Q. 294.0( 21 Votes )

The time required

Answer :

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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