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

For any reaction that follows the Arrhenius equation
$$k = A\,e^{-\dfrac{E_a}{RT}}$$
taking natural logarithms for two different temperatures $$T_1$$ and $$T_2$$ gives

$$\ln\dfrac{k_2}{k_1}= -\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$ $$-(1)$$

Given data
$$k_1 = 1.5\times10^{3}\,\text{s}^{-1},\; T_1 = 27^{\circ}\text{C}=300\,\text{K}$$
$$k_2 = 4.5\times10^{3}\,\text{s}^{-1},\; E_a = 60\,\text{kJ mol}^{-1}= 60000\,\text{J mol}^{-1}$$
$$R = 8.3\,\text{J K}^{-1}\text{ mol}^{-1}$$

First evaluate $$\ln\dfrac{k_2}{k_1}$$:
$$\dfrac{k_2}{k_1}=\dfrac{4.5}{1.5}=3$$
$$\ln 3 = 2.303\log 3 = 2.303 \times 0.48 \approx 1.105$$ $$-(2)$$

Substituting $$E_a$$, $$R$$ and equation $$(2)$$ into equation $$(1)$$:

$$1/T_2 - 1/T_1 = -\dfrac{R}{E_a}\,\ln\dfrac{k_2}{k_1}$$

$$1/T_2 = \dfrac{1}{300}\;-\;\dfrac{8.3}{60000}\times 1.105$$

Compute each term
$$\dfrac{1}{300}=3.333\times10^{-3}\,\text{K}^{-1}$$
$$\dfrac{8.3}{60000}=1.383\times10^{-4}\,\text{K}^{-1}$$
Product with $$1.105$$:
$$1.383\times10^{-4}\times1.105 \approx 1.522\times10^{-4}\,\text{K}^{-1}$$

Therefore
$$1/T_2 = 3.333\times10^{-3} - 1.522\times10^{-4}$$
$$1/T_2 = 3.181\times10^{-3}\,\text{K}^{-1}$$

Invert to find $$T_2$$:
$$T_2 = \dfrac{1}{3.181\times10^{-3}} \approx 3.14\times10^{2}\,\text{K} = 314\,\text{K}$$

Convert to Celsius:
$$T_2(^{\circ}\text{C}) = 314 - 273 = 41^{\circ}\text{C}$$

Hence, the required temperature is 41 °C (nearest integer).