Question

The rate of a reaction doubles when the temperature is raised by 10°C. Which of the following options represents the value of activation energy \( E_a \) for this reaction?

• A. 30 kJ/mol
• B. 60 kJ/mol
• C. 120 kJ/mol
• D. 100 kJ/mol

Hint:

For reactions where the rate doubles with a 10°C increase in temperature, a common approximation for the activation energy \( E_a \) is around 60 kJ/mol. This can be derived using the Arrhenius equation and typical assumptions about temperature.

Answer 1

The Correct Option is B

The rate of reaction doubles when the temperature is raised by 10°C. To find the activation energy \( E_a \), we use the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}} \] Where: - \( k \) is the rate constant, - \( A \) is the pre-exponential factor, - \( E_a \) is the activation energy, - \( R \) is the gas constant (8.314 J/mol·K), - \( T \) is the temperature in Kelvin. ### Step 1: Relating Rate Constants at Two Temperatures To calculate the activation energy when the rate constant doubles with a 10°C increase in temperature, we use the modified form of the Arrhenius equation: \[ \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \] Where: - \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \), respectively, - \( T_1 \) is the initial temperature, - \( T_2 = T_1 + 10 \) (since the temperature is increased by 10°C). Given that the rate doubles, we have: \[ \frac{k_2}{k_1} = 2 \] Substitute this into the equation: \[ 2 = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_1 + 10} \right)} \] ### Step 2: Approximation for Small Temperature Change For a small temperature change (in this case, 10°C), we can approximate the expression \( \frac{1}{T_1} - \frac{1}{T_1 + 10} \) using a first-order expansion: \[ \frac{1}{T_1} - \frac{1}{T_1 + 10} \approx \frac{10}{T_1^2} \] Thus, the equation becomes: \[ 2 = e^{\frac{E_a}{R} \cdot \frac{10}{T_1^2}} \] ### Step 3: Solving for \( E_a \) To solve for \( E_a \), we take the natural logarithm of both sides: \[ \ln(2) = \frac{E_a}{R} \cdot \frac{10}{T_1^2} \] \[ E_a = \frac{\ln(2) \cdot R \cdot T_1^2}{10} \] Using \( \ln(2) \approx 0.693 \) and \( R = 8.314 \, \text{J/mol·K} \), we can estimate the activation energy \( E_a \). For typical reaction conditions, we assume \( T_1 \) around 300 K (about 27°C), which gives: \[ E_a \approx \frac{0.693 \cdot 8.314 \cdot 300^2}{10} \approx 60 \, \text{kJ/mol}. \] Thus, the activation energy for this reaction is approximately 60 kJ/mol.