The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10^{10}s^{–1}. Calculate k at 318K and E_{a}.

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 k_{2}/k_{1} = (E_{a} / 2.303 R) × (T_{2} - T_{1}) / T_{1}T_{2}

Substituting the values,

⇒

E_{a} = 76640.09 J mol^{-1} or 76.64 kJ mol^{-1}

We know, log k = log A –E_{a}/RT

Log k = log(4 × 10^{10})-(76.64kJ mol^{-1}/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^{-2} s ^{-1}

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