Question

1. at 350.0 k, the rate constant for a reaction was found to be 0.0890 m⁻¹ s⁻¹. at 500.0 k, the rate constant was found to be 0.581 m⁻¹ s⁻¹. what is the activation energy, in kj/mol, for the reaction?

a) 104 kj/mol
b) 18.2 kj/mol
c) 52.5 kj/mol
d) 27.5 kj/mol
e) 72.0 kj/mol

Explanation:

Step1: Recall the Arrhenius equation

The Arrhenius equation in its two - point form is \(\ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\), where \(k_1\) and \(k_2\) are rate constants at temperatures \(T_1\) and \(T_2\) respectively, \(E_a\) is the activation energy, and \(R = 8.314\space J/(mol\cdot K)\)

Given \(k_1=0.0890\space M^{-1}s^{-1}\), \(T_1 = 350.0\space K\), \(k_2=0.581\space M^{-1}s^{-1}\), \(T_2 = 500.0\space K\)

First, calculate \(\ln\frac{k_2}{k_1}\):
\(\ln\frac{0.581}{0.0890}=\ln(6.528)\approx1.876\)

Next, calculate \(\frac{1}{T_1}-\frac{1}{T_2}\):
\(\frac{1}{350.0}-\frac{1}{500.0}=\frac{500 - 350}{350\times500}=\frac{150}{175000}=8.571\times10^{-4}\space K^{-1}\)

Step2: Solve for \(E_a\)

From \(\ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\), we can solve for \(E_a\)
\(E_a=R\times\ln\frac{k_2}{k_1}\div(\frac{1}{T_1}-\frac{1}{T_2})\)
Substitute \(R = 8.314\space J/(mol\cdot K)\), \(\ln\frac{k_2}{k_1}=1.876\), and \(\frac{1}{T_1}-\frac{1}{T_2}=8.571\times 10^{-4}\space K^{-1}\)
\(E_a=8.314\times1.876\div(8.571\times 10^{-4})\)
\(E_a=\frac{15.65}{8.571\times 10^{-4}}\space J/mol\approx18260\space J/mol = 18.26\space kJ/mol\)

Answer:

B. \(18.2\space kJ/mol\)