Chemistry - Chemical Kinetics Question with Solution | TestHub

Concept: Arrhenius Equation comparison, Temperature sensitivity of reactions with different activation energies.

Δ

Solution:

$\ln \left(\frac{k_1}{k_2}\right)=\ln \left(\frac{A_1}{A_2}\right)-\frac{E_{a1}-E_{a2}}{RT}$

Since $k_1=k_2$ at $T_0$ , 0 = $\ln \left(\frac{A_1}{A_2}\right)-\frac{\Delta E_a}{R{T}_0}⟹\ln \left(\frac{A_1}{A_2}\right)=\frac{\Delta E_a}{R{T}_0}$

Substitute back:

$$\ln \left(\frac{k_1}{k_2}\right)=\frac{\Delta E_a}{R}\left(\frac{1}{T_0}-\frac{1}{T}\right)$$

given $\Delta E_a=E_{a1}-E_{a2}$. Given $\Delta E_a>0$

If $T>T_0$:

Then $\left(\frac{1}{T_0}-\frac{1}{T}\right)>0$.

Since $\Delta E_a>0$ and $R>0$, it follows that $\ln \left(\frac{k_1}{k_2}\right)>0$.

This implies $\frac{k_1}{k_2}>1$, so $k_1>k_2$. Therefore, statement (A) is correct.

If $T<T_0$:

Then $\left(\frac{1}{T_0}-\frac{1}{T}\right)<0$.

Since $\Delta E_a>0$ and $R>0$, it follows that $\ln \left(\frac{k_1}{k_2}\right)<0$.

This implies $\frac{k_1}{k_2}<1$, so $k_1<k_2$. Therefore, statement (B) is incorrect.

The activation energy ($E_a$) is a measure of temperature sensitivity (it determines the slope of the Arrhenius plot, $\ln k$ vs. $1/T$). A larger $E_a$ means higher temperature sensitivity. Thus, statement (C) is correct.

slope of $\ln \left(\frac{k_1}{k_2}\right)$ versus $1/T$:

$$\ln \left(\frac{k_1}{k_2}\right)=\frac{\Delta E_a}{R{T}_0}-\frac{\Delta E_a}{R}\left(\frac{1}{T}\right)$$

This is in the form $y=c+mx$, where $y=\ln \left(\frac{k_1}{k_2}\right)$, $x=\frac{1}{T}$, and the slope $m=-\frac{\Delta E_a}{R}$.

Since $\Delta E_a>0$ and $R>0$, the slope $m$ is negative. Therefore, statement (D) is incorrect.

Final Answer: Options (A and C) are correct.