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The energy of activation for the following reaction is 10 kj per mole and K is $2.4 x 10^{- 5} s^{- 1}$ respectively at 300k . The temperature at which half life is equal to 2hrs will be _______K
$2 N_{2} O_{5} \to 4 N O_{2} + O_{2} i n C C l_{4} [\log 4 = 0.6021]$

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answer is 458.

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Detailed Solution

t_1/2= 2 hours = 2 x 60 x 60 sec = $\frac{0.693}{k^{'}}$

so, k'= 9.62 x 10^-5 s^-1

$\log_{e} (9.62 \times 10^{- 5}) = \log_{e} A - \frac{10000}{R T}$

On solving these 2 equations,

T = 458 K

the required temp.

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