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# The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10 ^ 10 s ^ –1. Calculate k at 318K and Ea.

4.29   The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is $4 \times 10 ^{10} s ^{-1}$ . Calculate k at 318K and Ea.

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We know that,

for a first order reaction-

$t = \frac{2.303}{k}\log\frac{a}{a-x}$

Case 1
At temp. = 298 K
$t = \frac{2.303}{k}\log\frac{100}{90}$
= 0.1054/k

Case 2
At temp = 308 K

$t' = \frac{2.303}{k}\log\frac{100}{75}$
= 2.2877/k'
As per the question
$t' = t$
K'/K = 2.7296

From Arrhenius equation,

= 76640.096 J /mol
=76.64 KJ/mol

k at 318 K
we have , T =318K
A= $4 \times 10^{10}$

Now $\log k = \log A- \frac{E_{a}}{2.303RT}$
After putting the calue of given variable, we get

$\log k = -1.9855$
on takingantilog we get,

k = antilog(-1.9855)

= 1.034 $\times 10^{-2}\ s^{-1}$

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