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# 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.

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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.

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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

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