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

Concept: For a first-order reaction: \[ t_{1/2}=\frac{0.693}{k} \] and Arrhenius equation is \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac1{T_1}-\frac1{T_2} \right) \]

Step 1: Find \(k_2\) from half-life. \[ 138.6 = \frac{0.693}{k_2} \] \[ k_2 = 5\times10^{-3}s^{-1} \]

Step 2: Apply Arrhenius equation. \[ k_1=2.5\times10^{-4} \] \[ T_1=573K \] \[ \frac{k_2}{k_1} = \frac{5\times10^{-3}} {2.5\times10^{-4}} =20 \] \[ \log20=1.30 \] \[ 1.30 = \frac{42000} {2.303\times2} \left( \frac1{573}-\frac1{T_2} \right) \] \[ 1.30 = 9115 \left( \frac1{573}-\frac1{T_2} \right) \] \[ \left( \frac1{573}-\frac1{T_2} \right) = 1.426\times10^{-4} \] \[ \frac1{T_2} = 0.001745-0.000143 \] \[ = 0.001602 \] \[ T_2 \approx625K \]

Step 3: Final conclusion. \[ \boxed{625K} \] Hence, \[ \boxed{\text{Option (D)}} \]