JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Chemical Kinetics

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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 vs 10³/T graph. The value of activation energy for the reaction is ______ kJ mol⁻¹. (Nearest integer) (Given: R = 8.31 J K⁻¹ mol⁻¹)

Answer: 154

Solution

The Arrhenius equation in logarithmic form is: ln k = ln A - Ea/(RT). When plotted as ln k vs 10³/T, the slope = -Ea/(R × 10³). From the graph, slope = -18.5. Therefore: -Ea/(8.31 × 10³) = -18.5. Ea = 18.5 × 8.31 × 10³ = 153735 J/mol = 153.7 kJ/mol ≈ 154 kJ/mol.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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