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Q.

The number of reactant molecules crossing over the energy barrier at 325K is e-x. If heat of reaction is 0.63 Kcal and activation energy of backward reaction is 0.02 Kcal. Find the value of “x”.
           (R=2 cal/mol/deg).
 

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Detailed Solution

ΔH=EfEb0.12=Ef0.02Ef=0.14Kcal=140cal

x no. of particles =eEaR.T

x=e1502×325 x=e-1=1

 

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