## Homework 15.65 part C

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

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104469125
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### Homework 15.65 part C

For the reversible, one step reaction 2A <-- --> B+C, the forward rate constant for the formation of B is 265 L/mol x min and the rate constant for the reverse action is 392 L/mol x min. The activation energy for the forward reaction is 39.7 kJ/mol and that of the reverse reaction is 25.4 kJ/mol.
What will be the effect of raising the temperature on the rate constant and the equilibrium.
The solutions manual says this will increase the rate constant with the higher activation barrier than it will the rate constant of the run with lower energy barrier. Then it says that the rate of the forward reaction will go up more than that for the reverse reaction.
Can anyone explain why that it?

MelanieAu1G
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### Re: Homework 15.65 part C

This is because the rate constant, k, increases exponentially as temperature, T, increases. Therefore, the rate constant for the forward reaction that correlates with the larger activation potential will increase more so than the smaller rate constant for the reverse reaction. This would also mean that the equilibrium will increase, because K (equilibrium) is equal to the forward rate constant divided by the reverse rate constant.

Emily_Lenh2A
Posts: 17
Joined: Sat Jul 09, 2016 3:00 am

### Re: Homework 15.65 part C

Because this is an endothermic reaction, as can be seen through the activation energies of the forward and reverse reactions, the forward reaction will take in energy/heat and the reverse reaction will release energy/heat. When you raise the temperature of the system, it tries to go back to equilibrium by going to the right (favoring the forward reaction), which takes in the excess heat.

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