Question

The rate constant of a reaction at 400 K is three times the value at 300 K. The activation energy of the reaction in kJ mol\(^{-1}\) is ................... (round off to 1 decimal place).
Given: Universal gas constant, R = 8.314 J mol\(^{-1}\)K\(^{-1}\)

Hint:

Always be careful with units in the Arrhenius equation. The gas constant R is usually given in J/mol-K, which means the calculated activation energy \(E_a\) will be in J/mol. Remember to convert to kJ/mol if the question asks for it.

Answer 1

Step 1: Understanding the Concept:
This problem relates the change in the rate constant of a reaction with temperature. This relationship is described by the Arrhenius equation. We can use the two-point form of the Arrhenius equation to solve for the activation energy (\(E_a\)).
Step 2: Key Formula or Approach:
The Arrhenius equation is \(k = A e^{-E_a/RT}\). The two-point form, which relates rate constants \(k_1\) and \(k_2\) at temperatures \(T_1\) and \(T_2\), is: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] We are given the information to solve for \(E_a\).
Step 3: Detailed Calculation:
Let's define the given values:

- \(T_1 = 300\) K

- \(T_2 = 400\) K

- The rate constant at 400 K (\(k_2\)) is three times the value at 300 K (\(k_1\)). So, \(k_2 = 3k_1\), which means \( \frac{k_2}{k_1} = 3 \).

- \(R = 8.314\) J mol\(^{-1}\)K\(^{-1}\)

Substitute these values into the Arrhenius equation: \[ \ln(3) = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{400}\right) \] First, calculate the terms in the parentheses: \[ \frac{1}{300} - \frac{1}{400} = \frac{4 - 3}{1200} = \frac{1}{1200} \] The value of \(\ln(3)\) is approximately 1.0986. Now the equation is: \[ 1.0986 = \frac{E_a}{8.314} \left(\frac{1}{1200}\right) \] Rearrange to solve for \(E_a\): \[ E_a = 1.0986 \times 8.314 \times 1200 \] \[ E_a \approx 9.135 \times 1200 \] \[ E_a \approx 10962.3 \text{ J mol}^{-1} \] The question asks for the activation energy in kJ mol\(^{-1}\). To convert, divide by 1000: \[ E_a = \frac{10962.3}{1000} \text{ kJ mol}^{-1} = 10.9623 \text{ kJ mol}^{-1} \] Final Calculation based on 11.0 kJ/mol \[ E_a = 10962.3 \text{ J mol}^{-1} \] \[ E_a = 10.9623 \text{ kJ mol}^{-1} \] Rounding off to 1 decimal place: \[ E_a = 11.0 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The activation energy of the reaction is 11.0 kJ mol\(^{-1}\).
Step 5: Why This is Correct:
The solution correctly applies the two-point Arrhenius equation to relate the change in rate constant to the change in temperature. The calculation yields an activation energy of approximately 10.96 kJ/mol, which rounds to 11.0 kJ/mol. This falls within the answer key range of 10.5 to 11.5.