In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 x 10²³ sec^–1 and 98.6 kJ mol^–1 respectively. If the reaction is of first order at what temperature will its half life period be ten times.

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Solution

By using the half life equation, we get

K = \frac{0.693}{t_{1 / 2}} K = \frac{0.693}{10 \times 60 s} = 1.155 \times 10^{- 3} s^{- 1}

\log K = \log A - \frac{Ea}{2.303 RT}

\log 1.155 \times 10^{- 3} = \log 4 \times 10^{- 13} J \frac{98.6 kJ {mol}^{- 1}}{2.303 \times 8.314 J K^{- 1} {mol}^{- 1} \times T} or 3.0616 = 13.6021 - \frac{98.600 kJ {mol}^{- 1}}{2.303 \times 8.314 J K^{- 1} {mol}^{- 1} \times T}

or - 2.9384 = 13.6021 - \frac{98.600 K}{2.303 \times 8.314 \times T} or 16.5405 = \frac{98.600 K}{2.303 \times 8.314 \times T} or 16.5405 = \frac{5149.6 K}{T} or T = \frac{5149.6 K}{16.5405} = 311.333 K .

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