Question

For reaction \(A \rightarrow P\), rate constant \(k = 1.5 \times 10^3\ s^{-1}\) at \(27^\circ C\). If activation energy for the above reaction is \(60\ kJ\ mol^{-1}\), then the temperature (in \(^{\circ}C\)) at which rate constant \(k = 4.5 \times 10^3\ s^{-1}\) is ______. (Nearest integer) \[ \text{Given: } \log 2 = 0.30,\ \log 3 = 0.48,\ R = 8.3\ J\ K^{-1}\ mol^{-1},\ \ln 10 = 2.3 \]

Answer 1

Correct Answer: 41

Concept: Arrhenius equation: \[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]
Step 1:Substitute values} \[ k_1 = 1.5 \times 10^3 \] \[ k_2 = 4.5 \times 10^3 \] \[ \frac{k_2}{k_1} = 3 \] \[ T_1 = 27^\circ C = 300K \] \[ E_a = 60 \times 10^3\ J \]
Step 2:Take logarithm} \[ \ln 3 = 1.1 \] Thus \[ 1.1 = \frac{60000}{8.3} \left(\frac{1}{300}-\frac{1}{T_2}\right) \] \[ \frac{60000}{8.3} \approx 7228 \]
Step 3:Solve} \[ 1.1 = 7228\left(\frac{1}{300}-\frac{1}{T_2}\right) \] \[ \frac{1}{300}-\frac{1}{T_2} = 1.52 \times 10^{-4} \] \[ \frac{1}{T_2} = 0.00333 - 0.000152 \] \[ \frac{1}{T_2} = 0.00318 \] \[ T_2 \approx 314K \]
Step 4:Convert to Celsius} \[ T_2 = 314 - 273 \] \[ = 41^\circ C \] \[ \boxed{41} \]