A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is again 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. [R = 8.314 J K^–1 mol^–1, log 4 = 0.602].

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Solution

BY using the half life equation, we get

t_{1 / 2} = 20 \min k_{1} = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{20} \min^{- 1} k_{2} = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5} \min^{- 1}

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

\log \frac{\frac{9.693}{5}}{\frac{0.693}{20}} = \frac{Ea}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{350})

Ea = \frac{19.147 \times 350 \times 300 \times \log 4}{50} = \frac{19.147 \times 7 \times 300 \times 0.6021}{1000} = 24.2 kJ {mol}^{- 1} .

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