Q22 / 75 · JEE Main 2026 (22 January Shift 1) · Chemistry
The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer)
$\mathrm{X} \longrightarrow \mathrm{Y} \quad \mathrm{k}_{1}=10^{6} e^{\frac{-30000}{\mathrm{~T}}}$
$\mathrm{P} \longrightarrow \mathrm{Q} \quad \mathrm{k}_{2}=10^{4} e^{\frac{-24000}{\mathrm{~T}}}$
Given: $\ln 10=2.303$
$\mathrm{X} \longrightarrow \mathrm{Y} \quad \mathrm{k}_{1}=10^{6} e^{\frac{-30000}{\mathrm{~T}}}$
$\mathrm{P} \longrightarrow \mathrm{Q} \quad \mathrm{k}_{2}=10^{4} e^{\frac{-24000}{\mathrm{~T}}}$
Given: $\ln 10=2.303$
Correct answer: 1303
Solution
To find the temperature at which $k_1 = k_2$,
set the rate constant equations equal: $10^6 e^{-30000/T} = 10^4 e^{-24000/T}$.
Dividing both sides by $10^4$: $100 = e^{-24000/T + 30000/T} = e^{6000/T}$.
Taking natural logarithm: $\ln(100) = \frac{6000}{T}$.
Since $\ln(100) = 2\ln(10) = 2 \times 2.303 = 4.606$: $4.606 = \frac{6000}{T}$.
Solving for T: $T = \frac{6000}{4.606} = 1302.5$ K.
Rounding to the nearest integer gives T = 1303 K.
set the rate constant equations equal: $10^6 e^{-30000/T} = 10^4 e^{-24000/T}$.
Dividing both sides by $10^4$: $100 = e^{-24000/T + 30000/T} = e^{6000/T}$.
Taking natural logarithm: $\ln(100) = \frac{6000}{T}$.
Since $\ln(100) = 2\ln(10) = 2 \times 2.303 = 4.606$: $4.606 = \frac{6000}{T}$.
Solving for T: $T = \frac{6000}{4.606} = 1302.5$ K.
Rounding to the nearest integer gives T = 1303 K.
Hence, the correct answer is 1303.