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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.

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.

Answers (1)

best_answer

We know that

for a first-order reaction-

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

Case 1
At temp. = 298 K
\begin{aligned} t & =\frac{2.303}{k} \log \frac{100}{90} \\ & =0.1054 / \mathrm{k}\end{aligned}

Case 2
At temp = 308 K

\begin{aligned} t^{\prime} & =\frac{2.303}{k} \log \frac{100}{75} \\ & =2.2877 / k^{\prime}\end{aligned}
As per the question 
t^{\prime}=t
K'/K = 2.7296

From Arrhenius's equation,

\log \frac{k^{\prime}}{k}=\frac{E_a}{2.303 \mathrm{R}}\left(\frac{T^{\prime}-T}{T T^{\prime}}\right)

\log (2.7296)=\frac{E_a}{2.303 \times 8.314}\left(\frac{308-298}{298 \times 308}\right)

E_a=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{308-298}


     = 76640.096 J /mol
     =76.64 KJ/mol

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

Now  \log k=\log A-\frac{E_a}{2.303 R T}
After putting the value of a given variable, we get 

\log k=-1.9855
on taking antilog we get, 

k = antilog(-1.9855)

  =1.034 \times 10^{-2} \mathrm{~s}^{-1}

Posted by

manish

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