JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

$A \to B$ (first reaction)

$C \to D$ (second reaction)

Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ____ $\times 1 0^{- 1}$ hour $^{- 1}$ (nearest integer).

Your answer

Solution

Step 1 — Arrhenius for reaction 1: $ln 2 = \frac{E _{a_{1}}}{R} (\frac{1}{300} - \frac{1}{500}) \Rightarrow E_{a_{1}} = \frac{ln 2 \cdot R \cdot 1500}{2}$ .

Step 2 — rate constants at 500 K: First-order half-life gives $(k_{1})_{500} = \frac{ln 2}{t _{1/2}} = \frac{ln 2}{2}$ hr $^{- 1}$ . Given $(k_{2})_{500} = 2 (k_{1})_{500} = ln 2$ hr $^{- 1}$ .

Step 3 — $E_{a_{2}}$ and Arrhenius for reaction 2: $E_{a_{2}} = E_{a_{1}} /2 = \frac{ln 2 \cdot R \cdot 1500}{4}$ .

$ln [\frac{( k _{2} ) _{500}}{( k _{2} ) _{300}}] = \frac{E _{a_{2}}}{R} (\frac{1}{300} - \frac{1}{500}) = \frac{ln 2 \cdot 1500}{4} \cdot \frac{200}{150000} = \frac{ln 2}{2}$ .

So $\frac{( k _{2} ) _{500}}{( k _{2} ) _{300}} = e^{l n 2/2} = 2$ .

$(k_{2})_{300} = \frac{ln 2}{2} \approx \frac{0.693}{1.414} \approx 0.49$ hr $^{- 1} = 4.9 \times 1 0^{- 1}$ hr $^{- 1}$ .

Nearest integer for the $\times 1 0^{- 1}$ multiplier: $5$ .

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