JEE Main 2026 Apr 6 Shift 2, Chemistry Q25: Factors Affecting The Rate of a Chemical Reaction

JEE Main2026Apr 6, Shift 2Chemistry

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T}$ . The activation energy of reaction is _____ kJ mol $^{- 1}$ . (Nearest integer)

Given: R = 8.3 J K $^{- 1}$ mol $^{- 1}$

Your answer

Solution

Compare with the Arrhenius form $k = A e^{- E_{a} / R T}$ . The exponents match when $\frac{E _{a}}{R} = 28000$ K.

$E_{a} = 28000 \times 8.3 = 232400$ J mol $^{- 1}$ $= 232.4$ kJ mol $^{- 1}$ .

Nearest integer: 232.

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q25: Factors Affecting The Rate of a Chemical Reaction

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