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Rate of Reaction

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Given that $K (s e c^{- 1} = 5 \times 10^{14} e^{- 124080 / R T})$ , where activation energy is expressed in joule. The temperature at which reaction has $t_{1 / 2}$ equal to 25 minute. Assume Ist order reaction in $^{o} C$ is

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Solution

$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{25 \times 60} = 4.62 \times 10^{- 4} s e c^{- 1}$ ......(1)
Also $K = 5 \times 10^{14} e^{- 124080 / R T}$ ......(2)
From (1) and (2)
$4.62 \times 10^{- 4} = 5 \times 10^{14} e^{- 124080 / R T}$
$e^{- 124080 / R T} = 9.24 \times 10^{- 19}$
$- 124080 / 8.314 T = l n 9.24 \times 10^{- 19} = - 41.53$
$T = 359 K = 86^{o} C$

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