# The time required

We know, time t = (2.303/k) × log([R]0/[R])

Where, k- rate constant

[R]° -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]0/0.9[R]0)

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]0/0.75[R]0)

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k2/k1 = (Ea / 2.303 R) × (T2 - T1) / T1T2

Substituting the values,

Ea = 76640.09 J mol-1 or 76.64 kJ mol-1

We know, log k = log A –Ea/RT

Log k = log(4 × 1010)-(76.64kJ mol-1/(8.314 × 318))

Log k = -1.986

k = 1.034 x 10-2 s -1

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