What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer: Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation: Consider an acid hydrolysis reaction of an ester like methyl acetate.
\[ \text{CH}_3\text{COOCH}_3(\text{aq}) + \text{H}_2\text{O(l)} \xrightarrow{\text{H(aq)}} \text{CH}_3\text{COOH}(\text{aq}) + \text{CH}_3\text{OH}(\text{aq}) \] Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k' [CH3COOCH3] x [H2O]
Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.
This is because solvent water being in a large excess, its concentration remains constant. Hence, [H2O] = constant = k"
Rate = k [CH3COOCH3] × [H2O]
= k [CH3COOCH3] x k"
= k' x k″ x [CH3COOCH3]
If k' x k" = k, then Rate = k [CH3COOCH3],
This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.
In simple words: Pseudo-first-order reactions are reactions that are truly of a higher order (like second order) but appear to be first order experimentally because one reactant is present in such large excess that its concentration effectively remains constant throughout the reaction.

🎯 Exam Tip: Understanding why these reactions are 'pseudo' is key. The constant concentration of one reactant simplifies the rate law, making it seem like a first-order reaction.

 

Question vii. What are the requirements for the colliding reactant molecules to lead to products?
Answer: Collisions of reactant molecules : The basic requirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.
However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.
Energy requirement (Activation energy) : The colliding molecules must possess a certain minimum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.
Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.
This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A - B + C → A + B - C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.
A-B+C → A+B-C (fruitful collision)
C+A → B- no reaction
In simple words: For a reaction to happen, reactant molecules must collide with sufficient energy (activation energy) and with the correct orientation. Inefficient collisions, lacking enough energy or proper alignment, will not form products.

🎯 Exam Tip: Remember the three key factors for effective collisions: collision frequency, activation energy, and proper orientation. All three must be met for a successful reaction.

 

Question viii. How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer: (i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO3 in the presence of small amount of MnO2 is very fast but very slow in the absence of MnO2.
\[ 2\text{KClO}_3(\text{s}) \xrightarrow{\text{MnO}_2} 2\text{KCl}(\text{s}) + 3\text{O}_2(\text{g}) \]
(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.
(v) Due to lowering of energy of activation, (Ea) the number of molecules possessing Ea increases, hence the rate of the reaction increases.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संभावित ऊर्जा आरेख उत्प्रेरित और अनुत्प्रेरित अभिक्रियाओं के लिए अभिक्रिया निर्देशांक (reaction coordinate) के विरुद्ध ऊर्जा (potential energy) को दर्शाता है। उत्प्रेरक की उपस्थिति में सक्रियण ऊर्जा (activation energy) कम हो जाती है, जिससे अभिक्रिया के लिए एक निचला ऊर्जा मार्ग (lower energy path) उपलब्ध होता है। (vi) The rate constant = \( k = \text{A x e}^{-\text{Ea/RT}} \) where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.
(vii) The catalyst does not change the extent of the reaction but hastens the reaction.
(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.
In simple words: A catalyst speeds up a reaction by providing an alternative reaction pathway with a lower activation energy, increasing the number of molecules that can react successfully. It's not consumed and doesn't change the final products.

🎯 Exam Tip: Focus on how a catalyst lowers activation energy and provides an alternative pathway, which is effectively depicted in a potential energy diagram.

 

Question ix. Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer: (a) By Arrhenius equation, \( k = \text{Ax e}^{-\text{Ea/RT}} \) where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As Ea increases, the rate constant and rate of the reaction decreases.
(b) As temperature increases Ea/RT decreases but due to negative sign, k and rate increase with the increase in temperature.
In simple words: The Arrhenius equation shows that higher activation energy leads to a slower reaction rate, while higher temperature leads to a faster reaction rate because more molecules have enough energy to overcome the activation barrier.

🎯 Exam Tip: Remember the inverse relationship between activation energy and rate constant, and the direct relationship between temperature and rate constant (due to the exponential term).

 

Question x. Derive the integrated rate law for first order reaction.
Answer: Consider following gas phase reaction,
\[ \text{A(g)} \longrightarrow \text{B(g)} + \text{C(g)} \]
At start \( \text{P}_0 \quad 0 \quad 0 \)
At time t \( \text{P}_0-\text{x} \quad \text{x} \quad \text{x} \)
Let initial pressure of A be \( \text{P}_0 \) at \( \text{t} = 0 \). If after time t the pressure of a A decreases by x then the partial pressures of the substances will be, \( \text{P}_{\text{A}} = \text{P}_0 - \text{x} \); \( \text{P}_{\text{B}} = \text{x} \) and \( \text{P}_{\text{C}} = \text{x} \)
Total pressure will be,
\( \text{P}_{\text{T}} = \text{P}_0 - \text{x} + \text{x} + \text{x} = \text{P}_0 + \text{x} \)

\( \implies \text{x} = \text{P}_{\text{T}} - \text{P}_0 \)
The partial pressures at time t will be,
\( [\text{A}]_{\text{t}} = \text{P}_0-\text{x} = \text{P}_0 - (\text{P}_{\text{T}}-\text{P}_0) = 2\text{P}_0 - \text{P}_{\text{T}} \)
\( [\text{A}]_0 = \text{P}_0 \)
\[ k = \frac{2.303}{t} \log_{10} \frac{[\text{A}]_0}{[\text{A}]_t} \]
\[ k = \frac{2.303}{t} \log_{10} \frac{\text{P}_0}{2\text{P}_0-\text{P}_{\text{T}}} \]
In simple words: The integrated rate law for a first-order reaction describes how the concentration of a reactant changes over time. For gas-phase reactions, this is expressed in terms of partial pressures, showing a logarithmic relationship between initial pressure, pressure at time 't', and the rate constant.

🎯 Exam Tip: Be careful with the pressure terms. Understand that \( \text{P}_{\text{T}} \) (total pressure) is used to find 'x' which then relates back to the partial pressure of A at time 't'.

 

Question xi. How will you represent first-order reactions graphically.
Answer: (1) A graph of rate of a reaction and concentration : The differential rate law for first-order reaction, A → Products is represented as, Rate = \( -\frac{d[\text{A}]}{d t}=k[\text{A}] \)

\( \implies \) Rate = k x \( [\text{A}]_{\text{t}} \) (y = mx). When the rate of a first order reaction is plotted against concentration, \( [\text{A}]_{\text{t}} \), a straight line graph is obtained.
With the increase in the concentration \( [\text{A}]_{\text{t}} \), rate R, increases. The slope of the line gives the value of rate constant k.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख प्रथम-कोटि अभिक्रिया के लिए दर (Rate) को अभिकर्मक की सांद्रता (Concentration) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जो मूलबिंदु से शुरू होकर ऊपर की ओर जाती है, जिसका ढलान 'k' है। यह दर्शाता है कि अभिक्रिया की दर सांद्रता के सीधे आनुपातिक है। (2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration \( [\text{A}]_{\text{t}} \), of the reactant decreases exponentially with time. The variation in the concentration can be represented as,
\( [\text{A}]_{\text{t}} = [\text{A}]_0 \text{e}^{-kt} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख प्रथम-कोटि अभिक्रिया के लिए अभिकर्मक की सांद्रता ([A]t) को समय (t) के विरुद्ध दर्शाता है। इसमें एक वक्र रेखा है जो समय के साथ सांद्रता में घातीय कमी (exponential decrease) दर्शाती है। where \( [\text{A}]_0 \) and \( [\text{A}]_{\text{t}} \) are initial and final concentrations the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.
(3) A graph of \( \log_{10} (a-x) \) against time t :
\[ k = \frac{2.303}{t} \log_{10} \left( \frac{a}{a-x} \right) \]

\( \implies \) \[ k = \frac{2.303}{t} [\log_{10} a - \log_{10} (a-x)] \]

\( \implies \frac{k \times t}{2.303} = \log_{10} a - \log_{10} (a-x) \)

\( \implies \log_{10} (a-x) = -\frac{k}{2.303} \times t + \log_{10} a \) (y = mx + c)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख प्रथम-कोटि अभिक्रिया के लिए \( \log_{10}(a-x) \) को समय (Time) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जिसका ढलान नकारात्मक है (\( -\frac{k}{2.303} \)) और यह दर्शाता है कि समय के साथ \( \log_{10}(a-x) \) का मान रैखिक रूप से घटता है। When \( \log_{10}(a-x) \) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.
(4) A graph of half-life period and concentration : The half-life period, \( t_{1/2} \) of a first order reaction is given by, where k is the rate constant.
For the given reaction at a constant temperature, \( t_{1/2} \) is constant and independent of the concentration of the reactant.
Hence when a graph of \( t_{1/2} \) is plotted against concentration, a straight line parallel to the concentration axis (slope = zero) is obtained.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख प्रथम-कोटि अभिक्रिया के लिए अर्ध-आयु काल (t½) को सांद्रता (Concentration) के विरुद्ध दर्शाता है। इसमें एक सीधी क्षैतिज रेखा है, जिसका ढलान शून्य है, यह दर्शाता है कि प्रथम-कोटि अभिक्रिया का अर्ध-आयु काल अभिकर्मक की प्रारंभिक सांद्रता से स्वतंत्र होता है। (5) A graph of \( \log_{10} \left(\frac{a}{a-x}\right) \) against time : The rate constant, for a first order reaction is represented as,
\[ k = \frac{2.303}{t} \log_{10} \left(\frac{a}{a-x}\right) = \frac{2.303}{t} \log_{10} \frac{[\text{A}]_0}{[\text{A}]_t} \] where \( [\text{A}]_0 \) and \( [\text{A}]_{\text{t}} \) are the respective initial and final concentrations of the reactant after time t.

\( \implies \log_{10} \left(\frac{a}{a-x}\right) = \frac{k}{2.303} \times t \) (y = mx)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख प्रथम-कोटि अभिक्रिया के लिए \( \log_{10} \left(\frac{a}{a-x}\right) \) को समय (t) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जो मूलबिंदु से शुरू होकर ऊपर की ओर जाती है, जिसका ढलान \( \frac{k}{2.303} \) है। When \( \log_{10}\left(\frac{a}{a-x}\right) \) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.
In simple words: First-order reactions can be graphically represented in several ways: rate versus concentration (straight line, positive slope), concentration versus time (exponential decay curve), log concentration versus time (straight line, negative slope), half-life versus concentration (horizontal straight line), and log ratio of concentrations versus time (straight line, positive slope).

🎯 Exam Tip: Be familiar with all common graphical representations of first-order reactions, as questions often involve interpreting or sketching these graphs.

 

Question xii. Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer: Consider following gas phase reaction,
\[ \text{A(g)} \longrightarrow \text{B(g)} + \text{C(g)} \]
At start \( \text{P}_0 \quad 0 \quad 0 \)
At time t \( \text{P}_0-\text{x} \quad \text{x} \quad \text{x} \)
Let initial pressure of A be \( \text{P}_0 \) at \( \text{t} = 0 \). If after time t the pressure of a A decreases by x then the partial pressures of the substances will be, \( \text{P}_{\text{A}} = \text{P}_0 - \text{x} \); \( \text{P}_{\text{B}} = \text{x} \) and \( \text{P}_{\text{C}} = \text{x} \)
Total pressure will be,
\( \text{P}_{\text{T}} = \text{P}_0 - \text{x} + \text{x} + \text{x} = \text{P}_0 + \text{x} \)

\( \implies \text{x} = \text{P}_{\text{T}} - \text{P}_0 \)
The partial pressures at time t will be,
\( [\text{A}]_{\text{t}} = \text{P}_0-\text{x} = \text{P}_0 - (\text{P}_{\text{T}}-\text{P}_0) = 2\text{P}_0 - \text{P}_{\text{T}} \)
\( [\text{A}]_0 = \text{P}_0 \)
\[ k = \frac{2.303}{t} \log_{10} \frac{[\text{A}]_0}{[\text{A}]_t} \]
\[ k = \frac{2.303}{t} \log_{10} \frac{\text{P}_0}{2\text{P}_0-\text{P}_{\text{T}}} \]
In simple words: For first-order gas-phase reactions, the integrated rate law can be derived using partial pressures instead of concentrations. The initial pressure (\( \text{P}_0 \)) and the total pressure at time 't' (\( \text{P}_{\text{T}} \)) are used to determine the rate constant 'k'.

🎯 Exam Tip: When dealing with gas-phase reactions, remember to convert concentrations to partial pressures using Dalton's law of partial pressures to apply the rate laws correctly.

 

Question xiii. What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer: Definition: Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.
Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \( = -\frac{d[\text{A}]}{dt} \)
By rate law,
Rate = k x \( [\text{A}]^0 \) = k

\( \implies - d[\text{A}] = k \times dt \)
If \( [\text{A}]_0 \) is the initial concentration of the reactant A at t = 0 and \( [\text{A}]_{\text{t}} \) is the concentration of A present after time t, then by integrating above equation,
\[ \int_{[\text{A}]_0}^{[\text{A}]_t} -d[\text{A}] = \int_{t=0}^{t=t} k dt \]
\[ - \left( [\text{A}]_{\text{t}} - [\text{A}]_0 \right) = k \left( t - 0 \right) \]

\( \implies [\text{A}]_0 - [\text{A}]_{\text{t}} = kt \)

\( \implies k = \frac{[\text{A}]_0 - [\text{A}]_{\text{t}}}{t} \)
This is the integrated rate law expression for rate constant for zero order reaction.

\( \implies k \times t = [\text{A}]_0 - [\text{A}]_{\text{t}} \)

\( \implies [\text{A}]_{\text{t}} = -kt + [\text{A}]_0 \)
For a zero order reaction :
The rate of reaction is R = k \( [\text{A}]^0 \) = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., \( \text{mol dm}^{-3} \text{ s}^{-1} \).
In simple words: A zero-order reaction's rate doesn't change with reactant concentration. Its integrated rate law shows a linear decrease in concentration over time, and the rate constant unit is concentration per unit time (e.g., mol dm\(^{-3}\) s\(^{-1}\)).

🎯 Exam Tip: Remember that the distinguishing feature of a zero-order reaction is its constant rate, independent of concentration. This is reflected in its linear concentration-time plot.

 

Question xiv. How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer: (a) By Arrhenius equation,
Rate constant = \( k = \text{A x e}^{-\text{Ea/RT}} \) where A is a fre-quency factor.

\( \implies \ln k = \ln \text{A} - \frac{\text{Ea}}{\text{RT}} \)

\( \implies 2.303 \log_{10} k = 2.303 \log_{10} \text{A} - \frac{\text{Ea}}{\text{RT}} \)

\( \implies \log_{10} k = \log_{10} \text{A} - \frac{\text{Ea}}{2.303 \text{RT}} \)

\( \implies \log_{10} k = - \left( \frac{\text{Ea}}{2.303 \text{R}} \right) \times \frac{1}{\text{T}} + \log_{10} \text{A} \) (y = mx + C)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख \( \log_{10} k \) को \( \frac{1}{\text{T}} \) (परम तापमान का व्युत्क्रम) के विरुद्ध दर्शाता है। इसमें एक सीधी रेखा है जिसका ढलान नकारात्मक है (\( -\frac{\text{Ea}}{2.303 \text{R}} \)), जिससे सक्रियण ऊर्जा (Ea) निर्धारित की जा सकती है। When \( \log_{10}k \) is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation Ea, is obtained as follows :
Slope = \( \frac{\text{Ea}}{2.303 \text{R}} \)

\( \implies \text{Ea} = 2.303 \text{R x Slope} \)
(b) For the given reaction, rate constants \( k_1 \) and \( k_2 \) are measured at two different temperatures \( \text{T}_1 \) and \( \text{T}_2 \) respectively. Then
\[ \log_{10} \frac{k_2}{k_1} = \frac{\text{Ea}(\text{T}_2-\text{T}_1)}{2.303\text{R}\times\text{T}_1\times\text{T}_2} \]
where Ea is the energy of activation.
Hence by substituting appropriate values, energy of activation Ea for the reaction is determined.
In simple words: Activation energy (Ea) can be found by plotting \( \log_{10} k \) against \( 1/\text{T} \) (graphical method, where Ea is derived from the slope) or by using the Arrhenius equation with rate constants measured at two different temperatures.

🎯 Exam Tip: Remember both methods for calculating activation energy. The graphical method provides a visual understanding, while the two-point method is practical for calculations.

 

Question xv. Explain graphically the effect of temperature on the rate of reaction.
Answer: (i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier, i. e. activation energy Ea increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (Ea) also increases, hence the rate of the reaction increases with increase in temperature.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेख टकराने वाले अणुओं के अंश (Fraction of collisions) को संघट्ट ऊर्जा (Collision energy) के विरुद्ध दो अलग-अलग तापमानों (T1 और T2, जहाँ T2 > T1) पर दर्शाता है। सक्रियण ऊर्जा (Ea) से अधिक ऊर्जा वाले अणुओं के लिए, उच्च तापमान (T2) पर छायांकित क्षेत्र बड़ा होता है, जो दर्शाता है कि उच्च तापमान पर प्रभावी संघट्टों का अंश बढ़ जाता है। (iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding Ea is greater at higher temperature T2 than at lower temperature T1. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy Ea represents fraction of collisions of activated molecules having energy Ea or greater.
In simple words: Increasing temperature raises the kinetic energy of molecules, leading to a larger fraction of molecules having energy equal to or greater than the activation energy. This increases effective collisions and thus the reaction rate. Graphically, the area under the Maxwell-Boltzmann distribution curve beyond Ea expands at higher temperatures.

🎯 Exam Tip: Relate increased temperature to increased kinetic energy, increased fraction of activated molecules (those with Ea or more), and ultimately, increased reaction rate. The Maxwell-Boltzmann distribution curve is a key visual aid.

 

Question xvii. For the reaction 2A + B → products, find the rate law from the following data.

[A]/M	[B]/M	rate/M s\(^{-1}\)
0.3	0.05	0.15
0.6	0.05	0.30
0.6	0.2	1.20

Answer: Solution:
Given : 2A + B → Products
Rates : \( R_1 = 0.15 \text{ Ms}^{-1} \) \( R_2 = 0.3 \text{ Ms}^{-1} \)
\( [\text{A}]_1 = 0.3 \text{ M} \) \( [\text{A}]_2 = 0.6 \text{ M} \)
\( [\text{B}]_1 = 0.05 \text{ M} \) \( [\text{B}]_2 = 0.05 \text{ M} \)
(i) If order of the reaction in A is x and in B is y then, by rate law,
Rate = \( R_1 = k [\text{A}]_1^{\text{x}} [\text{B}]_1^{\text{y}} \) and
\( R_2 = k [\text{A}]_2^{\text{x}} [\text{B}]_2^{\text{y}} \)

\( \implies \frac{R_2}{R_1} = \frac{k [\text{A}]_2^{\text{x}} [\text{B}]_2^{\text{y}}}{k [\text{A}]_1^{\text{x}} [\text{B}]_1^{\text{y}}} \)
\[ \frac{0.30}{0.15} = \left( \frac{0.6}{0.3} \right)^{\text{x}} \left( \frac{0.05}{0.05} \right)^{\text{y}} \]
\( 2 = 2^{\text{x}} \)

\( \implies \text{x} = 1 \). Hence, the reaction has order one in A.
(ii) Rates: \( R_1 = 0.3 \text{ Ms}^{-1} \) \( R_2 = 1.2 \text{ Ms}^{-1} \)
\( [\text{A}]_1 = 0.6 \text{ M} \) \( [\text{A}]_2 = 0.6 \text{ M} \)
\( [\text{B}]_1 = 0.05 \text{ M} \) \( [\text{B}]_2 = 0.2 \text{ M} \)

\( \implies \frac{R_2}{R_1} = \frac{k [\text{A}]_2^{\text{x}} [\text{B}]_2^{\text{y}}}{k [\text{A}]_1^{\text{x}} [\text{B}]_1^{\text{y}}} \)
\[ \frac{1.2}{0.3} = \left( \frac{0.6}{0.6} \right)^{\text{x}} \left( \frac{0.2}{0.05} \right)^{\text{y}} \]
\( 4 = (4)^{\text{y}} \)

\( \implies \text{y} = 1 \). Hence the reaction has order one in B.
The order of overall reaction = \( n = n_{\text{A}} + n_{\text{B}} = 1 + 1 = 2 \)
Rate = \( R = k[\text{A}] \times [\text{B}] \)

\( \implies k = \frac{R}{[\text{A}][\text{B}]} \)
\[ k = \frac{R_1}{[\text{A}]_1[\text{B}]_1} \]
\[ k = \frac{0.15 \text{ Ms}^{-1}}{0.3 \text{ M} \times 0.05 \text{ M}} \]

\( k = 10 \text{ M}^{-1}\text{s}^{-1} \)
(i) Rate law : Rate = k [A] x [B]
Rate constant = \( k = 10 \text{ M}^{-1}\text{s}^{-1} \)
Order of the reaction = 2
In simple words: By analyzing how the reaction rate changes when reactant concentrations are varied, we found that the rate is directly proportional to the concentration of A and also directly proportional to the concentration of B. This gives us the rate law: Rate = k[A][B], with an overall reaction order of 2 and a rate constant of 10 M\(^{-1}\)s\(^{-1}\).

🎯 Exam Tip: For determining rate laws from experimental data, identify experiments where one reactant's concentration changes while others remain constant. This simplifies the calculation of individual reaction orders.

4. Solve

Question i. In a first order reaction, the concentration of reactant decreases from 20 mmol dm-3 to 8 mmol dm-3 in 38 minutes. What is the half life of reaction? (28.7 min) Solution: Given: \( \text{[A]}_0 = 20 \text{ mmol dm}^{-3} \); \( \text{[A]}_t = 8 \text{ mmol dm}^{-3} \); \( t = 38 \text{ mm} \); \( t_{1/2} = ? \) \( k = \frac{2.303}{t} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \) \( = \frac{2.303}{38} \log_{10} \frac{20}{8} \) \( = \frac{2.303}{38} \times 0.3979 \) \( = 0.02411 \text{ min}^{-1} \)
\( t_{1/2} = \frac{0.693}{k} \) \( = \frac{0.693}{0.02411} = 28.74 \text{ min} \)
Answer: Half life period = 28.74 min
In simple words: We first calculate the rate constant (k) using the given initial and final concentrations and time. Then, we use this rate constant to find the half-life period for the first order reaction.

🎯 Exam Tip: Remember the integrated rate law for first order reactions and the formula for half-life. Ensure units are consistent throughout the calculation.

 

Question ii. The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min) Solution: Given: \( t_{1/2} = 1.7 \text{ hr} \); \( \text{[A]}_0 = 100 \); \( \text{[A]}_t = 100 - 20 = 80 \); \( t = ? \) \( t_{1/2} = \frac{0.693}{k} \)
\( \implies k = \frac{0.693}{t_{1/2}} \) \( = \frac{0.693}{1.7} \) \( = 0.4076 \text{ hr}^{-1} \)
\( k = \frac{2.303}{t} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \)
\( \implies t = \frac{2.303}{k} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \) \( = \frac{2.303}{0.4076} \log_{10} \frac{100}{80} \) \( = \frac{2.303}{0.4076} \log_{10} 1.25 \) \( = \frac{2.303 \times 0.09691}{0.4076} \) \( = 0.5476 \text{ hr} \) \( = 0.5476 \times 60 \text{ min} \) \( = 32.86 \text{ min}. \)
Answer: Time required = \( t = 32.86 \text{ min} \)
In simple words: First, we use the given half-life to calculate the rate constant. Then, using the integrated rate law, we find the time needed for 20% of the reactant to react, converting hours to minutes for the final answer.

🎯 Exam Tip: Pay attention to unit conversions (hours to minutes) and use the correct formulas for half-life and the integrated rate law for first order reactions. Logarithmic calculations should be precise.

 

Question iii. The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 °C is 3.7 × 10-5 s-1. What is the rate constant at 30°C? (R = 8.314 J/K mol) (7.4 × 10-5) Solution: Given: \( E_a = 104 \text{ kJ mol}^{-1} = 104 \times 10^3 \text{ J mol}^{-1} \) \( k_1 = 3.7 \times 10^{-5} \text{ s}^{-1} \); \( T_1 = 273 + 25 = 298 \text{ K} \) \( T_2 = 273 + 30 = 303 \text{ K} \); \( k_2 = ? \)
\( \log_{10} \frac{k_2}{k_1} = \frac{E_a (T_2 - T_1)}{2.303 \times R \times T_1 \times T_2} \)
\( \implies \log_{10} k_2 - \log_{10} k_1 = \frac{E_a (T_2 - T_1)}{2.303 \times R \times T_1 \times T_2} \)
\( \implies \log_{10} k_2 = \log_{10} k_1 + \frac{E_a (T_2 - T_1)}{2.303 \times R \times T_1 \times T_2} \) \( = \log_{10} (3.7 \times 10^{-5}) + \frac{104 \times 10^3 \times (303 - 298)}{2.303 \times 8.314 \times 298 \times 303} \) \( = -4.4318 + \frac{104 \times 10^3 \times 5}{2.303 \times 8.314 \times 298 \times 303} \) \( = -4.4318 + \frac{520 \times 10^3}{1733682.025} \) \( = -4.4318 + 0.300 \) \( = -4.1318 \)
\( k_2 = \text{Antilog } (-4.1318) \) \( = \text{Antilog } (5.8682 - 10) \) \( = 7.382 \times 10^{-5} \text{ s}^{-1} \)
Answer: \( k_2 = 7.382 \times 10^{-5} \text{ s}^{-1} \)
In simple words: We use the Arrhenius equation in its integrated form to relate the rate constants at two different temperatures. By substituting the given activation energy, initial rate constant, and temperatures, we calculate the rate constant at the new temperature.

🎯 Exam Tip: Convert activation energy to Joules per mole and temperature to Kelvin. Ensure careful calculation of logarithms and antilogarithms for accuracy.

 

Question iv. What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol) Solution: Given : \( k_2 = 2k_1 \), \( T_1 = 303 \text{ K} \); \( T_2 = 313 \text{ K} \); \( E_a = ? \)
\( \log_{10} \frac{k_2}{k_1} = \frac{E_a (T_2 - T_1)}{2.303 \times R \times T_1 \times T_2} \)
\( \implies E_a = \frac{2.303 \times R \times T_1 \times T_2}{(T_2 - T_1)} \times \log_{10} \frac{k_2}{k_1} \) \( = \frac{2.303 \times 8.314 \times 303 \times 313}{(313 - 303)} \times \log_{10} \frac{2k_1}{k_1} \) \( = \frac{2.303 \times 8.314 \times 303 \times 313}{10} \times \log_{10} 2 \) \( = \frac{2.303 \times 8.314 \times 303 \times 313}{10} \times 0.3010 \) \( = \frac{17.291 \times 94839 \times 0.3010}{10} \) \( = \frac{52119.789 \times 0.3010}{10} \) \( = \frac{15684.075}{10} \) \( = 54660 \text{ J} \) \( = 54.66 \text{ kJ} \)
Answer: Energy of activation = \( E_a = 54.66 \text{ kJ} \)
In simple words: We use the Arrhenius equation to determine the activation energy. Given that the rate constant doubles, we substitute \( k_2 = 2k_1 \) and the two temperatures into the equation to solve for \( E_a \).

🎯 Exam Tip: Ensure temperatures are in Kelvin. Recognize that "rate constant doubles" means the ratio \( k_2/k_1 \) is 2, simplifying the logarithmic term. Convert the final answer to kJ/mol if requested.

 

Question v. The rate constant of a reaction at 500°C is 1.6 × 103 M-1 s-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M-1 s-1) Solution: Given: \( k = 1.6 \times 10^3 \text{ M}^{-1}\text{s}^{-1} \); \( T = 273 + 500 = 773 \text{ K} \) Energy of activation = \( E_a = 56 \text{ kJ mol}^{-1} \) \( = 56000 \text{ J mol}^{-1} \) Frequency factor = \( A = ? \) By Arrhenius equation, \( k = A \times e^{-E_a/RT} \)
\( \implies \ln k = \ln A - \frac{E_a}{RT} \)
\( \implies 2.303 \log_{10} k = 2.303 \log_{10} A - \frac{E_a}{RT} \)
\( \implies \log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT} \)
\( \implies \log_{10} A = \log_{10} k + \frac{E_a}{2.303 RT} \) \( = \log_{10} (1.6 \times 10^3) + \frac{56000}{2.303 \times 8.314 \times 773} \) \( = 3.204 + \frac{56000}{14820.73} \) \( = 3.204 + 3.778 \) \( = 6.982 \)
\( \implies A = \text{Antilog } 6.982 \) \( = 9.59 \times 10^6 \text{ M}^{-1}\text{s}^{-1} \)
Answer: Frequency factor = \( A = 9.59 \times 10^6 \text{ M}^{-1}\text{s}^{-1} \)
In simple words: We use the Arrhenius equation to calculate the frequency factor. By plugging in the given rate constant, activation energy, and temperature (converted to Kelvin), we solve for 'A'.

🎯 Exam Tip: Ensure all units are consistent (J for energy, K for temperature). Remember to use the correct value of R (8.314 J/K mol) and to handle logarithms and antilogarithms carefully.

 

Question vi. Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion. Solution: Given: For 99.9% completion, if \( \text{[A]}_0 = 100 \), \( \text{[A]}_t = 100 - 99.9 = 0.1 \) For 90% completion, if \( \text{[A]}_0 = 100 \), \( \text{[A]}_t = 100 - 90 = 10 \)
\( k = \frac{2.303}{t} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \)
\( \implies t = \frac{2.303}{k} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \) If \( t_1 \) and \( t_2 \) are the times required for 99.9 % and 90 % completion of reaction respectively, then \( \frac{t_1}{t_2} = \frac{\frac{2.303}{k} \log_{10} \frac{100}{0.1}}{\frac{2.303}{k} \log_{10} \frac{100}{10}} \) \( = \frac{\log_{10} 1000}{\log_{10} 10} \) \( = \frac{\log_{10} 10^3}{\log_{10} 10} \) \( = \frac{3 \log_{10} 10}{\log_{10} 10} \) \( = 3 \)
\( \implies t_1 = 3t_2 \)
Answer: Time required for 99.9% completion of a first order reaction is three time the time required for 90% completion of the reaction.
In simple words: Using the first order integrated rate law, we calculate the time for 99.9% and 90% completion separately. By taking the ratio of these times, we prove that the time for 99.9% completion is three times that for 90% completion.

🎯 Exam Tip: Clearly define initial and final concentrations for each percentage completion. Show the derivation steps logically, ensuring proper use of logarithmic properties to simplify the ratio.

 

Question vii. A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min) Solution: Given: \( t = 40 \text{ min} \); \( \text{[A]}_0 = 100 \); \( \text{[A]}_t = 100 - 30 = 70 \); \( t_{1/2} = ? \)
\( k = \frac{2.303}{t} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \) \( = \frac{2.303}{40} \log_{10} \frac{100}{70} \) \( = \frac{2.303}{40} \log_{10} 1.4286 \) \( = \frac{2.303 \times 0.1549}{40} \) \( = 8.918 \times 10^{-3} \text{ min}^{-1} \)
\( t_{1/2} = \frac{0.693}{k} \) \( = \frac{0.693}{8.918 \times 10^{-3}} \) \( = 77.70 \text{ min} \)
Answer: Half life period = 77.70 min.
In simple words: First, we use the given time and percentage decomposition to find the rate constant of the first-order reaction. Then, we use this rate constant to calculate the reaction's half-life.

🎯 Exam Tip: Correctly calculate the final concentration ([A]t) after 30% decomposition. Ensure accurate logarithmic calculations to determine the rate constant before calculating the half-life.

 

Question viii. The rate constant for the first order reaction is given by \( \log_{10} k = 14.34 - 1.25 \times 10^4 T \). Calculate activation energy of the reaction. (239.3 kJ/mol) Solution: Given: \( \log_{10} k = 14.34 - \frac{1.25 \times 10^4}{T} \) (1) From Arrhenius equation we can write, \( \log_{10} k = \log_{10} A - \frac{E_a}{2.303 R T} \) (2) By comparing equations (1) and (2), \( \frac{E_a}{2.303 \times R} = 1.25 \times 10^4 \)
\( \implies E_a = 1.25 \times 10^4 \times 2.303 \times R \) \( = 1.25 \times 10^4 \times 2.303 \times 8.314 \) \( = 23.93 \times 10^4 = 239.3 \text{ kJ mol}^{-1} \) [Note : Frequency factor A may also be calculated as follows : \( \log_{10} A = 14.34 \) \( \implies A = \text{Antilog } 14.34 = 2.188 \times 10^4 \)]
Answer: Energy of activation = \( E_a = 239.3 \text{ kJ mol}^{-1} \).
In simple words: By comparing the given linear equation for \( \log_{10} k \) with the Arrhenius equation's logarithmic form, we can directly extract the activation energy. The slope of the line in the Arrhenius plot corresponds to \( -E_a/(2.303R) \).

🎯 Exam Tip: Recognize the form of the given equation and directly relate it to the Arrhenius equation's logarithmic form. Ensure the value of R is used correctly and that units for activation energy are in kJ/mol.

 

Question ix. What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10-9) Solution: Given : \( T = 300 \text{ K} \); \( E_a = 50 \text{ kJ mol}^{-1} \) \( = 50 \times 10^3 \text{ J mol}^{-1} \) The fraction of molecules undergoing fruitful collisions is \( f = e^{-E_a/RT} \) \( = e^{- (50 \times 10^3) / (8.314 \times 300)} \) \( = e^{- 50000 / 2494.2} \) \( = e^{-20.0465} \)
\( \implies \log_{10} f = \frac{-20.0465}{2.303} \) \( = -8.7045 \)
\( \implies f = \text{Antilog } (-8.7045) \) \( = \text{Antilog } (9.2955 - 10) \) \( = 1.974 \times 10^{-9} \) \( \approx 2 \times 10^{-9} \)
Answer: Fraction of molecules undergoing collision = \( 2 \times 10^{-9} \)
In simple words: We calculate the fraction of molecules with energy equal to or greater than the activation energy using the Boltzmann factor, \( e^{-E_a/RT} \). We substitute the given activation energy, temperature, and gas constant to find this fraction.

🎯 Exam Tip: Ensure activation energy is in Joules and temperature in Kelvin. Carefully perform the exponential calculation and its conversion to base 10 logarithm if needed, paying attention to negative exponents and antilogarithms.

 

Activity:

Question 1. If you wish to determine the reaction order and rate constant for the reaction, \( 2AB_2 \to A_2 + 2B_2 \).
(a) What data would you collect?
(b) How would you use these data to determine whether the reaction is zeroth or first order?
Answer:
(a) To determine the reaction order and rate constant, one would collect concentration-time data for the reactant \( AB_2 \). This typically involves measuring the concentration of \( AB_2 \) at various time intervals as the reaction proceeds. Alternatively, one could conduct a series of experiments varying the initial concentration of \( AB_2 \) and measuring the initial rate of reaction for each.
(b) To determine if the reaction is zeroth or first order:
**For zeroth order:**
- Plot concentration of \( AB_2 \) versus time. If a straight line with a negative slope is obtained, the reaction is zeroth order. The slope would be \( -k \).
- The half-life \( t_{1/2} \) for a zeroth order reaction is directly proportional to the initial concentration (\( t_{1/2} = \frac{[A]_0}{2k} \)). If \( t_{1/2} \) changes with initial concentration, it could be zeroth order.
**For first order:**
- Plot \( \ln[\text{AB}_2] \) (or \( \log_{10}[\text{AB}_2] \)) versus time. If a straight line with a negative slope is obtained, the reaction is first order. The slope would be \( -k \) (or \( -k/2.303 \)).
- The half-life \( t_{1/2} \) for a first order reaction is independent of the initial concentration (\( t_{1/2} = \frac{0.693}{k} \)). If \( t_{1/2} \) remains constant regardless of the initial concentration, it is a first order reaction.
In simple words: To find the reaction order and rate constant, you collect data on how reactant concentration changes over time. By plotting this data in specific ways (concentration vs. time for zeroth order, log of concentration vs. time for first order), you can see if it forms a straight line, indicating the order, and then use the slope to find the rate constant.

🎯 Exam Tip: Remember the graphical methods for determining reaction order (concentration vs. time, ln[concentration] vs. time, 1/[concentration] vs. time). Also, recall how half-life depends on initial concentration for different orders (independent for first order, dependent for zeroth and second order).

 

Question 2. The activation energy for two reactions are \( E_a \) and \( E'_a \) with \( E_a > E'_a \). If the temperature of reacting system increases from \( T_1 \) to \( T_2 \), predict which of the following is correct?
(a) \( \frac{k'_1}{k_1} = \frac{k'_2}{k_2} \)
(b) \( \frac{k'_1}{k_1} > \frac{k'_2}{k_2} \)
(c) \( \frac{k'_1}{k_1} < \frac{k'_2}{k_2} \)
(d) \( \frac{k'_1}{k_1} < 2 < \frac{k'_2}{k_2} \) \( k \) values are rate constants at lower temperatures and \( k' \) values are rate constants at higher temperatures.
Answer: (c) \( \frac{k'_1}{k_1} < \frac{k'_2}{k_2} \)
In simple words: When temperature increases, the rate constants for both reactions (k and k') will increase. However, the reaction with higher activation energy (\( E_a \)) will experience a more significant increase in rate constant for the same temperature rise compared to the reaction with lower activation energy (\( E'_a \)). Thus, the ratio of final to initial rate constant will be larger for the reaction with higher activation energy.

🎯 Exam Tip: Understand that reactions with higher activation energy are more sensitive to temperature changes. A larger \( E_a \) means a greater increase in rate constant for a given temperature rise, thus a larger ratio \( k_2/k_1 \).

 

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

Question 1. Write the expressions for rates of reaction for : \( 2N_2O_5(g) \to 4NO_2(g) + O_2(g) \)?
Answer:For the given reaction, Rate of reaction = \( R = - \frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} \) \( = + \frac{1}{4} \frac{d[\text{NO}_2]}{dt} \) \( = + \frac{d[\text{O}_2]}{dt} \)
In simple words: The rate of reaction is expressed by how quickly reactants disappear or products appear, adjusted by their stoichiometric coefficients. For reactants, it's negative, and for products, it's positive.

🎯 Exam Tip: Remember to divide the rate of change of concentration by the stoichiometric coefficient for each species and to use a negative sign for reactants and a positive sign for products.

 

Problem 6.1: (Textbook Page No 121)

Question 1. For the reaction, \( 3I^-(\text{aq}) + S_2O_8^{2-}(\text{aq}) \to I_3^-(\text{aq}) + 2SO_4^{2-}(\text{aq}) \) Calculate (a) the rate of formation of \( I_3^- \), (b) the rates of consumption of \( I^- \) and \( S_2O_8^{2-} \) and (c) the overall rate of reaction if the rate of formation of \( SO_4^{2-} \) is 0.022 moles dm-3 sec-1.
Answer:Reaction: \( 3I^-(\text{aq}) + S_2O_8^{2-}(\text{aq}) \to I_3^-(\text{aq}) + 2SO_4^{2-}(\text{aq}) \) Rate of reaction = \( R \) \( R = - \frac{1}{3} \frac{d[I^-]}{dt} = - \frac{d[S_2O_8^{2-}]}{dt} = + \frac{d[I_3^-]}{dt} = + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} \) Given: \( + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} = 0.022 \text{ mol dm}^{-3} \text{ s}^{-1} \)
(a) Rate of formation of \( I_3^- \): \( + \frac{d[I_3^-]}{dt} = + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} \) \( = 0.022 \text{ mol dm}^{-3} \text{ s}^{-1} \)
(b) Rates of consumption of \( I^- \) and \( S_2O_8^{2-} \): For \( I^- \): \( - \frac{1}{3} \frac{d[I^-]}{dt} = + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} \) \( - \frac{d[I^-]}{dt} = 3 \times 0.022 \) \( = 0.066 \text{ mol dm}^{-3} \text{ s}^{-1} \) For \( S_2O_8^{2-} \): \( - \frac{d[S_2O_8^{2-}]}{dt} = + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} \) \( = 0.022 \text{ mol dm}^{-3} \text{ s}^{-1} \)
(c) The overall rate of reaction is \( R \): \( R = + \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} = 0.022 \text{ mol dm}^{-3} \text{ s}^{-1} \)
In simple words: We use the stoichiometry of the balanced chemical equation to relate the rates of change of all reactants and products. Given one rate of formation, we can calculate all other rates based on their coefficients.

🎯 Exam Tip: Correctly write the rate expression based on stoichiometry, ensuring negative signs for reactants and positive signs for products. Use the given rate to find the overall reaction rate, which is then used to calculate other individual rates.

 

Try this..... (Textbook Page No 122)

Question 1. For the reaction : \( NO_2(g) + CO(g) \to NO(g) + CO_2(g) \), the rate of reaction is experimentally found to be proportional to the square of the concentration of \( NO_2 \) and independent that of CO. Write the rate law.
Answer:Since the rate of the reaction is proportional to \( [\text{NO}_2]^2 \) and \( [\text{CO}]^0 \), the rate law is \( R = k[\text{NO}_2]^2 \times [\text{CO}]^0 \) \( \implies R = k[\text{NO}_2]^2 \).
In simple words: The rate law describes how the reaction rate depends on reactant concentrations. Here, the rate is proportional to the square of \( NO_2 \) concentration but not affected by \( CO \) concentration, indicating its order with respect to \( CO \) is zero.

🎯 Exam Tip: Understand that reaction orders are determined experimentally, not from stoichiometry. "Proportional to the square" means order 2, and "independent" means order 0.

 

Try this..... (Textbook Page No 124)

Question 1. The reaction, \( CHCl_3(g) + Cl_2(g) \to CCl_4(g) + HCl(g) \) is first order in \( CHCl_3 \) and 1/2 order in \( Cl_2 \). Write the rate law and overall order of reaction.
Answer:Since the reaction is first order in \( CHCl_3 \) and 1/2 order in \( Cl_2 \), the rate law for the reaction will be, Rate \( = k[\text{CHCl}_3]^1 [\text{Cl}_2]^{1/2} \) The overall order (n) of the reaction will be, \( n = 1 + \frac{1}{2} = \frac{3}{2} \)
In simple words: The rate law shows how the reaction speed depends on reactant concentrations, here it's directly proportional to \( CHCl_3 \) and to the square root of \( Cl_2 \). The overall order is the sum of these individual orders.

🎯 Exam Tip: The overall order is simply the sum of the individual orders with respect to each reactant. Fractional orders are possible and derived from experimental data.

 

Use your brain power! (Textbook Page No 124)

Question 1. The rate of the reaction \( 2A + B \to 2C + D \) is \( 6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} \) when \( [\text{A}] = [\text{B}] = 0.3 \text{ mol dm}^{-3} \) If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:For the reaction, \( 2A + B \to 2C + D \), Rate \( = 6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} \) \( [\text{A}] = [\text{B}] = 0.3 \text{ mol dm}^{-3} \) \( n_A = 1 \) and \( n_B = 0 \) Rate \( = k[\text{A}]^{n_A} [\text{B}]^{n_B} \) \( = k[\text{A}]^1 [\text{B}]^0 \)
\( \implies k = \frac{\text{Rate}}{[\text{A}]} \) \( = \frac{6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}}{0.3 \text{ mol dm}^{-3}} \) \( = 2 \times 10^{-3} \text{ s}^{-1} \) Rate constant \( k = 2 \times 10^{-3} \text{ s}^{-1} \)
In simple words: Given the reaction rate and reactant concentrations, along with the orders for each reactant, we can directly calculate the rate constant by rearranging the rate law equation.

🎯 Exam Tip: Understand that any concentration term raised to the power of zero ([B]⁰) equals 1 and does not affect the rate. Ensure proper unit cancellation to get the correct units for the rate constant.

 

Problem 6.7) (Textbook Page No 126)

Question 1. A reaction occurs in the following steps :
(i) \( NO_2(g) + F_2(g) \to NO_2F(g) + F(g) \) (slow)
(ii) \( F(g) + NO_2(g) \to NO_2F(g) \) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate. Solution:
(a) The addition of two steps gives the overall reaction as \( 2NO_2(g) + F_2(g) \to 2NO_2F(g) \)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate \( = k[\text{NO}_2][\text{F}_2] \)
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.
In simple words: For a multi-step reaction, the overall reaction is the sum of all steps. The rate law is determined by the slowest step. Any species formed in an earlier step and consumed in a later step is a reaction intermediate.

🎯 Exam Tip: The slowest step in a reaction mechanism is the rate-determining step, and its stoichiometry directly yields the rate law. Reaction intermediates do not appear in the overall balanced equation.

 

Try this..... (Textbook Page No 126)

Question 1. A complex reaction takes place in two steps :
(i) \( NO(g) + O_3(g) \to NO_3(g) + O(g) \)
(ii) \( NO_3(g) + O(g) \to NO_2(g) + O_2(g) \) The predicted rate law is rate \( = k[\text{NO}][\text{O}_3] \). Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:Given steps:
(i) \( NO(g) + O_3(g) \to NO_3(g) + O(g) \)
(ii) \( NO_3(g) + O(g) \to NO_2(g) + O_2(g) \) Predicted rate law: rate \( = k[\text{NO}][\text{O}_3] \)
(a) The first step is slow and rate determining step since the rate depends on concentrations of \( NO(g) \) and \( O_3(g) \), which matches the predicted rate law.
(b) The overall reaction is the combination of two steps: \( NO(g) + O_3(g) \to NO_3(g) + O(g) \) \( NO_3(g) + O(g) \to NO_2(g) + O_2(g) \) Summing them: \( NO(g) + O_3(g) \to NO_2(g) + O_2(g) \)
(c) \( NO_3(g) \) and \( O(g) \) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).
In simple words: The rate-determining step is the one whose reactants' concentrations match the observed rate law. The overall reaction is the sum of all elementary steps, and intermediates are species formed in one step and consumed in another.

🎯 Exam Tip: The molecularity of the rate-determining step determines the overall rate law. Species that are produced and then consumed within a mechanism are intermediates and do not appear in the overall reaction or rate law.

 

Try this..... (Textbook Page No 129)

Question 1. The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:
(a) \( t_{1/2} = 0.5 \text{ min} = 30 \text{ s} \); \( \text{[A]}_0 = 100 \); \( \text{[A]}_t = 20 \);
\( \implies k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \text{ s}^{-1} \)
\( k = \frac{2.303}{t} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \)
\( \implies t = \frac{2.303}{k} \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} \) \( = \frac{2.303}{0.0231} \log_{10} \frac{100}{20} \) \( = \frac{2.303}{0.0231} \log_{10} 5 \) \( = \frac{2.303 \times 0.6990}{0.0231} \) \( = 69.69 \text{ s} \)
(b) Amount decomposed in 55 s:
\( \log_{10} \frac{\text{[A]}_0}{\text{[A]}_t} = \frac{t \times k}{2.303} \) \( = \frac{55 \times 0.0231}{2.303} \) \( = 0.5517 \)
\( \implies \frac{\text{[A]}_0}{\text{[A]}_t} = \text{Antilog } 0.5517 = 3.562 \) If \( \text{[A]}_0 = 100 \) then \( \text{[A]}_t = \frac{\text{[A]}_0}{3.562} = \frac{100}{3.562} = 28.07 \)
Amount remained = 28.07% Amount decomposed = \( 100 - 28.07 = 71.93\% \)
In simple words: For a first-order reaction, we first calculate the rate constant from its half-life. Then, we use the integrated rate law to find the time needed for the concentration to reduce to a certain percentage and the amount decomposed within a specific time.

🎯 Exam Tip: Convert all time units to be consistent (e.g., seconds). Remember to use the correct integrated rate law for first-order reactions and be careful with logarithmic and antilogarithmic calculations, especially when dealing with percentage reductions or decompositions.

 

Try this..... (Textbook Page No 123)

Question 1. For the reaction \( 2A + 2B \to 2C + D \), if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:Let the rate law be Rate \( = k[\text{A}]^x [\text{B}]^y \) When concentration of A = [2A] and [B] = constant, Rate \( R_2 = 4R_1 = k[2\text{A}]^x [\text{B}]^y \) Dividing by initial rate \( R_1 = k[\text{A}]^x [\text{B}]^y \): \( \frac{R_2}{R_1} = \frac{k[2\text{A}]^x [\text{B}]^y}{k[\text{A}]^x [\text{B}]^y} \) \( 4 = (2)^x \) \( \implies x = 2 \) When [A] = constant and concentration of B = [2B], Rate \( R_3 = 2R_1 = k[\text{A}]^x [2\text{B}]^y \) Dividing by initial rate \( R_1 = k[\text{A}]^x [\text{B}]^y \): \( \frac{R_3}{R_1} = \frac{k[\text{A}]^x [2\text{B}]^y}{k[\text{A}]^x [\text{B}]^y} \) \( 2 = (2)^y \) \( \implies y = 1 \) Hence order with respect to A is 2 and with respect to B is 1. By rate law, Rate \( = k[\text{A}]^2 [\text{B}]^1 \)
In simple words: We determine the reaction order for each reactant by analyzing how the reaction rate changes when their concentrations are individually doubled. If the rate quadruples when [A] doubles, the order with respect to A is 2. If the rate doubles when [B] doubles, the order with respect to B is 1. We then combine these to write the rate law.

🎯 Exam Tip: Systematically analyze the effect of concentration changes on rate to find individual orders. Remember that the overall order is the sum of these individual orders, and the rate law expresses this relationship.

 

Question 2. The rate law for the reaction \( A + B \to C \) is found to be rate \( = k[\text{A}]^2 \times [\text{B}] \). The rate constant of the reaction at 25 °C is 6.25 M-2 S-1. What is the rate of reaction when \( [\text{A}] = 1.0 \text{ mol dm}^{-3} \) and \( [\text{B}] = 0.2 \text{ mol dm}^{-3} \)?
Answer:Rate \( = k \times [\text{A}]^2 \times [\text{B}] \) \( = 6.25 \text{ M}^{-2}\text{s}^{-1} \times (1.0 \text{ mol dm}^{-3})^2 \times (0.2 \text{ mol dm}^{-3}) \) \( = 6.25 \times 1^2 \times 0.2 \text{ mol}^3 \text{ dm}^{-9} \text{ s}^{-1} \) \( = 6.25 \times 0.2 \text{ mol}^3 \text{ dm}^{-9} \text{ s}^{-1} \) \( = 1.25 \text{ mol dm}^{-3} \text{ s}^{-1} \)
In simple words: We calculate the reaction rate by plugging the given rate constant and concentrations of reactants A and B into the provided rate law equation.

🎯 Exam Tip: Substitute the values directly into the rate law. Ensure that the units for concentration and rate constant are consistent to obtain the correct units for the reaction rate.

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