CBSE Class 12 · 2024 · Chemical KineticsShort Answer

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The rate constant of a reaction quadruples when the temperature changes from 300 K to 320 K. Calculate the activation energy for this reaction. (log 2 = 0.30, log 4 = 0.60, 2.303 R = 19.15 J K

^{- 1}

mol

^{- 1}

)

Answer: E_a ≈ 55.1 kJ mol⁻¹

Solution

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

0.60 = \frac{E _{a}}{19.15} \times \frac{20}{300 \times 320}

E_{a} = \frac{0.60 \times 19.15 \times 96000}{20}

E_{a} = 5.51 \times 1 0^{4} J mo l^{- 1} \approx 55.1 kJ mo l^{- 1}

CBSE Class 12 Board · 2024 · Subjective

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