Question

The rate constants of a certain reaction at 500 K and 700 K are 0.02 sec^-1 and 0.07 sec^-1 respectively. Calculate \( E_a \) and \( A \). (2+2)

Hint:

Use the two-temperature Arrhenius equation \( \ln(k_2/k_1) = \frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2}) \) to get \( E_a \), then substitute back into \( k = A e^{-E_a/RT} \) to find A.

Answer 1

Given: \( k_1 = 0.02\ \text{s}^{-1} \) at \( T_1 = 500\ \text{K} \); \( k_2 = 0.07\ \text{s}^{-1} \) at \( T_2 = 700\ \text{K} \); \( R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \).

Step 1 (Concept): The temperature dependence of the rate constant is given by the two-temperature form of the Arrhenius equation:
\[ \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]
Step 2 (Ratio of rate constants):
\[ \frac{k_2}{k_1} = \frac{0.07}{0.02} = 3.5,\qquad \ln 3.5 = 1.2528 \]
Step 3 (Temperature term):
\[ \frac{1}{T_1}-\frac{1}{T_2} = \frac{1}{500}-\frac{1}{700} = 0.002 - 0.0014286 = 5.714\times10^{-4}\ \text{K}^{-1} \]
Step 4 (Solve for the activation energy):
\[ E_a = \frac{R\,\ln(k_2/k_1)}{\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)} = \frac{8.314 \times 1.2528}{5.714\times10^{-4}} = \frac{10.416}{5.714\times10^{-4}} \approx 18230\ \text{J mol}^{-1} \]
\[ \boxed{E_a \approx 18.23\ \text{kJ mol}^{-1}} \]
Step 5 (Frequency factor A): From \( k = A\,e^{-E_a/RT} \), so \( A = k_1\,e^{E_a/RT_1} \).
\[ \frac{E_a}{RT_1} = \frac{18230}{8.314 \times 500} = \frac{18230}{4157} = 4.385 \]
\[ A = 0.02 \times e^{4.385} = 0.02 \times 80.2 \approx 1.60\ \text{s}^{-1} \]
\[ \boxed{A \approx 1.60\ \text{s}^{-1}} \]
Check: Using \( T_2 \): \( E_a/RT_2 = 18230/5820 = 3.132 \), \( A = 0.07\times e^{3.132} = 0.07\times 22.9 \approx 1.60\ \text{s}^{-1} \); both temperatures give the same A, confirming the answer.