The temperature at which the rate constants of the given below two gaseous reactions become equal is \_\_\_\_ K. (Nearest integer). $\mathrm{X} \longrightarrow \mathrm{Y} \quad \mathrm{k}_{1}=10^{6} \mathrm{e}^{\frac{-30000}{\mathrm{~T}}}$ $\mathrm{P} \longrightarrow \mathrm{Q} \quad \mathrm{k}_{2}=10^{4} \mathrm{e}^{\frac{-24000}{\mathrm{~T}}}$ Given : $\ln 10=2.303$

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Chemical Kinetics

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

X ⟶ Y k_{1} = 1 0^{6} e^{\frac{- 30000}{T}}

P ⟶ Q k_{2} = 1 0^{4} e^{\frac{- 24000}{T}}

Given :

ln 10 = 2.303

Answer: 1303

Setting the two rate constants equal,

1 0^{6} e^{- 30000/ T} = 1 0^{4} e^{- 24000/ T}

. Dividing gives

1 0^{2} = e^{(30000 - 24000) / T} = e^{6000/ T}

. Taking natural logarithms,

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

, so

T = \frac{6000}{2 \times 2.303} = \frac{6000}{4.606} = 1302.6 K

. To the nearest integer the rate constants are equal at

1303 K

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer