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 pre-exponential factor for the reaction is $3.56 \times 10^{9} {s e c}^{- 1}$ , calculate its rate constant at $318$ $K$ and also the energy of activation.

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Solution

We know that,
$k = \frac{2.303}{t} {log}_{10} (\frac{a}{a - x})$
At $298$ $K$ , $x = 10$ , $a = 100$ ,
$k_{298} = \frac{2.303}{t_{1}} {log}_{10} \frac{100}{90}$ ....(i)
At $308$ $K$ , $a = 100$ , $x = 25$ , $(a - x) = 75$
$k_{308} = \frac{2.303}{t_{2}} {log}_{10} (\frac{100}{75})$ ...(ii)
$t_{1} = t_{2}$ , dividing equation (ii) by equation (i)
$∴ \frac{k_{308}}{k_{298}} = 2.73$
$log \frac{k_{308}}{k_{298}} = \frac{E}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$log 2.73 = \frac{E}{2.303 \times 8.314} (\frac{1}{308} - \frac{1}{298})$
$E = 76.622 k J / m o l$
Similarly, we can solve for $k_{318}$ which is equal to $9.22 \times 10^{- 4} s^{- 1}$ .

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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 pre-exponential factor for the reaction is 3.56×109 sec−1, calculate its rate constant at 318 K and also the energy of activation.