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 value of A is 4 x 10^10 s^1. Calculate k at 318 K and Ea.
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 value of A is 4 x 10^{10} s^{1}. Calculate k at 318 K and E_{a}.
For a first order reaction,
t = \(\frac{2.303}{k}\) log \(\frac{a}{ax}\)
At 298 K, t = \(\frac{2.303}{k}\) log \(\frac{100}{90}\)
= \(\frac{0.1054}{k}\)
At 308 K, t' = \(\frac{2.303}{k'}\)log \(\frac{100}{75}\)
= \(\frac{2.2877}{k'}\)
According to the question,
t = t'
⇒ \(\frac{0.1054}{k}\) = \(\frac{0.2877}{k'}\)
⇒ \(\frac{k'}{k}\) = 2.7296
From Arrhenius equation, we obtain
log \(\frac{k'}{k}\) = \(\frac{E_a}{2.303 R}\)\(\Big(\frac{T'T}{TT'}\Big)\)
log (2.7296) = \(\frac{E_a}{2.303 \times 8.314}\) \(\Big(\frac{308  298}{298 \times 308}\Big)\)
E_{a }= \(\frac{2.303 \times 8.314 \times 298 \times 308 \times log(2.7296)}{308  298}\)
= 76640.096 J mol^{1}
= 76.64 kJ mol^{1}
To calculate k at 318 K,
It is given that, A = 4 x 10^{10} s^{1}, T = 318K
Again, from Arrhenius equation, we obtain
log k = log A  \(\frac{E_a}{2.303 RT}\)
= log (4 x 10^{10})  \(\frac{76.64 \times 10^3}{2.303 \times 8.314 \times 318}\)
= (0.6021 + 10)12.5876
= 1.9855
Therefore, k = Antilog(1.9855)
= 1.034 x 10^{2} s^{1}

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