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 rea

Question

A $⟶$ B (first reaction)
C $⟶$ 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 __________ $\times 10^{- 1} {hour}^{- 1}$ [nearest integer] [2026]

Smart Achievers · Accepted Answer

(5)

$For A \overset{k_{1}}{\to} B$

$\ln (2) = \frac{E_{a 1}}{R} [\frac{1}{300} - \frac{1}{500}]$

$E_{a 1} = \frac{\ln 2 \times R \times 1500}{2}$

$E_{a 2} = \frac{E_{a 1}}{2} = \frac{\ln 2 \times R \times 1500}{4}$

${(k_{1})}_{at 500 K} = \frac{\ln 2}{2}$

${(k_{2})}_{at 500 K} = \ln 2$

$Now for C \overset{k_{2}}{\to} D$

$\ln [\frac{{(k_{2})}_{at 500 K}}{{(k_{2})}_{at 300 K}}] = (\frac{\ln 2 \times R \times 1500}{4}) \times \frac{1}{R} \times [\frac{1}{300} - \frac{1}{500}]$

${(k_{2})}_{at 300 K} = \frac{\ln 2}{\sqrt{2}} = 0.49$

${(k_{2})}_{at 300 K} = 4.9 \times 10^{- 1}$

$Ans is 5 .$

Solution

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Solution

Ans. (5) For A→k1B ln(2)=Ea1R[1300-1500] Ea1=ln2×R×15002 Ea2=Ea12=ln2×R×15004 (k1)at 500 K=ln22 (k2)at 500 K=ln2 Now for C→k2D ln[(k2)at 500 K(k2)at 300 K]=(ln2×R×15004)×1R×[1300-1500] (k2)at 300 K=ln22=0.49 (k2)at 300 K=4.9×10-1 Ans is 5.

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