CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 01

Very Short Answer Questions (1 Mark)

Teacher's Note:
1. Order is experimental and can be 0 or fractional; molecularity is theoretical and is 1, 2 or 3 for an elementary step.

Question 2. Define the half-life period of reaction (t_1/2).
Answer:
The time taken for half of the reaction to complete, i.e., the time in which the concentration of a reactant is reduced to half of its original value is called half-life period of the reaction.
t = t_1/2 when [R] = \( \frac{[R_0]}{2} \)

Teacher's Note:
1. For first order t_1/2 = 0.693/k (constant); for zero order t_1/2 = [R]₀/2k (depends on starting concentration).

Question 3. For a reaction R → P, half-life (t_1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction?
Answer:
Half-life of first order reaction is independent of the initial concentration of reactants.
t_1/2 = \( \frac{0.693}{k} \)
Hence, the reaction is of first order.

Teacher's Note:
1. A concentration-independent half-life is the quickest experimental test for first order kinetics.

Question 4. If half-life period of a first order reaction is x and 3/4^th life period of the same reaction is y, how are x and y related to each other?
Answer:
For a first order reaction, k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{x} \)
For \( \frac{3}{4} \)th life period, a − x = a − \( \frac{3}{4} \)a = \( \frac{1}{4} \)a
∴ k = \( \frac{2.303}{y} \) log \( \frac{a}{(a-x)} \); \( \frac{0.693}{x} \) = \( \frac{2.303}{y} \) log \( \frac{a}{a/4} \)
\( \frac{0.693}{x} \) = \( \frac{2.303}{y} \) 2 log 2; \( \frac{0.693}{x} \) = \( \frac{0.693\times2}{y} \)
y = 2x

Teacher's Note:
1. Three-fourth life means 75% has reacted and 25% remains, which takes exactly two half-lives.

Question 5. Draw a graph between concentration and time for a zero order reaction.
Answer:
[R] = −k₀t + [R]₀
The plot of [R] against t is a straight line with intercept [R]₀ on the concentration axis and slope = −k₀ (graph shown below).

Teacher's Note:
1. Label the intercept [R]₀ and the slope −k; examiners look for both on this sketch.

Question 6. Express the rate of the following reaction in terms of the formation of ammonia.
N_2(g) + 3H_2(g) → 2NH_3(g)
Answer:
\( \frac{-d[N_2]}{dt} \) = −\( \frac{1}{3} \) \( \frac{d[H_2]}{dt} \) = +\( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \)
In terms of formation of ammonia, Rate = +\( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \)

Teacher's Note:
1. Product terms are positive and are divided by their coefficient, here 2 for NH₃.

Question 7. If the rate constant of reaction is k = 3 × 10⁻⁴ s⁻¹, then identify the order of the reaction.
Answer:
First order reaction has s⁻¹ as the unit of the rate constant. Hence, the reaction is of first order.

Teacher's Note:
1. Read the order straight from the units of k: (mol L⁻¹)¹⁻ⁿ s⁻¹; s⁻¹ alone means n = 1.

Question 8. For the reaction 3H_2(g) + N_2(g) → 2NH_3(g), how are the rate of reaction expression \( \frac{-d[H_2]}{dt} \) and \( \frac{d[NH_3]}{dt} \) interrelated?
Answer:
−\( \frac{1}{3} \) \( \frac{d[H_2]}{dt} \) = +\( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \)

Teacher's Note:
1. H₂ is used up 1.5 times as fast as NH₃ is formed, matching the 3 : 2 ratio in the equation.

Question 9. Define elementary step in a reaction.
Answer:
Elementary step: Each step of a complex reaction is called an elementary step.

Teacher's Note:
1. Only for an elementary step can the rate law be written directly from the equation, and its order equals its molecularity.

Question 10. Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction.
Answer:
Rate expression is a way of expressing rate of reaction in terms of concentration of reactants, e.g., for a general reaction, aA + bB → cC + dD; Rate = k[A]^x[B]^y
Rate constant (k) is equal to the rate of reaction when molar concentration of reactant is unity. Its units depends upon the order of reaction.

Teacher's Note:
1. Rate changes as the reaction proceeds; the rate constant stays fixed at a given temperature.

Question 11. For the assumed reaction X₂ + 3Y₂ → 2XY₃, write the rate of equation in terms of rate of disappearance of Y₂. [HOTS]
Answer:
Rate = \( -\frac{d[X_2]}{dt} = -\frac{1}{3}\frac{d[Y_2]}{dt} = +\frac{1}{2}\frac{d[XY_3]}{dt} \)
Rate of disappearance of Y₂ = \( -\frac{d[Y_2]}{dt} = -3\frac{d[X_2]}{dt} = +\frac{3}{2}\frac{d[XY_3]}{dt} \)

Teacher's Note:
1. Write the general rate expression first, then multiply through by 3 to isolate the Y₂ term.

Short Answer Questions (2 Marks)

Question 12. For a reaction A + B → P, the rate law is given by, r = k[A]^1/2[B]². What is the order of this reaction?
Answer:
Rate law, r = k[A]^1/2[B]²
Order of reaction is sum of the powers of concentration terms,
∴ Order of reaction = \( \frac{1}{2} \) + 2 = \( \frac{5}{2} \) = 2.5

Teacher's Note:
1. Add the exponents in the rate law, not the coefficients in the equation (which are 1 and 1 here).

Question 13. For a chemical reaction R → P, the variation in the concentration [R] vs. time (t) plot is given as
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?

Answer:
(i) The reaction is of zero order.
(ii) Slope of the curve = −k = \( \frac{d[R]}{dt} \)

Teacher's Note:
1. A straight line on a [R] vs t plot (not ln[R] vs t) is the signature of zero order.

Question 14. For a reaction 2H₂O₂ \( \xrightarrow[\text{alkaline medium}]{I^-} \) 2H₂O + O₂ the proposed mechanism is as given below:
(1) H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
(2) H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate determining step?
Answer:
(i) Rate = k[H₂O₂][I⁻]
(ii) Overall order of reaction is 2.
(iii) Step (1) being the slow step is the rate determining step of the reaction.

Teacher's Note:
1. I⁻ appears in the rate law even though it is a catalyst, because it takes part in the slow step and is regenerated in the fast step.

Question 15. For a first order reaction, the time taken to reduce initial concentration by a factor of 1/4 is 10 minutes. What will be the time required to reduce initial concentration by a factor of 1/16?
Answer:
Let initial concentration (a) = 1, then final concentration (a − x) = 1/4
∴ t_1/4 = \( \frac{2.303}{k} \) log \( \frac{1}{1/4} \) ...(i)
Similarly, t_1/16 = \( \frac{2.303}{k} \) log \( \frac{1}{1/16} \) ...(ii)
Dividing equation (ii) by (i), \( \frac{t_{1/16}}{10} \) = \( \frac{\log 16}{\log 4} \) ⇒ t_1/16 = 10 × \( \frac{1.204}{0.6021} \) = 20 minutes

Teacher's Note:
1. 1/4 is two half-lives and 1/16 is four half-lives, so the time simply doubles.

Question 16. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{[R_0]}{[R_t]} \)
For 99% completion of reaction: t = t_0.99, [R]₀ = 1, [R]_t = (1 − 0.99) = 0.01 = 10⁻²
t_0.99 = \( \frac{2.303}{k} \) log \( \frac{1}{10^{-2}} \) = \( \frac{2.303}{k} \) log 10² = \( \frac{2.303}{k} \) × 2 ...(i)
For 90% completion of reaction: t = t_0.90, [R]₀ = 1, [R]_t = (1 − 0.9) = 0.1 = 10⁻¹
t_0.90 = \( \frac{2.303}{k} \) log \( \frac{1}{10^{-1}} \) = \( \frac{2.303}{k} \) log 10 = \( \frac{2.303}{k} \) ...(ii)
Comparing equations (i) and (ii), t_0.99 = 2 × t_0.90

Teacher's Note:
1. Take [R]₀ = 1 to keep the algebra short; 99% complete means 1% (0.01) remains, not 0.99.

Question 17. What is meant by rate of reaction? Differentiate between average rate and instantaneous rate of reaction.
Answer:
Change in concentration i.e., either (decrease in concentration of reactant or increase in concentration of product) per unit time is called rate of reaction.
Rate of reaction = \( \frac{C_2 - C_1}{t_2 - t_1} \) = \( \frac{\Delta C}{\Delta t} \)
The ratio of change of concentration of reactants to the time consumed in that change is called average rate of reaction.
r_av = \( \frac{\Delta x}{\Delta t} \) = −\( \frac{C_2 - C_1}{t_2 - t_1} \)
The rate of reaction at a particular instant (time) is called instantaneous rate of reaction.
r_ins = \( \frac{dx}{dt} \); dx = small change in concentration, dt = small time interval

Teacher's Note:
1. Average rate is the slope of a chord between two points; instantaneous rate is the slope of the tangent at one point.

Question 18. For a reaction A + B → P, the rate is given by Rate = k[A][B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
Answer:
(i) From the rate law equation, order of reaction w.r.t. B is 2. Hence, if concentration of B is doubled, rate will become four times.
(ii) If A is present in large excess, rate of reaction will be independent of concentration of A and hence, order of reaction will be 2.

Teacher's Note:
1. When one reactant is in excess, drop its term from the rate law and add up only the remaining powers.

Question 19. The thermal decomposition of HCO₂H is a first order reaction with a rate constant of 2.4 × 10⁻³ s⁻¹ at a certain temperature. Calculate how long will it take for three-fourth of initial quantity of HCO₂H to decompose.
(log 0.25 = −0.6021)
Answer:
For a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)
Given k = 2.4 × 10⁻³ s⁻¹, [R] = \( \frac{[R]_0}{4} \), t = ?
Substituting these values in the equation, we get t = \( \frac{2.303}{2.4\times10^{-3}\ \text{s}^{-1}} \) log \( \frac{[R]_0}{[R]} \)
t = \( \frac{2.303}{2.4\times10^{-3}\ \text{s}^{-1}} \) log 4 = \( \frac{2.303}{2.4\times10^{-3}} \) × 0.6021 s
t = 577.7 s = 578 s

Teacher's Note:
1. If three-fourth decomposes, one-fourth is left, so [R]₀/[R] = 4 (not 4/3).

Question 20. For a reaction: 2NH_3(g) \( \xrightarrow{\text{Pt}} \) N_2(g) + 3H_2(g); Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer:
(i) The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure.
In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.
However, two molecules of ammonia react to give products thus, the molecularity is two.
Order = zero; molecularity = two.
(ii) For a zero order reaction, unit of rate constant is mol L⁻¹ s⁻¹.

Teacher's Note:
1. This is a standard example where order (0) and molecularity (2) differ.

Question 21. For the reaction, 2N₂O_5(g) → 4NO_2(g) + O_2(g), the rate of formation of NO_2(g) is 2.8 × 10⁻³ M s⁻¹. Calculate the rate of disappearance of N₂O_5(g).
Answer:
Given, \( \frac{d[NO_2]}{dt} \) = 2.8 × 10⁻³ M s⁻¹
According to rate law expression, −\( \frac{1}{2} \) \( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{4} \) \( \frac{d[NO_2]}{dt} \) = \( \frac{d[O_2]}{dt} \)
∴ −\( \frac{1}{2} \) \( \frac{d[N_2O_5]}{dt} \) = \( \frac{1}{4} \) × 2.8 × 10⁻³
\( \frac{-d[N_2O_5]}{dt} \) = \( \frac{1}{2} \) × 2.8 × 10⁻³ = 1.4 × 10⁻³ M s⁻¹

Teacher's Note:
1. N₂O₅ disappears at half the rate NO₂ forms, since 2 mol N₂O₅ give 4 mol NO₂.

Question 22. For a reaction, 2H₂O₂ → 2H₂O + O₂ (in presence of I⁻, alkaline medium)
The proposed mechanism is given below:
(1) H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
(2) H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast)
(i) Write rate law of the reaction.
(ii) Write overall order of reaction.
(iii) Out of step (1) and (2), which one is rate determining step? [CBSE Delhi Set-1, 2019]
Answer:
(i) The rate law for the reaction is: rate = −\( \frac{d[H_2O_2]}{dt} \) = k[H₂O₂][I⁻]
(ii) Order = 2. This reaction is first order with respect to both H₂O₂ and I⁻; the order of the reaction is determined from the slowest step of the reaction mechanism.
(iii) Step 1. The first reaction is slow, so this is the rate determining step.

Teacher's Note:
1. Write the rate law from the slow step only; I⁻ appears in it even though it is a catalyst that does not appear in the overall equation.

Question 23. For a reaction the rate law expression is represented as follows: Rate = k[A][B]^1/2
(i) Interpret whether the reaction is elementary or complex. Give reason to support your answer.
(ii) Write the unit of rate constant for this reaction if concentration of A and B is expressed in moles/L. [CBSE SQP 2020-21]
Answer:
(i) This is a complex reaction. Order of reaction is a fraction value, i.e., 1.5. Molecularity cannot be a fraction. The reaction occurs in steps, so it is a complex reaction where order and molecularity have different values.
(ii) Rate = k[A][B]^1/2
Unit of rate constant (k) = \( \frac{\text{Unit of rate}}{\text{Unit of [A]}\times\text{Unit of [B]}^{1/2}} \) = \( \frac{mol\ L^{-1}\,s^{-1}}{(mol\ L^{-1})(mol\ L^{-1})^{1/2}} \)
= mol^−1/2 L^1/2 s⁻¹

Teacher's Note:
1. A fractional overall order is a sure sign of a complex (multi-step) reaction, since an elementary step must have a whole-number molecularity.

Teacher's Note:
1. Experiments 1 and 3 change only [Q] and the rate does not change at all – that immediately tells you the order in Q is zero.
2. Experiment 4 is a useful check: doubling [P] from Exp. 1 doubles the rate, as Rate = k[P] predicts.

Question 25. For a reaction: H₂ + Cl₂ →(hν) 2HCl; Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k. [CBSE Outside Delhi Set 1, 2016]
Answer:
(i) Zero order reaction; molecularity is 2 (bimolecular reaction).
(ii) mol L⁻¹ s⁻¹

Teacher's Note:
1. Rate = k means rate does not depend on any concentration (zero order), but two molecules (H₂ and Cl₂) still collide in the step, so molecularity is 2.

Short Answer Questions (3 Marks)

Question 26. What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L⁻¹ mol s⁻¹ (ii) L mol⁻¹ s⁻¹
Answer:
Order of a reaction: It is the sum of the power of reactant in the rate law expression.
(i) L⁻¹ mol s⁻¹ – Zero order reaction
(ii) L mol⁻¹ s⁻¹ – Second order reaction

Teacher's Note:
1. Match units to order with (mol L⁻¹)¹⁻ⁿ s⁻¹: n = 0 gives mol L⁻¹ s⁻¹, n = 2 gives L mol⁻¹ s⁻¹.

Question 27. For a chemical reaction R → P, the variation in the concentration, ln [R] vs. time (s) plot is given as
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
(iii) Write the unit of the rate constant for this reaction.

Answer:
(i) The reaction is of 1^st order.
(ii) For first order reaction, ln[R] = −kt + ln [R]₀
comparing eqn. y = m × x + c, we get a straight line with slope = −k and intercept equal to ln[R]₀.
(iii) Unit of rate constant for first order reaction: \( \frac{\text{mol L}^{-1}}{\text{s}} \) × \( \frac{1}{(\text{mol L}^{-1})^{1}} \) = s⁻¹

Teacher's Note:
1. Compare with SA-I 12: the same straight line means zero order if the y-axis is [R], but first order if it is ln[R].

Question 28. For a reaction, the rate law is: Rate = k[A][B]^1/2. Can this reaction be an elementary reaction?
Answer:
For an elementary reaction, order should be equal to molecularity and further molecularity should be integral. For the given reaction, order of reaction = 1 + 1/2 = 3/2. Since molecularity cannot be fractional, therefore, for the given reaction, order is not equal to molecularity. Hence given reaction cannot be an elementary reaction.

Teacher's Note:
1. Any fractional order is proof that a reaction is complex (multi-step).

Question 29. A reaction is first order in A and second order in B.
(i) Write differential rate equation.
(ii) How is rate affected when concentration of B is tripled?
(iii) How is rate affected when concentration of both A and B is doubled?
(iv) What is molecularity of a reaction?
Answer:
(i) Differential rate equation of reaction is \( \frac{dx}{dt} \) = k[A]¹[B]² = k[A][B]²
(ii) When conc. of B is tripled, it means conc. of B becomes [3 × B]
∴ New rate of reaction, \( \frac{dx'}{dt} \) = k[A][3B]² = 9k[A][B]² = 9 \( \left(\frac{dx}{dt}\right) \)
i.e., the rate of reaction will become 9 times.
(iii) When conc. of A is doubled and that of B is also doubled, then conc. of A becomes [2A] and that of B becomes [2B].
∴ Now rate of reaction, \( \frac{dx'}{dt} \) = k[2A][2B]² = 8k[A][B]²
i.e., the rate of reaction will become 8 times.
(iv) Molecularity of a reaction is the number of reacting particles which collide simultaneously to bring about the chemical change. It is a theoretical concept.

Teacher's Note:
1. Multiply the factor for each reactant raised to its own order: 2¹ × 2² = 8.

Teacher's Note:
1. Average rate has units of concentration per time, while the pseudo first order k has units s⁻¹; keep them distinct.

Question 31. The rate of decomposition of ammonia is found to depend upon the concentration of NH₃ according to the equation −\( \frac{d[NH_3]}{dt} \) = \( \frac{k_1[NH_3]}{1 + k_2[NH_3]} \)
What will be the order of reaction when
(i) concentration of NH₃ is very high?
(ii) concentration of NH₃ is very low?
Answer:
The given rate law equation can be written as −\( \frac{d[NH_3]}{dt} \) = \( \frac{k_1}{1/[NH_3] + k_2} \)
(i) If [NH₃] is very high, 1/[NH₃] becomes negligible
∴ −\( \frac{d[NH_3]}{dt} \) = \( \frac{k_1}{k_2} \) = k
i.e., rate becomes independent of concentration. Hence, it is of zero order.
(ii) If [NH₃] is very small, 1/[NH₃] will be very large (≫ k₂), so that k₂ can be neglected in comparison to 1/[NH₃]. Hence −\( \frac{d[NH_3]}{dt} \) = \( \frac{k_1}{1/[NH_3]} \) = k₁[NH₃]
Thus, reaction is of 1st order.

Teacher's Note:
1. This rate law describes NH₃ decomposition on a metal surface: first order when the surface is sparsely covered, zero order once it is saturated.

Question 32. A certain reaction takes 5 minutes for initial concentration 0.5 mol L⁻¹ to become 0.25 mol L⁻¹ and another 5 minutes to becomes 0.125 mol L⁻¹. What is the order and specific rate constant of the reaction?
Answer:
The given data is 0.5 mol L⁻¹ \( \xrightarrow{5\ \text{min}} \) 0.25 mol L⁻¹ \( \xrightarrow{5\ \text{min}} \) 0.125 mol L⁻¹
Half life period is independent of initial concentration of the reactant, hence reaction is of first order.
For first order reaction, k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{5\ \text{min}} \) = 0.138 min⁻¹

Teacher's Note:
1. Two successive halvings in equal times show a constant half-life, which identifies first order without any log calculation.

Teacher's Note:
1. The concentration halves every 300 s, so t_1/2 ≈ 300 s can be read straight from the table as a check.

Long Answer Questions (5 Marks)

Teacher's Note:
1. Each doubling of [N₂O₅] doubles the rate, so the reaction is first order and k = rate/[N₂O₅] for any row.

Question 35. The half-life period of a first order reaction is 30 minutes. Calculate the specific reaction rate of the reaction. What fraction of the reactant remains after 70 minutes?
Answer:
For a first order reaction k = \( \frac{0.6932}{t_{1/2}} \) = \( \frac{0.6932}{30} \) = 0.0231 min⁻¹
Suppose the reaction is A → Products; t = 0: a, 0; t = 70 min: (a − x), x
∴ Fraction of the reactant remains unreacted = \( \frac{(a-x)}{a} \)
Now, k = \( \frac{2.303}{t} \) log \( \left(\frac{a}{a-x}\right) \) or 0.0231 = \( \frac{2.303}{70} \) log \( \left(\frac{a}{a-x}\right) \)
log \( \left(\frac{a}{a-x}\right) \) = \( \frac{0.0231\times70}{2.303} \) = 0.7021
or \( \frac{a}{a-x} \) = antilog 0.7021 = 5.036
∴ \( \frac{a-x}{a} \) = \( \frac{1}{5.036} \) = 0.1985 ≈ 0.2

Teacher's Note:
1. Check with half-lives: 70 min is 2.33 half-lives, and (1/2)^2.33 ≈ 0.2.

Teacher's Note:
1. A constant k from different time intervals is the standard proof of first order; the concentration also halves every 30 s.

Question 37. For a homogeneous gas phase reaction A_(g) → B_(g) + C_(g) + D_(g), the pressure of the reaction mixture increases by 40% in 20 minute. Calculate rate constant of a reaction.
Answer:
The given reaction is A_(g) → B_(g) + C_(g) + D_(g)
Initial pressure: P_o, 0, 0, 0; After 20 min: P_o − x, x, x, x
Total pressure = P_o − x + x + x + x = P_o + 2x
P_o + 2x = P_o + P_o × \( \frac{40}{100} \); \( \frac{2x}{1} \) = \( \frac{P_{o}}{2.5} \) or P_o = 5x
∴ x = \( \frac{P_{o}}{5} \)
Now, [R]₀ = P_o; [R] = P_o − x = P_o − \( \frac{P_{o}}{5} \) = 0.8 P_o
Using formula, k = \( \frac{2.303}{t} \) log₁₀ \( \frac{[R]_0}{[R]} \)
or, k = \( \frac{2.303}{20\ \text{min}} \) log₁₀ \( \frac{P_{o}}{0.8P_{o}} \) or, k = \( \frac{2.303}{20\ \text{min}} \) log 1.25
or, k = \( \frac{2.303\times0.0969}{20\ \text{min}} \)
k = 1.115 × 10⁻² min⁻¹

Teacher's Note:
1. One mole of gas gives three, so each unit of A reacted raises the total pressure by 2x; a 40% rise means only 20% of A has reacted.