Question

The order of reaction A → Product can be given by the expression(s) [where r = rate of reaction, $[A]_1$ and $[A]_2$ are concentration at $t_1$ and $t_2$ respectively]

• A. $\frac{ln\,r_2 - ln\,r_1}{ln\,[A]_2 - ln\,[A]_1}$
• B. $\frac{ln\,[A]_2 - ln\,[A]_1}{ln\,[t_{1/2}]_2 - ln\,[t_{1/2}]_1}$
• C. $\frac{ln(\frac{-d[A]}{kdt})}{ln[A]}$
• D. $\frac{ln(\frac{r}{k})}{ln[A]}$

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The rate law for a reaction A → Product is $r = k[A]^n$.

1. Option A: Taking natural log of the rate law, $ln\,r = ln\,k + n\,ln\,[A]$. For two data points $(r_1, [A]_1)$ and $(r_2, [A]_2)$, subtracting the equations $ln\,r_2 - ln\,r_1 = n(ln\,[A]_2 - ln\,[A]_1)$ yields $n = \frac{ln\,r_2 - ln\,r_1}{ln\,[A]_2 - ln\,[A]_1}$.
2. Option B: The half-life for $n \neq 1$ is $t_{1/2} \propto [A]_0^{1-n}$. Taking natural log, $ln\,t_{1/2} = C' + (1-n)ln\,[A]_0$. For two data points, $ln\,[t_{1/2}]_2 - ln\,[t_{1/2}]_1 = (1-n)(ln\,[A]_2 - ln\,[A]_1)$. This leads to $\frac{ln\,[A]_2 - ln\,[A]_1}{ln\,[t_{1/2}]_2 - ln\,[t_{1/2}]_1} = \frac{1}{1-n}$, not 'n'.
3. Option C and D: From the rate law $r = k[A]^n$, divide by $k$ to get $\frac{r}{k} = [A]^n$. Taking natural log, $ln(\frac{r}{k}) = n\,ln[A]$. Thus, $n = \frac{ln(\frac{r}{k})}{ln[A]}$. Option C uses $r = -d[A]/dt$, so $\frac{-d[A]}{kdt} = \frac{r}{k}$, making C equivalent to D.