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The rate constant for the first order decompoistion of a certain reaction is described by the equation `log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)

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The rate constant for the first order decompoistion of a certain reaction is described by the equation logk(s-1)=14.34- 1.25×104K T (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life periof be 256min? Select the correct answer from above options

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Correct Answer - a. Ea=239.339kJmol-1 b.T=669K a. According to Arrhenius equation, k=Ae-Ea/RT logk=logA- Ea 2.303RT ......(i) Comparing with the griven equation logk=14.34-1.25×104K/T...(ii) ∴ Ea 2.303RT = 1.25×104K T orEa=2.303R×1.25×104K =2.303×(8.314JK-1mol-1)×1.25×104K =239.34kJmol-1 b. When t1/2=256min,k= 0.693 t1/2min = 0.693 256×60s =4.51×10-5s-1 Substitute the value of k in Eq.(ii), log(4.51×10-5s-1)=14.34- 1.25×104K T log(-5+0.6542)=14.34- 1.25×104K T or 1.25×104K T =18.6858 orT=669K

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