Question

Solve

The energy of activation for a first-order reaction is 104 kJ/mol. The rate constant at 25°C is 3.7 × 10^–5 s ^–1. What is the rate constant at 30°C? (R = 8.314 J/K mol)

Answer 1

Given:

Activation energy (E_a) = 104 kJ mol^-1 = 104 × 10³ J mol^-1

Rate constant (k₁) = ${\displaystyle 3.7\times {10}^{-5}{\text{s}}^{-1}}$

Temperatures; T₁ = 25 + 273 = 298 K, T₂ = 30 + 273 = 303 K

R = 8.314 J K^-1 mol^-1

To find:

Rate constant (k₂) at 30°C

Formula:

${\displaystyle {\text{log}}_{10}\frac{{\text{k}}_2}{{\text{k}}_1}=\frac{{\text{E}}_{\text{a}}}{2.303\text{R}}\left(\frac{{\text{T}}_2-{\text{T}}_1}{{\text{T}}_2{\text{T}}_1}\right)}$

Calculation: 

${\displaystyle {\text{log}}_{10}\frac{{\text{k}}_2}{3.7\times {10}^{-5}{\text{s}}^{-1}}=\frac{104\times {10}^3\text{J} {\text{mol}}^{-1}}{2.303\times 8.314\text{J} {\text{K}}^{-1}{\text{mol}}^{-1}}\left(\frac{303\text{K}-298\text{K}}{303\text{K}\times 298\text{K}}\right)}$

∴ ${\displaystyle {\text{log}}_{10} \frac{{\text{k}}_2}{3.7\times {10}^{-5}{\text{s}}^{-1}}=\frac{104000}{2.303\times 8.314}\times \frac{5}{303\times 298}}$

∴ ${\displaystyle {\text{log}}_{10}\frac{{\text{k}}_2}{3.7\times {10}^{-5}{\text{s}}^{-1}}=0.301}$

∴ ${\displaystyle \frac{{\text{k}}_2}{3.7\times {10}^{-5}{\text{s}}^{-1}}}$ = antilog(0.301) = 2.00

k₂ = ${\displaystyle 2.00\times 3.7\times {10}^{-5}{\text{s}}^{-1}=7.4\times {10}^{-5}{\text{s}}^{-1}}$

The rate constant of the reaction is ${\displaystyle 7.4\times {10}^{-5}{\text{s}}^{-1}}$.