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ChemistryInteger+4 / -1Question 85 of 90

JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 85

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting ln⁡k\ln k vs 103T\dfrac{10^3}T graph. The value of activation energy for the reaction is ______ kJ mol−1\mathrm{kJ\,mol^{-1}}. (Nearest integer)

(Given : R=8.31 J K−1mol−1R=8.31\ \mathrm{J\,K^{-1}mol^{-1}})

Plot of ln k (y-axis) against 10³/T (x-axis): a straight line falling from left to right, labelled Slope = −18.5.
Answer and worked solution

Official answer: 154

Arrhenius. ln⁡k=ln⁡A−EaRT=ln⁡A−Ea1000R⋅103T\ln k=\ln A-\dfrac{E_a}{RT}=\ln A-\dfrac{E_a}{1000R}\cdot\dfrac{10^3}T.

Slope. −Ea1000R=−18.5-\dfrac{E_a}{1000R}=-18.5, so Ea=18.5×1000×8.31≈153.7×103E_a=18.5\times1000\times8.31\approx153.7\times10^3 J mol−1^{-1}.

Nearest integer 154 kJ mol−1^{-1}.

Answer: 154

Topics:Arrhenius equation
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