CBSE Class 12 Chemistry Chemical Kinetics Questions Answers

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Class 12 Chemistry: Chemical Kinetics Questions and Answers

1 MARK QUESTIONS

What do you understand by the rate determining step of a reaction?

Answer: The slowest step in a reaction is known as the rate determining step.

Find the molecularity of following reaction. RCOOR’ + H₂O →^H+ RCOOH + R’OH

Answer: Molecularity of reaction = 2 (there are two reactant molecules in the balanced equation).

The rate constant of a reaction is 5.0 × 10⁻⁵ L mol⁻¹ min⁻¹. What is the order of the reaction?

Answer: The unit L mol⁻¹ min⁻¹ is the unit of a second order rate constant, so the order of the reaction is 2.

Why rate of the reaction does not remain constant throughout?

Answer: The rate of reaction depends upon the concentration of reactants, which keeps on decreasing with time. Hence the rate does not remain constant.

What is the order of reaction whose rate constant has the same units as the rate of reaction?

Answer: Zero order.

Write Arrhenius equation.

Answer: \(k = Ae^{-E_a/RT}\), where k = rate constant, E_a = activation energy, A = Arrhenius factor (frequency factor), R = gas constant and T = temperature.

Define rate constant or specific reaction rate.

Answer: When the molar concentration of each reactant is unity, the rate of reaction is called the specific reaction rate (rate constant).

The reaction A + 3B → 2C obeys the rate equation. Rate = k [A]^1/2 [B]^3/2. What is the order of this reaction?

Answer: Order = \(\tfrac{1}{2} + \tfrac{3}{2} = 2\).

What are the units of rate constant for a first order reaction?

Answer: s⁻¹.

Give one example of a reaction where order and molecularity are equal.

Answer: An elementary reaction, which is carried out in only one step, has order equal to molecularity, e.g. 2HI → H₂ + I₂ (order = molecularity = 2).

What do you understand by half-life period of a reaction?

Answer: The time taken for half of the reaction to complete (or for the concentration of the reactant to fall to half of its initial value) is known as the half-life period of that reaction.

Hydrolysis of ethyl acetate with NaOH is a reaction of second order while with HCl, it is of first order. Why?

Answer: The rate of hydrolysis with NaOH depends upon the concentration of both CH₃COOC₂H₅ and NaOH, while with HCl (a catalyst, whose concentration remains constant) it depends only upon the concentration of ethyl acetate.

What is the effect of adding catalyst on the free energy (ΔG) change of a reaction?

Answer: The free energy of the reactants and the products remains the same. Hence there is no change in ΔG on adding a catalyst.

The rate of a reaction whose rate law is Rate = k [B]ⁿ becomes double on doubling the concentration of B. Find the value n.

Answer: n = 1.

A reaction is 50% complete in 2 hrs. And 75% complete in 4 hrs. What is the order of reaction?

Answer: The time for 75% completion (4 h) is twice the half-life (2 h), so the half-life is constant. It is a first order reaction.

Give one example of a first order reaction.

Answer: Decomposition of N₂O₅: 2N₂O₅ → 4NO₂ + O₂ (also the decomposition of H₂O₂). (No answer was given for this in the source; written to complete the key.)

Define threshold energy.

Answer: Threshold energy is the minimum energy which the colliding molecules must possess so that the collision between them may be effective. Threshold energy = activation energy + energy possessed by the reactant molecules.

The reaction A + B → C has zero order. What is the rate equation?

Answer: Rate = k [A]⁰ [B]⁰ = k.

Which reactions proceed with constant rate?

Answer: Zero order reactions.

How does catalyst affect rate of reaction?

Answer: A catalyst increases the rate of reaction by lowering the activation energy. It gives an alternate path to the reaction.

For a chemical reaction represented by R → P the rate of reaction is denoted by −d[R]/dt or +d[P]/dt. State the significance of plus and minus sign.

Answer: The (−) sign represents a decrease in concentration of the reactant with time, while the (+) sign represents an increase in concentration of the product.

Express the rate of reaction in terms of disappearance of hydrogen and appearance of ammonia in the given reaction. N₂(g) + 3H₂(g) → 2NH₃(g)

Answer: \(\text{Rate} = -\tfrac{1}{3}\dfrac{d[H_2]}{dt} = +\tfrac{1}{2}\dfrac{d[NH_3]}{dt}\)

Why rate of reaction does not remain constant throughout?

Answer: It is because the concentration of reactants goes on decreasing with time.

Write the unit of first order rate constant of a gaseous reaction if the partial pressure of gaseous reactant is given in bar.

Answer: s⁻¹.

What will be the order of reaction, if the rate of reaction does not depend on the concentration of any of the reactant?

Answer: Zero order.

For the elementary step of a chemical reaction: H₂ + I₂ → 2HI, rate of reaction = k[H₂][I₂]. What is the (i) molecularity and (ii) order of the reaction.

Answer: (i) Molecularity = 2. (ii) Order = 1 + 1 = 2. (The source key gave the order as 1, which does not match the rate law in the question; corrected to 2.)

For a chemical reaction A → B. The rate of the reaction is given as Rate = k [A]ⁿ. The rate of the above reaction quadruples when the concentration of A is doubled. What is the value of n?

Answer: \(2^n = 4\), so n = 2.

Mention one example of zero order reaction.

Answer: Decomposition of NH₃ on a platinum catalyst at high pressure and 1130 K: 2NH₃(g) → N₂(g) + 3H₂(g).

What is the value of the order of reaction of radioactive decay?

Answer: First order.

Express the relation between the half life period of a reactant and initial concentration for a reaction of n^th order.

Answer: \(t_{1/2} \propto \dfrac{1}{[R]_0^{\,n-1}}\), where n is the order of the reaction.

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of reaction?

Answer: First order (the half-life is constant, 2 h).

Suggest an appropriate reason for the observation: “On increasing temperature of the reacting system by 10 degrees, the rate of reaction almost doubles or even sometimes becomes five folds.”

Answer: Increasing the temperature increases the fraction of molecules which collide with energy greater than E_a.

For a chemical reaction, activation energy is zero and at 300 K rate constant is 5.9 × 10⁻⁵ s⁻¹, what will be the rate constant at 400 K?

Answer: If E_a = 0, then \(k = Ae^{0} = A\), so k does not depend on temperature. The rate constant at 400 K is 5.9 × 10⁻⁵ s⁻¹.

Two reactions occurring at the same temperature have identical values of E_a. Does this ensure that also they will have the same rate constant? Explain.

Answer: No, because the rate constant depends on the nature of the reactants and also on the pre-exponential factor A, not only on E_a.

The rate constant of a reaction is given by the expression k = Ae^−E_a/RT. Which factor in this expression should register a decrease so that the reaction proceeds rapidly?

Answer: E_a should decrease.

For a chemical reaction rate constant k = 5.3 × 10⁻⁴ mol L⁻¹ s⁻¹, what will be the order of reaction?

Answer: Zero order (the unit of k is the same as the unit of rate).

Write the rate law and order for the following reaction: AB₂ + C₂ → AB₂C + C (slow); AB₂ + C → AB₂C (fast)

Answer: The rate depends on the slow step: Rate = k [AB₂][C₂]; Order = 1 + 1 = 2.

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to 3 times how will it affect the rate of formation of Y.

Answer: Rate = k[X]², so the rate will increase 3² = 9 times.

When rate of reaction becomes equal to specific reaction rate.

Answer: When the concentration of each reactant is unity.

87.5% of the substance disintegrated in 45 minutes (first order reaction). What is its half life.

Answer: 87.5% disintegration leaves 12.5% = 1/8 = (1/2)³, so 3 half-lives = 45 min. Half-life = 15 minutes.

2 MARKS QUESTIONS

What is (i) Rate law expression? (ii) Rate determining step?

Answer: (i) The rate law is the expression of the rate of reaction in terms of the concentrations of the reactants, each raised to a power which is found experimentally. (ii) The slowest step of a reaction is the rate determining step.

This reaction is of first order and rate constant of reaction is 5.7 × 10⁻³ s⁻¹. Find the value of t_1/2.

Answer: \(t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{5.7\times10^{-3}\ \text{s}^{-1}} = 121.6\ \text{s}\)

Define (i) Activation energy (ii) Collision frequency

Answer: (i) Activation energy is the minimum extra energy which the reactant molecules must acquire (over their average energy) so that they can form the products; E_a = threshold energy − average energy of the reactants. (ii) Collision frequency is the number of collisions taking place per second per unit volume of the reaction mixture.

For a reaction A → B, the rate of reaction doubles when concentration of A is increased by 4. What is the order of reaction?

Answer: Taking this as “increased four times”: Rate ∝ [A]ⁿ, so \(2 = 4^n\), i.e. \(n = \tfrac{1}{2}\). The order of the reaction is 1/2. (The source wording “increased by 4” is read as 4 times.)

Time required to decompose SO₂Cl₂ to half of its initial amount is 1 hour. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Answer: \(k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{1\ \text{h}} = 0.693\ \text{h}^{-1}\) (= 1.925 × 10⁻⁴ s⁻¹)

The conversion of molecules from A to B follows second order kinetics. If the concentration of A is increased to five times, how will it affect the rate of formation of B?

Answer: Rate = k[A]². On increasing [A] five times, the rate becomes 5² = 25 times.

The rate of decomposition of N₂O₅ is 2.4 × 10⁻⁴ mol L⁻¹ s⁻¹ when [N₂O₅] is 0.36 M. What is the rate law and value of k for this first order reaction?

Answer: Rate law: Rate = k[N₂O₅]. \(k = \dfrac{\text{Rate}}{[N_2O_5]} = \dfrac{2.4\times10^{-4}}{0.36} = 6.67\times10^{-4}\ \text{s}^{-1}\)

In a first order reaction, 75% of reactants disappeared in 1.386 hrs. Calculate the rate constant of the reaction.

Answer: \(k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]} = \dfrac{2.303}{1.386}\log\dfrac{100}{25} = \dfrac{2.303\times0.6021}{1.386} \approx 1.0\ \text{h}^{-1}\) (= 2.78 × 10⁻⁴ s⁻¹)

Rate of formation of product for second order reaction is 9.5 × 10⁻⁵ mol L⁻¹ s⁻¹. The initial concentration of reactant was found to be 0.01 mol/L. (a) Write rate law for second order reaction. (b) Calculate the rate constant for the given second order reaction.

Answer: (a) Rate = k[A]². (b) \(k = \dfrac{\text{Rate}}{[A]^2} = \dfrac{9.5\times10^{-5}}{(0.01)^2} = 0.95\ \text{L mol}^{-1}\text{s}^{-1}\)

What do you understand by effective collision? What are the factors affecting the effective collision?

Answer: A collision in which the molecules have energy equal to or more than the threshold energy and collide with the proper orientation, so that the products are formed, is an effective collision. It depends upon (i) the energy of the colliding molecules (they should have at least the threshold energy) and (ii) proper orientation of the molecules at the time of collision.

3 MARKS QUESTIONS

(i) Define half life period. (ii) The half-life period of reaction is 10 minutes. How long it will take for concentration of reactant to be reduced to 10% of original?

Answer: (i) The half-life period is the time in which the concentration of a reactant falls to half of its initial value. (ii) \(k = \dfrac{0.693}{10} = 0.0693\ \text{min}^{-1}\). \(t = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{0.0693}\times 1 = 33.2\) minutes.

Explain the term Activation energy and threshold energy? How temperature affects the rate of reaction?

Answer: Threshold energy is the minimum energy that the colliding molecules must have for the collision to be effective. Activation energy is the extra energy that the reactant molecules must gain over their average energy to reach the threshold: E_a = threshold energy − average energy of reactants. When the temperature rises, the number of molecules having energy equal to or greater than the activation energy increases greatly, so the number of effective collisions rises and the rate increases. For every 10 °C rise the rate generally doubles; this is given by the Arrhenius equation \(k = Ae^{-E_a/RT}\).

What do you understand by integrated rate equation? Deduce the integrated rate equation for first order reaction?

Answer: The integrated rate equation gives the relation between the concentration of the reactant and time. For a first order reaction R → P: \(\text{Rate} = -\dfrac{d[R]}{dt} = k[R]\). Rearranging, \(\dfrac{d[R]}{[R]} = -k\,dt\). Integrating, \(\ln[R] = -kt + I\). At t = 0, [R] = [R]₀, so \(I = \ln[R]_0\). Hence \(\ln\dfrac{[R]_0}{[R]} = kt\), or \(k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}\).

A chemical reaction is of second order w.r.t. a reactant. How will the rate of reaction be affected if the concentration of this reactant is: (a) Doubled; (b) Reduced to 1/8th.

Answer: Rate = k[A]². (a) When doubled: rate = k(2[A])² = 4 k[A]², i.e. the rate becomes 4 times. (b) When reduced to 1/8: rate = k([A]/8)² = k[A]²/64, i.e. the rate becomes 1/64 of the original.

From the following data for a chemical reaction between A and B at 300 K: [A] = 2.5 × 10⁻⁴, [B] = 3 × 10⁻⁵, rate = 5 × 10⁻⁴; [A] = 2.5 × 10⁻⁴, [B] = 6 × 10⁻⁵, rate = 4 × 10⁻³; [A] = 1 × 10⁻³, [B] = 6 × 10⁻⁵, rate = 1.6 × 10⁻². Calculate (i) the order of reaction with respect to A and with respect to B. (ii) the rate constant at 300 K

Answer: Let Rate = k[A]^x[B]^y. Experiments 1 and 2 (A constant, B doubled): the rate becomes \(\dfrac{4\times10^{-3}}{5\times10^{-4}} = 8\) times, so \(2^y = 8\), y = 3. Experiments 2 and 3 (B constant, A becomes 4 times): the rate becomes \(\dfrac{1.6\times10^{-2}}{4\times10^{-3}} = 4\) times, so \(4^x = 4\), x = 1. (i) The order is 1 with respect to A and 3 with respect to B (overall order 4). (ii) \(k = \dfrac{\text{Rate}}{[A][B]^3} = \dfrac{5\times10^{-4}}{(2.5\times10^{-4})(3\times10^{-5})^3} = 7.4\times10^{13}\ \text{L}^3\text{mol}^{-3}\text{s}^{-1}\).

(i) Distinguish between elementary and complex reaction? (ii) What are the factors affecting the rate of chemical reaction?

Answer: (i) An elementary reaction takes place in a single step, and for it the order equals the molecularity. A complex reaction takes place in a number of steps (a sequence of elementary reactions); its overall order is decided by the slowest step and need not equal its molecularity. (ii) The rate depends on the concentration of the reactants, temperature, presence of a catalyst, the nature of the reactants, the surface area of solid reactants, and light (for photochemical reactions).

The decomposition of phosphine 4PH₃(g) → P₄(g) + 6H₂(g) has rate law; Rate = k[PH₃]. The rate constant is 6.0 × 10⁻⁴ s⁻¹ at 300 K and activation energy is 3.05 × 10⁵ J mol⁻¹. Calculate the value of the rate constant at 310 K. (R = 8.314 J K⁻¹mol⁻¹).

Answer: \(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right) = \dfrac{3.05\times10^5}{2.303\times8.314}\times\dfrac{10}{300\times310} = 1.713\). So \(\dfrac{k_2}{k_1} = 51.6\) and \(k_2 = 6.0\times10^{-4}\times51.6 \approx 3.1\times10^{-2}\ \text{s}^{-1}\).

The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s⁻¹) e^{−28000 K/T}. Calculate E_a.

Answer: Comparing with \(k = Ae^{-E_a/RT}\): \(\dfrac{E_a}{R} = 28000\ \text{K}\). \(E_a = 28000 \times 8.314 = 2.33\times10^5\ \text{J mol}^{-1} = 232.8\ \text{kJ mol}^{-1}\).

Show that for a first order reaction, time required for 99% completion is twice for the time required for the completion of 90% of reaction.

Answer: For 99% completion: \(t_{99} = \dfrac{2.303}{k}\log\dfrac{100}{1} = \dfrac{2.303\times2}{k} = \dfrac{4.606}{k}\). For 90% completion: \(t_{90} = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{k}\). Hence \(t_{99}/t_{90} = 2\), i.e. \(t_{99} = 2\,t_{90}\).

The rate of reaction triples when the temperature changes from 20°C to 50°C. Calculate the energy of activation. [R = 8.314 J K⁻¹mol⁻¹, log 3 = 0.48]

Answer: T₁ = 293 K, T₂ = 323 K, k₂/k₁ = 3. \(\log 3 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{323-293}{293\times323}\). So \(0.48 = \dfrac{E_a}{19.15}\times3.17\times10^{-4}\), giving \(E_a \approx 2.9\times10^{4}\ \text{J mol}^{-1} \approx 29\ \text{kJ mol}^{-1}\).

5 MARKS QUESTIONS

(a) For a reaction A + B → Products, the rate law is given by r = k [A]^1/2 [B]². What is the order of reaction? (b) The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y? (c) Write two points of difference between order and molecularity of a reaction.

Answer: (a) Order = \(\tfrac{1}{2} + 2 = \tfrac{5}{2}\) = 2.5. (b) Rate = k[X]²; if [X] becomes 3 times, the rate becomes 3² = 9 times. (c) 1. Order is determined experimentally and can be zero, a fraction or a whole number; molecularity is a theoretical quantity and is always a whole number (never zero or a fraction). 2. Order is for the overall reaction (elementary or complex); molecularity is meaningful only for an elementary step.

(a) Define (i) Activation energy (ii) Collision frequency. (b) The rate constants of a reaction at 500 K & 700 K are 0.02 s⁻¹ & 0.07 s⁻¹ respectively. Calculate the value of E_a & A.

Answer: (a) Activation energy: the minimum extra energy the reactant molecules must acquire over their average energy to form the products. Collision frequency: the number of collisions per second per unit volume of the reaction mixture. (b) \(\log\dfrac{0.07}{0.02} = \dfrac{E_a}{2.303\times8.314}\times\dfrac{700-500}{500\times700}\). \(0.5441 = \dfrac{E_a}{19.147}\times5.714\times10^{-4}\), so \(E_a = 1.82\times10^{4}\ \text{J mol}^{-1} = 18.2\ \text{kJ mol}^{-1}\). Now, \(\log k = \log A - \dfrac{E_a}{2.303RT}\): at 500 K, \(\log A = \log 0.02 + \dfrac{18231}{19.147\times500} = -1.699 + 1.904 = 0.205\), so \(A \approx 1.6\ \text{s}^{-1}\).

(i) A reaction is first order in A & second order in B. (a) Write differential rate equation. (b) How is the rate affected on increasing the concentration of B three times? (c) How is the rate affected when conc of both A & B is doubled? (ii) Name the factors which affect the rate of reaction.

Answer: (i)(a) \(-\dfrac{d[R]}{dt} = k[A][B]^2\). (b) The rate becomes 3² = 9 times. (c) The rate becomes 2 × 2² = 8 times. (ii) Concentration of reactants, temperature, catalyst, nature of reactants, surface area, and light.

(a) Deduce the expression of half-life for zero order reaction. (b) Show that time required for 99.9% completion of the first order reaction is 10 times of t_1/2 for first order chemical reaction.

Answer: (a) For a zero order reaction, \(-\dfrac{d[R]}{dt} = k\), so \([R] = [R]_0 - kt\). At \(t = t_{1/2}\), \([R] = \dfrac{[R]_0}{2}\), so \(\dfrac{[R]_0}{2} = [R]_0 - kt_{1/2}\), giving \(t_{1/2} = \dfrac{[R]_0}{2k}\). (b) For 99.9% completion: \(t = \dfrac{2.303}{k}\log\dfrac{100}{0.1} = \dfrac{2.303\times3}{k} = \dfrac{6.909}{k}\). Since \(t_{1/2} = \dfrac{0.693}{k}\), \(\dfrac{t}{t_{1/2}} = \dfrac{6.909}{0.693} \approx 10\). So \(t_{99.9\%} \approx 10\,t_{1/2}\).

(a) For a reaction A + B → Products, the rate law is given by r = k [A]^1/2 [B]². What is the order of reaction? (b) The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Answer: (a) Order = \(\tfrac{1}{2} + 2 = \tfrac{5}{2}\) = 2.5. (b) Rate = k[X]²; if [X] becomes 3 times, the rate becomes 3² = 9 times.

9. s-1

10. For elementary reaction, order and molecularity are same. These reactions are carried out.