Question

The rate constant, k for a first order reaction, $ \text{C}_2\text{H}_5\text{I}(g) \rightarrow \text{C}_2\text{H}_4(g) + \text{HI}(g) $ is $ x \, \text{s}^{-1} $ at 600 K and $ 4x \, \text{s}^{-1} $ at 700 K. The energy of activation of the reaction (in kJ mol$ ^{-1} $) is (log 4 = 0.6, R = 8.3 J K$ ^{-1} $ mol$ ^{-1} $)

• A. 48.16
• B. 58.16
• C. 38.16
• D. 28.16

Hint:

The Arrhenius equation relates rate constant and temperature. Use the form involving two temperatures to find activation energy.

Answer 1

The Correct Option is A

Activation Energy Calculation

To find the activation energy (\(E_a\)) for the first-order reaction, we use the Arrhenius equation in its logarithmic form, which relates the rate constants at two different temperatures:

\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Given:

• Rate constant at \(T_1 = 600 \, \text{K}\): \(k_1 = x \, \text{s}^{-1}\)
• Rate constant at \(T_2 = 700 \, \text{K}\): \(k_2 = 4x \, \text{s}^{-1}\)
• Gas constant: \(R = 8.3 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}\)
• \(\log 4 = 0.6\)
• We need \(E_a\) in \(\text{kJ} \, \text{mol}^{-1}\).

Step 1: Calculate \(\ln\left(\frac{k_2}{k_1}\right)\)

\[ \frac{k_2}{k_1} = \frac{4x}{x} = 4 \]

\[ \ln 4 = 2.303 \times \log 4 = 2.303 \times 0.6 = 1.3818 \]

Step 2: Calculate the temperature term

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{600} - \frac{1}{700} \]

\[ = \frac{700 - 600}{600 \times 700} = \frac{100}{420,000} = \frac{1}{4,200} \, \text{K}^{-1} \]

Step 3: Substitute into the Arrhenius equation

\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

\[ 1.3818 = \frac{E_a}{8.3} \times \frac{1}{4,200} \]

Step 4: Solve for \(E_a\)

\[ E_a = 1.3818 \times 8.3 \times 4,200 \]

\[ = 1.3818 \times 34,860 = 48,168.708 \, \text{J} \, \text{mol}^{-1} \]

\[ E_a = \frac{48,168.708}{1,000} = 48.16 \, \text{kJ} \, \text{mol}^{-1} \]

Final Answer:

The activation energy is approximately 48.16 kJ mol\(^{-1}\)