CBSE Class 12 · 2007 · Foreign · Chemical KineticsShort Answer

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The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate the activation energy of such a reaction. (R = 8.314 J mol

^{- 1}

^{- 1}

, log 2 = 0.3010)

Answer: E_a ≈ 52.9 kJ mol⁻¹

Solution

lo g 2 = \frac{E _{a}}{2.303 R} (\frac{308 - 298}{298 \times 308})

0.3010 = \frac{E _{a}}{19.147} \times \frac{10}{91784}

E_{a} = \frac{0.3010 \times 19.147 \times 91784}{10}

E_{a} = 5.29 \times 1 0^{4} J mo l^{- 1} = 52.9 kJ mo l^{- 1}

CBSE Class 12 Board · 2007 · Foreign · Subjective

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