CBSE Class 12 · 2008 · Foreign · Chemical KineticsShort Answer

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The rate of a reaction increases to four times when the prevailing temperature is raised from 300 K to 320 K. Calculate the energy of activation of this reaction, assuming that it does not change with temperature. (R = 8.314 J mol

^{- 1}

^{- 1}

)

Answer: E_a ≈ 55.3 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}})

lo g 4 = 0.6021 = \frac{E _{a}}{19.147} \times \frac{20}{300 \times 320}

E_{a} = \frac{0.6021 \times 19.147 \times 96000}{20}

E_{a} = 5.53 \times 1 0^{4} J mo l^{- 1} = 55.3 kJ mo l^{- 1}

CBSE Class 12 Board · 2008 · Foreign · Subjective

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