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### 15.65, part c

Posted: Wed Mar 14, 2018 4:31 pm
15.65 For the reversible, one-step reaction 2 A ---> B + C,
the forward rate constant for the formation of B is 265 L/(mol*min) and the rate constant for the reverse reaction is 392 L/(mol*min). The activation energy for the forward reaction is 39.7 kJ mol 1 and that of the reverse reaction is 25.4 kJ/mol .
(c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

I am confused on the answer to part c. Can someone please explain how to get to the right answer? Thanks

### Re: 15.65, part c

Posted: Wed Mar 14, 2018 6:44 pm
From parts a and b, we can see that this reaction has K<1, and thus favors the reactants at equilibrium. We can also see that the forward reaction is endothermic. Raising the temperature of an endothermic reaction will favor more products, and thus will increase K. As for the rate constants, they will also both increase at higher temperature, but k will increase more than k', simply because the activation energy of the forward process is higher than that of the reverse process.

### Re: 15.65, part c

Posted: Thu Mar 15, 2018 3:56 pm
can you answer this question only in terms of activation energies?

### Re: 15.65, part c

Posted: Thu Mar 15, 2018 10:23 pm
In the problem they give you Ea for the forward and reverse processes, 39.7 kj/mol*min and 25.4 kj/mol*min. You can see that the reaction is endothermic because endothermic reactions have larger activation energies in the forward process than in the reverse process, because it requires more energy to break bonds. Endothermic reactions shift towards products when temperature is raised because their Ea is more sensitive to increased temp in the forward direction than the reverse reaction, therefore increasing k more than k'. Since K=k/k', this also increases the equilibrium constant, K. Hope this makes sense in terms of activation energy!