Q25 / 90 · JEE Main 2022 (24 Jun Shift 1) · Chemistry

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700-1000 K. The data has been analysed by plotting lnk vs 103 T graph. The value of activation energy for the reaction is kJmol-1. (Nearest integer) (Given : R=8.31 JK-1 mol-1)

Figure

Correct answer: 154

Solution

k=Ae-EaRT

lnk=lnA-EaRT

lnk=lnA+-Ea1000R1000T

Slope =-Ea1000R=-18.5

Ea=18.5×1000×8.31

=153.735×103 J

=154 KJ

Hence, the correct answer is 154.

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