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1.The reaction,a+b------>c+d,proceeds to right hand upto 99.9%.Find the value of equilibrium constant. Ans:10^6 2.Kp for a reaction at 25`C is 10 atm.The activation energy for forward and reverse reactions are 12 and 20kJ mol respectively.Find Kc for the reaction at 40`C. Ans:3.33*10^-1M I wanted help with the method to solve it.Could you give the method in brief.Thanks!

1.The reaction,a+b------>c+d,proceeds to right hand upto 99.9%.Find the value of equilibrium constant.


   Ans:10^6


 


2.Kp for a reaction at 25`C is 10 atm.The activation energy for forward and reverse reactions are 12 and 20kJ mol respectively.Find Kc for the reaction at 40`C.


Ans:3.33*10^-1M


 


I wanted help with the method to solve it.Could you give the method in brief.Thanks!

Grade:11

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
6 years ago
the value of equilibrium constant will be

[c][d]/[a][b]
=99.9*99.9/.1*.1
=10^6

2

delta h for the reaction is Ef-Eb=12-20=-8kj/mol
Kp for the reaction is 10atm therefore delta n=1
now Kp=Kc(RT)deltan
=Kc=10/.0821*298=.4M
delta H for the reaction is -8Kj/mol
Kc at 40 c
log(k40/k25) =deltaH/2.303R(1/T1-1/T2)
K40 =.34m

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