CBSE Class 12 · 2018 · Chemical KineticsShort Answer

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A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 J K

^{- 1}

mol

^{- 1}

)

Answer: E_a ≈ 27.7 kJ mol⁻¹

Solution

Half-lives give the ratio of rate constants:

\frac{k _{2}}{k _{1}} = \frac{40}{20} = 2

lo g 2 = \frac{E _{a}}{2.303 R} (\frac{320 - 300}{300 \times 320})

0.3010 = \frac{E _{a}}{19.147} \times \frac{20}{96000}

E_{a} = \frac{0.3010 \times 19.147 \times 96000}{20}

E_{a} = 2.77 \times 1 0^{4} J mo l^{- 1} = 27.7 kJ mo l^{- 1}

CBSE Class 12 Board · 2018 · Subjective

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