The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K, if the value of A is 4 x 10¹⁰ s^–1. Calculate k at 318 k and E_a.

Solution

For the first order reaction

(i) t_{10 %} = \frac{2.303}{k} \log \frac{100}{90} = \frac{2.303}{k} \log \frac{100}{9}

(ii) t_{25 %} = \frac{2.303}{k} \log \frac{100}{75} = \frac{2.303}{k_{1}} \log \frac{4}{3}

According to problem,

t_{10 %} = t_{25 %}

Therefore,

\frac{2.303}{k_{1}} \log \frac{10}{9} = \frac{2.303}{k_{2}} \log \frac{4}{3} \frac{1}{k_{1}} \times [\log 10 \times \log 9] = \frac{1}{k_{2}} \times [0.6021 - 0.4771] \frac{1}{k_{1}} \times [1 - 0.9542 = \frac{1}{k_{2}} \times [0.602 - 0.4771]] \frac{k_{1}}{k_{2}} = \frac{0.6021 - 0.4771}{1 - 0.9542} = \frac{0.125}{0.0458} = 2.7

Form the Arrhenius equation, we obtain

\log \frac{k_{1}}{k_{2}} = \frac{Ea}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] \log 2.7 = \frac{Ea}{2.304 \times 8.314 R} [\frac{1}{298} - \frac{1}{308}] 0.4314 = \frac{Ea}{2.304 \times 8.314 R} \times \frac{(308 - 298)}{298 \times 308} or 0.4314 = \frac{Ea}{2.304 \times 8.314 R} \times \frac{10}{298 \times 308} Ea = \frac{0.4314 \times 2.304 \times 8.314 \times 298 \times 308}{10} = 75.847 kJ {mol}^{- 1} .

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