Question

At \(300^\circ C\), decomposition of azomethane follows first order kinetics. Rate constant for this reaction at this temperature is \(2.5\times10^{-4}\,s^{-1}\). If the activation energy of the reaction is \(42\,kcal\,mol^{-1}\), what is the temperature (in K) at which the half-life of the reaction is \(138.6\) seconds? \[ (R=2\,cal\,K^{-1}mol^{-1},\; \log20=1.30) \]

• A. 725
• B. 425
• C. 525
• D. 625

Hint:

For first-order reactions: \[ t_{1/2}=\frac{0.693}{k} \] Convert half-life into rate constant first, then use the Arrhenius equation.

Answer 1

The Correct Option is D

Step 1: Connect half-life to rate constant.
For a first order reaction the half-life is \[ t_{1/2} = \frac{0.693}{k}. \] We use this to find the new rate constant at the unknown temperature.
Step 2: Find the second rate constant.
Given $t_{1/2} = 138.6$ s, \[ k_2 = \frac{0.693}{138.6} = 5\times10^{-3} \ \text{s}^{-1}. \]
Step 3: Recall the Arrhenius two-temperature form.
\[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right). \] Here $k_1 = 2.5\times10^{-4}$ at $T_1 = 300^\circ C = 573$ K.
Step 4: Find the ratio of rate constants.
\[ \frac{k_2}{k_1} = \frac{5\times10^{-3}}{2.5\times10^{-4}} = 20, \qquad \log 20 = 1.30. \]
Step 5: Solve for $1/T_2$.
With $E_a = 42000$ cal and $R = 2$, \[ 1.30 = \frac{42000}{2.303\times2}\left(\frac{1}{573} - \frac{1}{T_2}\right) = 9115\left(\frac{1}{573} - \frac{1}{T_2}\right). \] So $\dfrac{1}{573} - \dfrac{1}{T_2} = 1.426\times10^{-4}$, giving $\dfrac{1}{T_2} = 0.001745 - 0.000143 = 0.001602$.
Step 6: Find the temperature.
\[ T_2 = \frac{1}{0.001602} \approx 625 \ \text{K}. \] So \[ \boxed{T_2 \approx 625 \ \text{K}} \]