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Question

At 380C, the half-life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ/mol1. Calculate the time required for 75% decomposition at 450.
Use : [16531723]=1.48×104,e14.8=2.67×106.

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Solution

Given that ,
Decomposition of H2O2 is first order reaction.
at, T=380=380+273=653K
t1/2=360 min
Ea=200kJ/mol

We know that Arrhenius equation for calculation of activation energy is given by:-
K=A eEa/RT (4)
where,
K= rate constant of reaction at temperature (TK)
A= pre exponential factor (Arrhenius constant)
Ea= universal gas constant
T= temperature in K

Let us suppose at temperture T1K and T2K the rate of constants of the reaction are K1 and K2 respectively,so,
K1=A eEa/RT1 (ii)
K2=A eEa/RT2 (iii)

Now, (iii) and (ii)
K2K1=eEa/RT2eEa/RT1

Taking log on both sides:-
log(K2K1)=logeEaR(1T21T1)

Now, as given, t1/2=360 min at T=653K
we have for first order reaction,
K1=0.693t1/2=0.693360 min1
T2=450C=450+273=723K

Putting the values of K1,Ea,R,T2,T1:-

log(K2K1)=EaK(1T21T1)
log(K20.693×360)=200×10008.314×[17231653]
log(K2×3600.693)=+200×10008.314×1.48×104
log(K2×3600.693)=+200×1488314

Taking antilog on both sides

K2×3600.693=Antilog(3.56)
K2=0.693360×35.16
=0.068 min1

Now, at T=723K, we have K2=0.068 min1
Now, for first order reaction we have:-
K2t=log([A]0[A]t)

where, [A]0= conc. of reactant at t=0
[A]t= conc. of reactant at t=t

Given, [A]t[A]0=0.25 (Since 75 % of reactant is decomposed)
t=1K2ln(10.25)
=10.068ln(4)
t=20 min.

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