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 1010 s-1. Calculate k at 318 K and Ea.
For a first order reaction,
t = \(\frac{2.303}{k}\) log \(\frac{a}{a-x}\)
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) = \(\frac{E_a}{2.303 \times 8.314}\) \(\Big(\frac{308 - 298}{298 \times 308}\Big)\)
Ea = \(\frac{2.303 \times 8.314 \times 298 \times 308 \times log(2.7296)}{308 - 298}\)
= 76640.096 J mol-1
= 76.64 kJ mol-1
To calculate k at 318 K,
It is given that, A = 4 x 1010 s-1, T = 318K
Again, from Arrhenius equation, we obtain
log k = log A - \(\frac{E_a}{2.303 RT}\)
= log (4 x 1010) - \(\frac{76.64 \times 10^3}{2.303 \times 8.314 \times 318}\)
= (0.6021 + 10)-12.5876
= -1.9855
Therefore, k = Antilog(-1.9855)
= 1.034 x 10-2 s-1