Reactant Concentration Variation with Time for Three Reactions

To determine the correct statement, let us analyze the given concentration-time ($[R]$ vs $t$) graph for the three reactions starting from the same initial concentration $[R]_0$:

1. Analysis of Curve 1:

   • Curve 1 is a straight line with a negative slope. This corresponds to a zero-order reaction, for which the rate law is given by: $[R] = [R]_0 - k_1 t$
   • For a zero-order reaction, the unit of the rate constant $k_1$ is: $\text{Unit of } k_1 = \frac{\text{Concentration}}{\text{Time}} = \text{mol L}^{-1}\text{s}^{-1}$
   • Therefore, Option C is incorrect because $\text{s}^{-1}$ is the SI unit for a first-order rate constant, not zero-order.
   • Thermal decomposition of $\text{HI}$ on a gold surface is a classic example of a zero-order reaction, which is represented by curve 1, not curve 2. Thus, Option D is incorrect.

2. Analysis of Curves 2 and 3:

   • For a reaction of order $n$, the general rate equation is: $-\frac{d[R]}{dt} = k [R]^n$
   • Integrating this equation for $n = 1$ gives: $[R](t) = [R]_0 e^{-kt}$
   • For $n \neq 1$ ($n > 0$), integration gives: $\frac{1}{n-1}\left(\frac{1}{[R]^{n-1}} - \frac{1}{[R]_0^{n-1}}\right) = kt$
   • In both cases, for a fixed order $n$ and identical initial concentration $[R]_0$, a larger rate constant $k$ causes the reactant concentration $[R]$ to decrease more rapidly over time $t$.
   • Looking at the graph, at any given time $t > 0$: $[R]_3 < [R]_2$
   • This shows that the concentration of reactant decreases faster in reaction 3 than in reaction 2. Thus, if both reactions have the same order, the rate constant of reaction 3 must be larger than that of reaction 2 ($k_3 > k_2$).
   • Therefore, Option B is correct.

3. Evaluation of Option A:

   • Reaction 1 is linear (zero-order) while reactions 2 and 3 are non-linear curves. Hence, all three reactions do not have the same order, making Option A incorrect.

Correct Answer: B