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 determined by the time required for a specific fraction of the reaction to complete. The relationship for the rate constant \(k\) in a first-order reaction is given by: \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] Where: - \([A_0]\) represents the initial concentration.
- \([A]\) denotes the concentration at time \(t\).
- \(k\) is the rate constant.
For this problem, the Arrhenius equation will be employed to link rate constants at two distinct temperatures. The Arrhenius equation is expressed as: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Where: - \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\), respectively.
- \(E_a\) 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: Calculate the rate constants \(k_1\) and \(k_2\). The reaction reaches 50% completion at both temperatures. For a first-order reaction, the time to achieve 50% completion (half-life, \(t_{1/2}\)) is related to the rate constant by: \[ t_{1/2} = \frac{0.693}{k} \] Rearranging this equation to solve for \(k\): \[ k_1 = \frac{0.693}{t_{1/2,1}} = \frac{0.693}{20 \, \text{min}} = 0.03465 \, \text{min}^{-1} \] \[ k_2 = \frac{0.693}{t_{1/2,2}} = \frac{0.693}{5 \, \text{min}} = 0.1386 \, \text{min}^{-1} \] Step 2: Apply the Arrhenius equation. The Arrhenius equation can now be used to determine the activation energy (\(E_a\)): \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substituting the known values: \[ \ln \left( \frac{0.1386}{0.03465} \right) = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{350} \right) \] Simplifying the logarithmic term: \[ \ln(4.0) = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{350} \right) \] \[ 0.602 = \frac{E_a}{8.314} \times \left( \frac{50}{105000} \right) \] \[ 0.602 = \frac{E_a}{8.314} \times 0.0004762 \] Solving for \(E_a\): \[ E_a = \frac{0.602 \times 8.314}{0.0004762} = 10,574.7 \, \text{J/mol} \] Converting to kilojoules per mole: \[ E_a = 10.57 \, \text{kJ/mol} \] The activation energy for this reaction is approximately 10.57 kJ/mol. \vspace{10pt}