Question

For a first-order reaction, the rate constant increases by a factor of \(16\) when the temperature is increased from \(300\,K\) to \(340\,K\). The activation energy of the reaction is closest to:

• A. \(28\,kJ\,mol^{-1}\)
• B. \(57\,kJ\,mol^{-1}\)
• C. \(85\,kJ\,mol^{-1}\)
• D. \(114\,kJ\,mol^{-1}\)

Hint:

Whenever rate constants at two temperatures are given, use the logarithmic form of the Arrhenius equation directly instead of calculating the pre-exponential factor.

Answer 1

The Correct Option is B

Concept: The Arrhenius equation relates the rate constants at two temperatures as \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] This equation allows direct calculation of activation energy from temperature-dependent rate constant data.

Step 1: Substitute the given values. \[ \frac{k_2}{k_1}=16 \] \[ T_1=300\,K \] \[ T_2=340\,K \] Therefore, \[ \ln(16) = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{340} \right) \]

Step 2: Calculate the temperature term. \[ \frac{1}{300}-\frac{1}{340} = \frac{40}{102000} \] \[ = 3.92\times10^{-4} \] Also, \[ \ln(16)=2.773 \]

Step 3: Calculate activation energy. \[ E_a = \frac{2.773\times8.314} {3.92\times10^{-4}} \] \[ = 5.88\times10^4\,J\,mol^{-1} \] \[ = 58.8\,kJ\,mol^{-1} \] Nearest option: \[ \boxed{57\,kJ\,mol^{-1}} \]