The temperature at which the rate constants of the given below two gaseous reactions become equal is ______ K. (Nearest integer)

$$X \rightarrow Y $$ $$ k_{1}=10^{6}e^{\frac{-30000}{T}}$$

$$P \rightarrow Q $$ $$ k_{2}=10^{4}e^{\frac{-24000}{T}}$$

Given: ln 10 = 2.303

Find the temperature at which $$k_1 = k_2$$ for $$k_1 = 10^6 e^{-30000/T}$$ and $$k_2 = 10^4 e^{-24000/T}$$.

$$10^6 e^{-30000/T} = 10^4 e^{-24000/T}$$

$$\frac{10^6}{10^4} = \frac{e^{-24000/T}}{e^{-30000/T}} = e^{(-24000+30000)/T} = e^{6000/T}$$

$$10^2 = e^{6000/T}$$

$$\ln(10^2) = \frac{6000}{T}$$

$$2\ln 10 = \frac{6000}{T}$$

$$2 \times 2.303 = \frac{6000}{T}$$

$$T = \frac{6000}{4.606} = 1302.6 \approx 1303 \text{ K}$$

The correct answer is 1303 K.