CBSE Class 12 Chemistry Chemical Kinetics Case Study Questions Set 01

Answer: (c) 230.3 s

Explanation:
1. For a first order reaction: t_1/2 = \( \frac{0.693}{k} \), k = 3.01 × 10⁻³ s⁻¹
2. ∴ t_1/2 = \( \frac{0.693}{3.01\times10^{-3}} \) = 230.3 s

Teacher's Note:
1. "Decompose half of the reactant" is just another way of asking for t_1/2.

Answer: (a) 990 s

Explanation:
1. Half life (t_1/2) of a first order reaction is given as: t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{7.0\times10^{-4}} \) = 990 s

Teacher's Note:
1. The initial concentration 0.080 M is a distractor; first order half-life does not depend on it.

Answer: (b) It is dependent on rate of the reaction.

Explanation:
1. For a first order reaction t_1/2 = \( \frac{0.693}{k} \)
2. therefore t_1/2 depends solely on k (the rate constant), not on the rate, which changes as concentration changes.

Teacher's Note:
1. Keep "rate" and "rate constant" apart: the rate falls as the reaction proceeds, but k (and hence t_1/2) stays fixed at a given temperature.

Answer: (c) 24.086 min

Explanation:
1. Let the concentrations of the reactant after 10 min and 20 min be C₁ and C₂ respectively.
2. ∴ Rate after 10 min = 0.04 × 60 mol L⁻¹ min⁻¹ = k.C₁ and rate after 20 min = 0.03 × 60 mol L⁻¹ min⁻¹ = k.C₂
3. ∴ \( \frac{C_1}{C_2} \) = \( \frac{0.04\times60}{0.03\times60} \) = \( \frac{4}{3} \)
4. Let the reaction starts after 10 minutes. k = \( \frac{2.303}{10} \) log \( \frac{C_1}{C_2} \) = \( \frac{2.303}{10} \) log \( \frac{4}{3} \) = 0.02878 min⁻¹
5. ∴ t_1/2 = \( \frac{0.6932}{k} \) = \( \frac{0.6932}{0.02878} \) = 24.086 min

Teacher's Note:
1. For first order, rate ∝ concentration, so the ratio of rates equals the ratio of concentrations; you never need the actual concentrations.

Answer: (b) Graph (b)

Explanation:
1. For first order reactions, t_1/2 ∝ [A]₀⁰, i.e., t_1/2 = \( \frac{0.693}{k} \). Thus t_1/2 is independent of initial concentration. Hence plot of t_1/2 vs [A]₀ will be a horizontal line.

Teacher's Note:
1. Compare: for zero order t_1/2 = [R]₀/2k, so its t_1/2 vs [R]₀ plot is a rising line through the origin.

Answer: (c) A is true, but R is false.

Explanation:
1. For first order reaction, Rate₁ = k[A₁]
2. According to question, [A₂] = [2A₁]
3. ∴ Rate₂ = k[2A₁] ⇒ Rate₂ = 2 Rate₁
4. For a given reaction, rate constant is constant and independent of the concentration of reactant.

Teacher's Note:
1. Rate changes with concentration; the rate constant changes only with temperature (or a catalyst).

Answer: (c) A is true, but R is false.

Explanation:
1. For a first order reaction k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \)
2. k = \( \frac{2.303}{t_{1/2}} \) log \( \frac{a}{a - a/2} \) = \( \frac{2.303}{t_{1/2}} \) log \( \frac{a}{a/2} \) = \( \frac{2.303}{t_{1/2}} \) log 2
3. Therefore half-life period t_1/2 = \( \frac{2.303}{k} \) log 2.
4. Thus t_1/2 is independent of initial concentration of reactant for first order reaction.

Teacher's Note:
1. (2.303/k) log 2 is exactly 0.693/k, since 2.303 × 0.3010 = 0.693.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For a first order reaction, [A] = [A]₀e^−kt ...(i)
2. According to eq. (i), concentration of reactant decreases exponentially.
3. The reason is also true (rate = k[A]), but it is a general statement and not the explanation of the exponential decrease.

Teacher's Note:
1. The exponential form comes from integrating the first order rate law, [A] = [A]₀ e^−kt.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a first order reaction, t_1/2 is inversely proportional to k, it does not depend on the initial concentration of the reactant.

Teacher's Note:
1. No concentration term appears in t_1/2 = 0.693/k, which is exactly why it is independent of initial concentration.

Answer: (b) first

Explanation:
1. From the expression −\( \frac{d[A]}{dt} \) = k[A][B]²
2. when B is present in large excess, rate will be independent upon the change in conc. of B, therefore order of reaction will be one.

Teacher's Note:
1. The excess reactant term k[B]² becomes a new constant k′, leaving rate = k′[A].

Answer: (c) 10

Explanation:
1. The total number of reactant molecules participating in a chemical reaction is known as its molecularity, hence the molecularity = 6 + 3 + 1 = 10.

Teacher's Note:
1. This follows the passage definition (sum of reactant coefficients). In reality ten particles never collide at once; such a reaction goes through several elementary steps, each with molecularity 1, 2 or 3.

Answer: (c) Molecularity and order always have same values for a reaction.

Explanation:
1. Molecularity may or may not be equal to the order of a reaction.

Teacher's Note:
1. Order equals molecularity only for elementary reactions; for complex reactions they often differ (e.g., pseudo first order hydrolysis).

Answer: (d) \( \frac{13}{12} \)

Explanation:
1. Order of reaction = \( \frac{1}{2} \) + \( \frac{1}{3} \) + \( \frac{1}{4} \) = \( \frac{6+4+3}{12} \) = \( \frac{13}{12} \)

Teacher's Note:
1. Overall order is the sum of all the powers in the rate law; use the LCM (12) to add the fractions.

Answer: (c) 3

Explanation:
1. 2NO_(g) + Cl_2(g) → 2NOCl_(g)
2. Molecularity = 3

Teacher's Note:
1. Molecularity counts reacting particles in the step: 2 NO + 1 Cl₂ = 3.

Answer: (b) r = k[NO]²[Cl₂]

Explanation:
1. Let the rate of this reaction, r = k[NO]^m[Cl₂]ⁿ
2. then \( \frac{r_1}{r_2} \) = \( \frac{0.60}{1.20} \) = \( \frac{k(0.15)^m(0.15)^n}{k(0.15)^m(0.30)^n} \) or, \( \frac{1}{2} \) = \( \left(\frac{1}{2}\right)^n \) ⇒ n = 1
3. Again from \( \frac{r_2}{r_3} \) = \( \frac{1.20}{2.40} \) = \( \frac{k(0.15)^m(0.30)^n}{k(0.30)^m(0.15)^n} \) or \( \frac{1}{2} \) = \( \left(\frac{1}{2}\right)^m \) · \( \frac{2}{1} \) or \( \frac{1}{4} \) = \( \left(\frac{1}{2}\right)^m \) ⇒ m = 2
4. Hence, expression for rate law is r = k[NO]²[Cl₂]¹

Teacher's Note:
1. Doubling [Cl₂] (expt 1 → 2) doubles the rate, and doubling [NO] (expt 1 → 3) quadruples it: first order in Cl₂, second order in NO.

Answer: (d) 3

Explanation:
1. As the order w.r.t. NO is 2 and order w.r.t. Cl₂ is 1, hence the overall order is 3.

Teacher's Note:
1. Here order and molecularity happen to be equal (3), as expected for an elementary termolecular step.

Answer: (c) 177.77 M⁻² min⁻¹

Explanation:
1. Substituting the values of experiment 1 in rate law expression
2. 0.60 M min⁻¹ = k(0.15 M)²(0.15 M)¹
3. or k = \( \frac{0.60\ \text{M min}^{-1}}{0.0225\times0.15\ \text{M}^3} \) = 177.77 M⁻² min⁻¹

Teacher's Note:
1. For a third order reaction k has units M⁻² min⁻¹; that alone rules out options (b) and (d).

Answer: (d) 2.77 M min⁻¹

Explanation:
1. r = 177.7 M⁻² min⁻¹ × (0.25 M)²(0.25 M) = 2.77 M min⁻¹

Teacher's Note:
1. Once k and the orders are known, any unknown rate is a direct substitution into r = k[NO]²[Cl₂].

Answer: (a) s⁻¹

Explanation:
1. Unit of rate constant for a reaction of n^th order = (conc.)¹⁻ⁿ time⁻¹
2. For a first order reaction, n = 1
3. Unit of rate constant = (mol L⁻¹)^{1 − 1} s⁻¹ = s⁻¹

Teacher's Note:
1. Units of k at a glance: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹.

Answer: (d) 0.0693 × 3 M min⁻¹

Explanation:
1. 12 M → 6 M (after one t_1/2) → 3 M (after two t_1/2); t_1/2 = 10 min
2. k = \( \frac{0.693}{10} \) = 0.0693 min⁻¹
3. As t_1/2 is 10 min, after 20 minutes the concentration will be 3 M.
4. Hence, Rate = 0.0693 × 3 M min⁻¹

Teacher's Note:
1. Use the concentration at that moment (3 M), not the initial 12 M, when calculating an instantaneous rate.

Answer: (b) 76.4 minutes

Explanation:
1. t_1/2 = 23 minutes
2. t_1/2 = \( \frac{0.693}{k} \) ⇒ k = \( \frac{0.693}{t_{1/2}} \) ⇒ k = \( \frac{0.693}{23} \) min⁻¹
3. For 90% completion, t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \); t = \( \frac{2.303\times23}{0.693} \) log \( \frac{100}{100-90} \)
4. t = 76.4 minutes

Teacher's Note:
1. Useful shortcut: for first order, t_90% = 3.32 × t_1/2.

Answer: (a) 3.47 × 10⁻⁴ M/min

Explanation:
1. For the first order reaction, k = \( \frac{2.303}{t} \) log \( \left(\frac{a}{a-x}\right) \)
2. a = 0.1 M, a − x = 0.025 M, t = 40 min
3. k = \( \frac{2.303}{40} \) log \( \frac{0.1}{0.025} \) = \( \frac{2.303}{40} \) log 4 = 0.0347 min⁻¹
4. Thus, rate = k[A]
5. rate = 0.0347 × 0.01 M min⁻¹ = 3.47 × 10⁻⁴ M min⁻¹

Teacher's Note:
1. 0.1 → 0.025 M is two half-lives in 40 min, so t_1/2 = 20 min and k = 0.693/20 = 0.0347 min⁻¹: a quick check.

Answer: (a) 6.25%

Explanation:
1. t_1/2 = \( \frac{0.693}{k} \) ⇒ k = \( \frac{0.693}{60} \) = 0.01155 min⁻¹
2. k = \( \frac{2.303}{t} \) log \( \left(\frac{a}{a-x}\right) \). Let the initial amount (a) be 100.
3. 0.01155 min⁻¹ = \( \frac{2.303}{240\ \text{min}} \) log \( \left(\frac{100}{a-x}\right) \)
4. \( \frac{0.01155\ \text{min}^{-1}\times240\ \text{min}}{2.303} \) = log \( \left(\frac{100}{a-x}\right) \)
5. 1.204 = log 100 − log (a − x); 1.204 = 2 − log (a − x)
6. log (a − x) = 2 − 1.204 = 0.796
7. (a − x) = 6.25%

Teacher's Note:
1. 240 min is exactly 4 half-lives: 100 → 50 → 25 → 12.5 → 6.25%, no logarithms needed.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Rate of a reaction is the rate of change of concentration of reactant or product with time.
2. Mathematical expression for rate of reaction, R → P: Rate = −\( \frac{\Delta C_R}{\Delta t} \) = \( \frac{\Delta C_P}{\Delta t} \)

Teacher's Note:
1. Rate can be expressed using any reactant (with a minus sign) or any product (with a plus sign); both forms give the same positive value after dividing by coefficients.

Answer: (c) A is true, but R is false.

Explanation:
1. For the given chemical reaction, order is the sum of powers of the reactants which is calculated experimentally, i.e., 1/2 + 1 = 1.5
2. In an elementary reaction, order can be the sum of stoichiometric coefficients of the reactants but the given reaction is a complex reaction.

Teacher's Note:
1. Add the powers in the rate law, not the coefficients in the equation: here 1/2 + 1 = 1.5, while the coefficients would wrongly give 3.

Answer: (d) A is false, but R is true.

Explanation:
1. For a general reaction aA + bB → cC + dD, Rate = k[A]^x × [B]^y, where x + y = n = order of the reaction.
2. Unit of k = [mol L⁻¹]¹⁻ⁿ s⁻¹. Thus, k is dependent on the order of the reaction, so the unit of k for different orders have different units. For zero order, n = 0, thus unit of k is mol L⁻¹ s⁻¹.

Teacher's Note:
1. Remember the three common units: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Rate of reaction is the speed with which a reaction takes place, which can occur at different speeds.
2. The more often molecules collide with each other, the faster the reaction proceeds; and the more forcefully molecules collide with each other, the more likely they are to react and the faster the reaction proceeds.

Teacher's Note:
1. Ionic reactions in solution are almost instantaneous, while rusting takes months – the same idea of rate covers both.

Answer: (b) First order

Explanation:
1. Since the rate of the reaction is proportional to the concentration of the reactant, i.e., ethene, it is a first order reaction.

Teacher's Note:
1. “Rate proportional to concentration” means rate ∝ [R]¹, i.e. first order; the unit s⁻¹ of k confirms it.

Answer: (c) Rate = k[C₂H₄]

Explanation:
1. C₂H₄ (ethene) + H₂ →(catalyst) C₂H₆ (ethane)
2. Here, hydrogen is taken in excess so the reaction is first order reaction. Rate law equation, Rate = k[C₂H₄]

Teacher's Note:
1. A rate law contains only reactant concentrations – any option with the product (ethane) in it can be rejected straight away.

Answer: (a) 2.772 × 10¹⁴ s

Explanation:
1. For first order reaction, t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{2.5\times10^{-15}\ s^{-1}} \) = 2.772 × 10¹⁴ s

Teacher's Note:
1. Dividing by a very small k gives a very large half-life: 10⁻¹⁵ in the denominator becomes 10⁺¹⁵, so the exponent must be positive.

Answer: (c) 0.2197 min⁻¹

Explanation:
1. t = 5 min, \( \frac{[R]_0}{[R]} \) = 3
2. For first order reaction, k = \( \frac{2.303}{t} \) log \( \frac{[R]_0}{[R]} \) = \( \frac{2.303}{5} \) log 3 = \( \frac{2.303}{5} \) × 0.4771 = 0.2197 min⁻¹

Teacher's Note:
1. “Reduces to one-third” means [R] = [R]₀/3, so the ratio [R]₀/[R] = 3 (not 1/3).

Answer: (b) −k

Explanation:
1. For this first order reaction, the plot of ln[R] vs t is a straight line with intercept ln[R]₀; Slope = −k.

Teacher's Note:
1. The line slopes downward because [R] falls with time; that is why the slope is negative.