1.

What is the effect of positive catalyst on the energy of activation?

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SOLUTION :(a) Consider a first order reaction RTOP Rate = `-d[R]/dt=k[R](d[R])/[r]`=-kdt
Integrating the above equation In[R] =-kt +1
When t = 0, [R] = [R]
Equation (1) becomes In[R] = 1 Therefore equation (1) becomes
in[r]=-kt+in[R]
`k=1/(t) In[R]/[R] `
`k= 2.303/(t) “LOG” [R]_(0)/[R]`
(b) Energy of activation decreases


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