Question

R₁ and R₂ are two reactions having identical pre-exponential factors. The activation energy of R₁ exceeds R₂ by $10{\mathrm{kJmol}}^{-1}$.k₁ and k₂ are two rate constants for reaction R₁ and R₂ at$300\mathrm{K}$, then the ratio of $\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}$ will be?( R=$8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}$)

• A. $6$
• B. $4$
• C. $8$
• D. $12$

Answer 1

The correct option is B

$4$

Explanation for the correct options:

B. $4$

Step1:

• According to Arrhenius Equation $\mathrm{K}={\mathrm{Ae}}^{-\mathrm{Ea}/\mathrm{RT}}$
• Where $\mathrm{A}$ is pre-exponential, $\mathrm{K}$ is Rate constant, ${\mathrm{E}}_{\mathrm{a}}$ is the Activation energy, $\mathrm{T}$ is temperature and $\mathrm{R}$ is gas constant.
• In the log form, it can be written as $\mathrm{lnK}=\mathrm{lnA}-\frac{\mathrm{Ea}}{\mathrm{RT}}$
• As it is given that R₁ exceeds R₂ it will be=${\mathrm{E}}_{{\mathrm{a}}_1}-{\mathrm{E}}_{{\mathrm{a}}_2}=10{\mathrm{kJmol}}^{-1}$

Step 2:

• For R₁ it will be=$\ln \hspace{0.17em}{\mathrm{K}}_1=\ln \mathrm{A}-\frac{\mathrm{Ea}}{\mathrm{RT}}$__(equation i)
• For R₂ it will be=${\mathrm{lnK}}_2=\mathrm{lnA}-\frac{\mathrm{Ea}}{\mathrm{RT}}$___(equation ii)

Step 3:

• Subtract equations (I) and (II)
• ${\mathrm{lnK}}_2-{\mathrm{K}}_1=\frac{-{\mathrm{Ea}}_2}{\mathrm{RT}}+\frac{{\mathrm{Ea}}_1}{\mathrm{RT}}$
• $\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{1}{\mathrm{RT}}\left[{\mathrm{Ea}}_1-{\mathrm{Ea}}_2\right]$
• $\ln \frac{K_2}{K_1}=\frac{1}{8.314\times 300}\times 10\times {10}^3$$=4.016$$\simeq 4$

Hence option (B) is correct., the ratio of $\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}$ will be $4$.