Question

The temperature dependence of some reactions can be non-arrhenius, in the sense that activation energy $E_a$ is function of temperature. But the activation energy can still be defined as

$E_a=R{T}^2\left(\frac{\text{d lnk}}{\text{dT}}\right)$.

Assume if activation energy of one such reaction varies as $2T^3$ $(T\to$ temperature$)$. $k_1$ and $k_2$ are rate constants at temperature $T_1$ and $T_2$ respectively, where $T_2>T_1.$ On the basis of this information select the correct statement(s).

• A. On increasing temperature from $T_1$ to $T_2,$ rate constant increases
• B. On increasing temperature, rate of reaction increases
• C. In such case straight line is not obtained when $\ln k$ is plotted against $\frac{\text{1}}{\text{T}}$
• D. On increasing temperature from $T_1$ to $T_2,$ rate constant decreases

Answer 1

Answer: (A), (B), (C)

In such case straight line is not obtained when $\ln k$ is plotted against $\frac{\text{1}}{\text{T}}$

$\frac{\text{d}\left(\ln \text{k}\right)}{\text{dT}}=\frac{E_a}{\text{R}{\text{T}}^{\text{2}}}$

$E_a=2T^3$

$\underset{k_1}{\overset{{\text{k}}_2}{\int }}\text{d}\ln \text{k}=\underset{T_1}{\overset{T_2}{\int }}\frac{\text{2T}}{\text{R}}\text{dT}$

$R\ln \left(\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\right)=T_2^2–T_1^2$

as $T_2>T_1$ then $k_2>k_1$

Means rate increase