Mix Examples-Chemical Kinetics Questions in English

(A) For zero order reaction: $t_{1/2} = \frac{a}{2k}$ $\Rightarrow 10 = \frac{100}{2k}$ $\Rightarrow k = 5 \ mol \ L^{-1} \ min^{-1}$. After $t = 20 \ min$,amount reacted $x = kt = 5 \times 20 = 100 \ mol$. Remaining reactant $(a-x) = 100 - 100 = 0 \ mol$.
For first order reaction: Number of half-lives $n = \frac{20}{10} = 2$. Remaining reactant $(a-x) = \frac{a}{2^n} = \frac{100}{2^2} = 25 \ mol$.
For second order reaction: $t_{1/2} = \frac{1}{ak}$ $\Rightarrow 10 = \frac{1}{100k}$ $\Rightarrow k = 10^{-3} \ L \ mol^{-1} \ min^{-1}$. Using integrated rate law $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$,we get $\frac{1}{[A]_t} = \frac{1}{100} + (10^{-3} \times 20) = 0.01 + 0.02 = 0.03$. Thus,$[A]_t = \frac{1}{0.03} = 33.33 \ mol$.

(C) $1$. For Statement $I$: Using the Arrhenius equation,$\log(k_2/k_1) = \frac{E_a}{2.303R} \left(\frac{T_2-T_1}{T_1 T_2}\right)$.
Given $E_a = 12.6 \text{ kcal/mol} = 12600 \text{ cal/mol}$,$R = 1.987 \text{ cal K}^{-1} \text{ mol}^{-1} \approx 2 \text{ cal K}^{-1} \text{ mol}^{-1}$,$T_1 = 298 \text{ K}$,$T_2 = 308 \text{ K}$.
$\log(2) = 0.301 = \frac{12600}{2.303 \times 2} \left(\frac{10}{298 \times 308}\right) \approx \frac{12600}{4.606} \times \frac{10}{91784} \approx 2735.5 \times 0.000109 \approx 0.298 \approx 0.3$.
Since $\log(2) \approx 0.3$,Statement $I$ is true.
$2$. For Statement $II$: For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$. This shows that $t_{1/2}$ is independent of the initial concentration $[A]_o$. Therefore,the graph of $t_{1/2}$ versus $[A]_o$ should be a horizontal line,not a straight line passing through the origin. Thus,Statement $II$ is false.

(B) $1$. For a zero-order reaction: The rate constant is $k = [A]_0 / t_{100\%}$. Also,$t_{1/2} = [A]_0 / (2k)$,which implies $[A]_0 = 2kt_{1/2}$. Substituting this into the first equation: $k = (2kt_{1/2}) / t_{100\%}$,which simplifies to $t_{100\%} = 2t_{1/2}$.
$2$. For a first-order reaction: The concentration at time $t$ is given by $[A]_t = [A]_0 e^{-kt}$. For $100\%$ completion,$[A]_t = 0$. This implies $e^{-kt} = 0$,which occurs only when $t \to \infty$. Thus,theoretically,a first-order reaction never reaches $100\%$ completion in finite time.