CBSE Class 12 Chemistry Chemical Kinetics Case Study Questions Set 02

Case Study 1

Concentration dependence of rate is called differential rate equation. Integrated differential equations give relation between directly measured experimental data i.e. concentration at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. The first-order reaction has a rate constant 1.15 × 10⁻³ s⁻¹.

Case Study 2

The rate of a chemical reaction is expressed either in terms of decrease in the concentration of a reactant per unit time or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.

Question 6. Express the rate of the following reaction in terms of different reactants and products.
4NH₃(g) + 5O₂(g) → 4NO₂(g) + 6H₂O(g)
Answer:
Rate of reaction = \( -\frac{1}{4}\frac{d[NH_3]}{dt} = -\frac{1}{5}\frac{d[O_2]}{dt} = +\frac{1}{4}\frac{d[NO_2]}{dt} = +\frac{1}{6}\frac{d[H_2O]}{dt} \)

Teacher's Note:
1. Reactant terms carry a minus sign and product terms a plus sign; each is divided by its own coefficient.

Question 7. Why do pieces of wood burn faster than a log of wood of the same mass?
Answer:
Pieces of wood have larger surface area than the log of wood of the same mass. Greater the surface area, faster is the reaction.

Teacher's Note:
1. The same idea explains why powdered reactants react faster than lumps.

Question 8. Why does the rate of any reaction generally decrease during the course of the reaction?
Answer:
The rate of reaction depends on the concentration of reactants. As the reaction progresses, reactants start getting converted to products so the concentration of reactants decreases hence the rate of reaction decreases.

Teacher's Note:
1. This is why the concentration–time curve becomes flatter as time passes.

Question 9. Why is molecularity applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
Answer:
A complex reaction proceeds through several elementary reactions. Number of molecules involved in each elementary reaction may be different, i.e., the molecularity of each step may be different. Therefore, discussion of the molecularity of overall complex reaction is meaningless. On the other hand, order of a complex reaction is determined by the slowest step in its mechanism and is not meaningless even in the case of complex reactions.

Teacher's Note:
1. For a complex reaction, the molecularity of its slowest (rate-determining) step and the order of the overall reaction are often the same.

Question 10. The kinetics of the reaction
mA + nB + pC → m′X + n′Y + p′Z
obey the rate expression dx/dt = k[A]^m[B]ⁿ
Calculate total order and molecularity of the reaction.
Answer:
The total order of reaction = m + n
The molecularity of the reaction = m + n + p

Teacher's Note:
1. Order counts only the concentration terms in the rate law (C does not appear), while molecularity counts all species colliding in the elementary reaction.

Case Study 3

Temperature has a marked effect on the rate of reaction. For most of the reactions, the rate of reaction becomes nearly double for every 10 degree rise in temperature. The effect of temperature is usually expressed in terms of temperature coefficient. The quantitative dependence of reaction rate on temperature was first explained by Swante Arrhenius. Arrhenius proposed a simple equation known as Arrhenius equation
k = Ae^−E_a/RT
This equation provides a relationship between the rate constant (k) of a reaction and the temperature of the system. A is the Arrhenius factor or pre-exponential factor, E_a is the activation energy and R is the gas constant.

Question 11. Define ‘activation energy’ of a reaction. [CBSE (AI) 2011]
Answer:
The energy required to form the intermediate called activated complex is known as activation energy.
Activation energy = Threshold energy − Average energy of the reactants

Teacher's Note:
1. Activation energy is always measured from the energy level of the reactants up to the top of the barrier.

Question 12. How does a catalyst affect the rate of a reaction? Explain with respect to the Arrhenius equation.
Answer:
A catalyst decreases the activation energy. According to Arrhenius equation, lower the activation energy, greater will be the rate constant and thus the rate of reaction increases.

Teacher's Note:
1. Since E_a sits in the exponent, even a small decrease in E_a increases k by a large factor.

Question 13. Can a reaction have zero activation energy? Justify.
Answer:
No, E_a ≠ 0.
If E_a = 0, then according to Arrhenius equation,
k = Ae^−E_a/RT
k = Ae⁰ = A,
i.e., Rate constant = Collision frequency
This means every collision results into a chemical reaction which cannot be true.

Teacher's Note:
1. Use the equation itself to argue: putting E_a = 0 makes k equal to A, which would mean every collision is effective.

Question 14. The plot of log k vs X is linear with slope = −E_a/2.303R. What is X?
Answer:
1/T

Teacher's Note:
1. From log k = log A − E_a/2.303RT, the variable multiplying the slope is 1/T.

Question 15. What is the fraction of molecules having energy greater than activation energy, E_a?
Answer:
e^−E_a/RT at temperature T.

Teacher's Note:
1. This fraction increases with temperature, which is why the rate rises so sharply on heating.

Case Study 4

We have maximum number of reactions of first order. The question arises how we test experimentally that it is a reaction of first order. The general integrated expression for the reaction of first order \( \left(k = \frac{2.303}{t}\log\frac{a}{a-x}\right) \) is modified based on the nature of reaction. For example, in the decomposition of N₂O₅ in CCl₄ solution, \( \left(N_2O_5 \rightarrow N_2O_4\,(2NO_2) + \frac{1}{2}O_2\right) \), O₂ gas is collected at different times and the formula for k is modified accordingly. Further, there are many reactions which are not truly of first order but under certain conditions, they become reactions of first order. These are called pseudo-first order reactions, e.g. acid-catalysed hydrolysis of ethyl acetate or inversion of cane-sugar (sucrose). Radioactive disintegration is also a reaction of first order and the expressions for rate constant (here called disintegration constant or decay constant) or for the half-life or average life are same as for other reactions of first order.

Question 16. For a zero order reaction, calculate the ratio of the times taken for \( \frac{3}{4} \)th (75%) of the reaction to complete and half (50%) of the reaction to complete.
Answer:
For a zero order reaction, \( t_{1/2} = \frac{[A_0]}{2k} \) and \( t_{3/4} = \frac{3}{4}\frac{[A_0]}{k} \)
\( \therefore\; \frac{t_{3/4}}{t_{1/2}} = \frac{3}{4}\frac{[A_0]}{k} \times \frac{2k}{[A_0]} = \frac{3}{2} \) = 1.5

Teacher's Note:
1. For zero order, time is simply proportional to the amount reacted (x = kt), so 75% versus 50% gives 0.75/0.50 = 1.5 directly.

Question 17. What is the order of reaction if a straight line plot is obtained when half-life period (t_1/2) of a reaction is plotted against 1/[A₀] where [A₀] is the initial concentration of the reactant ?
Answer:
Second order because \( t_{1/2} \propto \frac{1}{[A_0]^{n-1}} \)
or \( t_{1/2} \propto \frac{1}{[A_0]} \) only when n = 2

Teacher's Note:
1. A straight line of t_1/2 against 1/[A₀] means direct proportionality to 1/[A₀], which is the second order pattern.

Question 18. On doubling the concentration of the reactant, half-life period of the reaction is also doubled. What is the order of reaction ?
Answer:
Zero order because for a zero order reaction \( t_{1/2} = \frac{[A_0]}{2k} \), i.e., t_1/2 ∝ [A₀]

Teacher's Note:
1. Half-life rising with concentration is unique to zero order; for first order it stays the same and for second order it falls.

Teacher's Note:
1. Always pick a pair of experiments in which only one concentration changes; experiments I and II cannot be compared directly because both change.

Case Study 5

For a reaction of the type aA → Products, if n is the order of the reaction, the rate law expression is written as
Rate = \( -\frac{d[A]}{dt} = k[A]^n \)
For reactions of zero order, n = 0. Likewise, for reactions of 1st, 2nd, 3rd order etc. n = 1, 2, 3 etc. The expression thus written are the differential equations. For many reasons, these are not convenient equations for solving problems. We integrate these equations to get suitable expressions which make the task of solving problems easier. For reactions of different orders (0, 1, 2), we get expression for the rate constant k in terms of time and the initial concentration of the reactant. Using the experimental data, we can confirm the order of reaction. Further, from these expressions, it is useful to know the half-life period of a reaction, i.e., the time in which half of the substance is reacted. A general expression is obtained for half-life period in terms of initial concentration of the reactant and order of reaction. Using these expressions, the experimental data of any reaction can be used to determine the rate law, rate constant and order of reaction.

Question 20. For the decomposition of N₂O₅ in CCl₄ solution, if V_t represents volumes of O₂ gas at different times and V_∞ represents the final volume of O₂ collected when decomposition is complete, what expression will be written for k in terms of V_t and V_∞.
Answer:
\( k = \frac{2.303}{t}\log\frac{V_\infty}{V_\infty - V_t} \)

Teacher's Note:
1. V_∞ is proportional to the initial amount of N₂O₅ and (V_∞ − V_t) to the amount still left at time t.

Question 21. How does hydrolysis of ethyl acetate in acidic medium differ from that in the alkaline medium from the point of view of chemical kinetics ?
Answer:
In acid medium, hydrolysis of ethyl acetate is a reaction of first order whereas in alkaline medium, it is of second order.

Teacher's Note:
1. In acid, water (in excess) is the other reactant, so the reaction is pseudo first order; in alkali, OH⁻ is consumed and its concentration enters the rate law.

Question 22. One fourth of a reaction is completed in 32 minutes. What is the half life period of the reaction ?
Answer:
Taking the reaction to be of first order, \( k = \frac{2.303}{32}\log\frac{a}{a - a/4} = \frac{2.303}{32}\log\frac{4}{3} \) = \( \frac{2.303 \times 0.1249}{32} \) = 8.99 × 10⁻³ min⁻¹
\( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.99\times10^{-3}} \) = 77 min

Teacher's Note:
1. One fourth completed means three fourth remains, so use the ratio 4/3; using 4 by mistake would give the three-fourth-life instead.

Question 23. Half-life period of a radioactive element is 100 seconds. How much time will it take to lose its activity by 90% ?
Answer:
\( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{100} \) s⁻¹
When 90% activity is lost, 10% remains: \( t = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303 \times 100}{0.693} \) = 332.3 seconds

Teacher's Note:
1. Losing 90% means 10% remains, so the ratio is 100/10, giving log 10 = 1.

Case Study 6

According to collision theory, a reaction takes place because the reactant molecules collide with each other. But every collision is not an effective collision, i.e., does not result into a chemical reaction. For a collision to be effective, the colliding molecules must have energy more than a particular value. This value is called ‘threshold energy’. Thus, reactant molecules have to absorb a certain amount of energy to reach the threshold value. This extra energy absorbed is called ‘activation energy’. This quantity is important in deciding the effect of temperature on the rate of reaction. Quantitatively, the studies were made by Arrhenius. He put forward an equation giving the effect of temperature on the rate constant of a reaction. This equation is called ‘Arrhenius equation’. It is generally observed that for every 10° rise in temperature, the rate of reaction is nearly doubled. Besides temperature, another factor affecting the rate of reaction is the presence of a catalyst. A catalyst lowers the activation energy barrier and hence the rate becomes faster.

Question 24. The rate constant of a first order reaction is 10⁻³ min⁻¹ at 27°C. The temperature coefficient of this reaction is 2. What will be the rate constant of this reaction at 17°C ?
Answer:
Suppose rate constant at 17°C = k
Then as rate is doubled for 10°C rise in temperature, rate at 27°C = 2k = 10⁻³ min⁻¹ (Given). Hence,
k = \( \frac{10^{-3}}{2} \) min⁻¹ = 5 × 10⁻⁴ min⁻¹

Teacher's Note:
1. Going down in temperature, divide by the temperature coefficient; going up, multiply.

Question 25. Which of the following property/properties changes/change with temperature and why ?
ΔG, ΔH, E_a, K
Answer:
ΔG and K change with temperature.
ΔG changes because ΔG = ΔH − TΔS contains T. K (rate constant k as well as equilibrium constant) changes because it depends exponentially on temperature (k = Ae^−E_a/RT; ΔG° = −RT ln K).
ΔH and E_a are practically independent of temperature.

Teacher's Note:
1. Activation energy is fixed for a given path; it is lowered only by a catalyst, not by heating.

Case Study 7

From experimental studies, it has been found that the rate of reaction may not depend upon the stoichiometric coefficients of the reactants in the balanced chemical equation. This means that a balanced chemical equation does not give a true picture of how reaction takes place. Thus, it is very rare that the reaction may be completed in one step. The reactions taking place in one step are called ‘elementary reactions’. The reactions which do not take place in one step are called ‘complex reactions.’ The different steps in which the complex reaction takes place is called the ‘mechanism of the reaction’. Each step of the mechanism of the reaction is an elementary reaction. According to ‘collision theory’, in order that the reaction may take place, the atoms, ions or molecules of the reactants must come together simultaneously and collide with one another. The number of atoms, ions or molecules that must collide with one another simultaneously so as to result into a chemical reaction is called the ‘molecularity of the reaction’. Further, the different elementary reactions of a complex reaction do not take place at the same rate. The overall rate of the reaction depends upon the ‘slowest step’. Hence, the slowest step is called the ‘rate determining step’. A series of step reactions or elementary reactions proposed to account for the overall reaction is called the mechanism of the reaction.

Case Study 8

Besides the feasibility and extent of reaction, it is equally important to know the rate at which the reaction takes place, the factors controlling the rate of a chemical reaction. For example, which parameters determine as to how rapidly food gets spoiled ? How to design the rapidly setting material for dental filling ? What controls the rate at which fuel burns in an auto-engine ? Thermodynamics predicts only the feasibility of a reaction. It can only tell diamond can change into graphite but in reality the conversion is so slow that the change is not perceptible at all. Chemical kinetics is the branch of chemistry which helps to study the rates of reactions, conditions by which the reaction rates can be altered and effect of factors like temperature, pressure and catalyst on the rate of reaction. There are two ways to express the rate of reaction viz average rate over the time or instantaneous rate (at a particular instant of time). The rate of reaction can be expressed in terms of any reactant or product. As rate of reaction should have the same value in terms of any reactant or product, the change in concentration per unit time has to be divided by its stoichiometric coefficient in the balanced chemical equation.