For two chemical reactions A and B, if the difference between their activation energy is 20 kJ at 300 K (R = 8.3 J K^–1 mol^–1). Determine In K₂/K₁

The correct answer is 8.

To find the value of the natural logarithm of the ratio of the rate constants for the two chemical reactions, we use the Arrhenius equation. The Arrhenius equation relates the rate constant ($k$) of a reaction to its activation energy ($E_a$) and temperature ($T$):

$k=A{e}^{-\frac{E_a}{RT}}$

where:
• $A$ is the pre-exponential frequency factor (assumed to be constant and equal for both reactions),
• $R$ is the universal gas constant,
• $T$ is the absolute temperature in Kelvin.

For the two reactions, we can write their respective rate constants as:

$k_1=A{e}^{-\frac{E_{a1}}{RT}}$

and

$k_2=A{e}^{-\frac{E_{a2}}{RT}}$

Dividing the two equations gives:

$\frac{k_2}{k_1}=e^{\frac{E_{a1}-E_{a2}}{RT}}$

Taking the natural logarithm ($\ln$) on both sides, we get:

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

Given values in the question:
• Difference in activation energy, $\Delta E_a=20\text{kJ}=20000{\text{J mol}}^{-1}$
• Gas constant, $R=8.3{\text{J K}}^{-1}{\text{mol}}^{-1}$
• Temperature, $T=300\text{K}$

Substitute these parameters into the derived relation:

$\ln \left(\frac{k_2}{k_1}\right)=\frac{20000}{8.3\times 300}$

$\ln \left(\frac{k_2}{k_1}\right)=\frac{20000}{2490}\approx 8.03$

Rounding to the nearest integer value as given in the options yields 8.