Question

A first-order reaction has a rate constant of \(2.0\times10^{-3}\ s^{-1}\) at \(300\ K\) and \(8.0\times10^{-3}\ s^{-1}\) at \(330\ K\). Assuming the Arrhenius equation to be valid, the activation energy of the reaction is closest to:

• A. \(28.5\ kJ\,mol^{-1}\)
• B. \(38.0\ kJ\,mol^{-1}\)
• C. \(57.0\ kJ\,mol^{-1}\)
• D. \(76.0\ kJ\,mol^{-1}\)

Hint:

For Arrhenius numericals: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac1{T_1} - \frac1{T_2} \right) \] Always calculate the \(k_2/k_1\) ratio first. Most CUET questions become straightforward after this step.

Answer 1

The Correct Option is B

Concept: This question combines two of the most important areas of Chemical Kinetics:
• First-order reactions
• Arrhenius equation The Arrhenius equation relates the rate constant to temperature and activation energy. For two temperatures: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] where: \[ k_1,k_2 \] are the rate constants, \[ T_1,T_2 \] are absolute temperatures, and \[ E_a \] is the activation energy.

Step 1: Write the given data carefully. \[ k_1 = 2.0\times10^{-3}\ s^{-1} \] \[ T_1 = 300K \] \[ k_2 = 8.0\times10^{-3}\ s^{-1} \] \[ T_2 = 330K \]

Step 2: Calculate the ratio of rate constants. \[ \frac{k_2}{k_1} = \frac{8\times10^{-3}} {2\times10^{-3}} \] \[ =4 \] Therefore: \[ \log\left(\frac{k_2}{k_1}\right) = \log4 \] \[ =0.602 \]

Step 3: Calculate the temperature factor. \[ \left( \frac{1}{300} - \frac{1}{330} \right) \] \[ = \frac{330-300} {300\times330} \] \[ = \frac{30}{99000} \] \[ = 3.03\times10^{-4} \]

Step 4: Substitute into Arrhenius equation. \[ 0.602 = \frac{E_a} {2.303\times8.314} \times 3.03\times10^{-4} \] \[ 0.602 = \frac{E_a}{19.147} \times 3.03\times10^{-4} \] \[ E_a = \frac{0.602\times19.147} {3.03\times10^{-4}} \] \[ = 3.80\times10^4 J\,mol^{-1} \] \[ = 38.0 kJ\,mol^{-1} \]

Step 5: Interpretation of activation energy. Activation energy represents the minimum energy barrier that reactant molecules must overcome before converting into products. A larger activation energy means:
• Slower reaction at a given temperature
• Greater temperature dependence
• Fewer molecules possessing sufficient energy

Step 6: Final conclusion. \[ \boxed{E_a = 38.0\ kJ\,mol^{-1}} \] Hence: \[ \boxed{\text{Option (B)}} \]