Question

A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \] 

Hint:

The Arrhenius equation relates the rate constant of a reaction to the activation energy, allowing us to calculate activation energy when rate constants are known at different temperatures.

Answer 1

For a first-order reaction, the rate constant \(k\) is related to the time taken for a certain fraction of the reaction to be completed. The equation for the rate constant \(k\) for a first-order reaction is: \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] Where: - \([A_0]\) is the initial concentration,
- \([A]\) is the concentration after time \(t\), and
- \(k\) is the rate constant.
However, for this problem, we will use the Arrhenius equation to relate the rate constants at two different temperatures. The Arrhenius equation is: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Where: - \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\),
- \(E_a\) is the activation energy,
- \(R\) is the universal gas constant (8.314 J mol\(^{-1}\) K\(^{-1}\)),
- \(T_1\) and \(T_2\) are the temperatures in Kelvin.
Step 1: Determine the rate constants \(k_1\) and \(k_2\). The reaction is 50 complete at both temperatures. For a first-order reaction, the time taken to reach 50 completion is related to the rate constant by the equation: \[ t_{1/2} = \frac{0.693}{k} \] This equation relates the half-life (\(t_{1/2}\)) to the rate constant (\(k\)) for a first-order reaction. Thus: \[ k_1 = \frac{0.693}{t_{1/2,1}} = \frac{0.693}{20 \, \text{min}} = 0.03465 \, \text{min}^{-1} \] \[ k_2 = \frac{0.693}{t_{1/2,2}} = \frac{0.693}{5 \, \text{min}} = 0.1386 \, \text{min}^{-1} \] Step 2: Use the Arrhenius equation. Now we can use the Arrhenius equation to solve for the activation energy (\(E_a\)): \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substitute the known values into the equation: \[ \ln \left( \frac{0.1386}{0.03465} \right) = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{350} \right) \] Simplify the logarithm: \[ \ln(4.0) = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{350} \right) \] \[ 0.602 = \frac{E_a}{8.314} \times \left( \frac{50}{300 \times 350} \right) \] Simplifying further: \[ 0.602 = \frac{E_a}{8.314} \times 0.0004762 \] Solving for \(E_a\): \[ E_a = \frac{0.602 \times 8.314}{0.0004762} = 10,574.7 \, \text{J/mol} = 10.57 \, \text{kJ/mol} \] Thus, the activation energy for the reaction is approximately 10.57 kJ/mol. \vspace{10pt}