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- Apr 17, 2012
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So the equilibrium constant is only affected by temperature.
I understand that if I have an exothermic reaction, and I remove heat, the reaction goes forward. This is because the removal of thermal energy from the system INCREASES the kf/kr ratio to give a greater K.
However, what if I had an exothermic reaction and I increased the temperature? I know that the reaction would proceed in reverse, but is this due to a decrease in the equilibrium constant, or due to the system compensating for the heat in some other way?
How can temperature increase favor the reverse reaction by increasing one rate over the other? Isn't it according to Arrhenius equation that temperature will always increase rates of reaction, so the ratio should be unchanged?
I understand that if I have an exothermic reaction, and I remove heat, the reaction goes forward. This is because the removal of thermal energy from the system INCREASES the kf/kr ratio to give a greater K.
However, what if I had an exothermic reaction and I increased the temperature? I know that the reaction would proceed in reverse, but is this due to a decrease in the equilibrium constant, or due to the system compensating for the heat in some other way?
How can temperature increase favor the reverse reaction by increasing one rate over the other? Isn't it according to Arrhenius equation that temperature will always increase rates of reaction, so the ratio should be unchanged?