JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

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

$X \to Y k_{1} = 1 0^{6} e^{- 30000/ T}$

$P \to Q k_{2} = 1 0^{4} e^{- 24000/ T}$

Given : $ln 10 = 2.303$

Your answer

Solution

Set $k_{1} = k_{2}$ : $1 0^{6} e^{- 30000/ T} = 1 0^{4} e^{- 24000/ T}$ .

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

Take logs: $\frac{6000}{T} = ln (100) = 2 ln 10 = 2 \times 2.303 = 4.606$ .

$T = \frac{6000}{4.606} \approx 1302.64$ K, rounded to 1303 K.

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