Question

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 $ . If the value of $A$ is $4 \times 10^{10} s ^{-1} .$ The value of $E_{a}$ is $(\log 2.728=0.436)$

• A. $76.623\, kJ / mol$
• B. $100\, kJ / mol$
• C. $94\, kJ / mol$
• D. $52.261\, kJ / mol$

Answer 1

Answer: (A)

Apply, $k t=2.303 \log \frac{(A)_{0}}{(A)}$

$k _{1} \times t =2.303 \log \frac{( A )_{0}}{90( A )_{0}} \times 100$...(1)

$k _{2} \times t =2.303 \log \frac{( A )_{0}}{75( A )_{0}} \times 100$...(2)

$\frac{k_{2}}{k_{1}}=2.728$

Now,

$\log \frac{ k _{2}}{ k _{1}}=\frac{ E _{ a }}{2.303 R } \frac{\left( T _{2}- T _{1}\right)}{ T _{1} T _{2}}$

$\log (2.78)=\frac{ E _{ a }}{2.303 \times 8.314 \times 10^{-3}}\left(\frac{308 -298}{308 \times 298}\right)$

$E_{a}=76.623\, kJ / mol$