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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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M manish

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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