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### Le Chatelier's Principle: Change in temperature

Posted: Sat Jan 12, 2019 11:03 am
Hello,

I'm a bit confused about how changing temperature changes K. I understand that if the reaction requires heat to form products, then heating will favor products and if the reaction releases heat to form products, then heating will favor reactions, but how does this affect K? Wouldn't the reaction adjust to reach equilibrium, so K would remain constant?

Thanks!

### Re: Le Chatelier's Principle: Change in temperature

Posted: Sat Jan 12, 2019 11:29 am
To answer this question, there needs to be an understanding of the thermodynamics behind it.

G = H – T*S, where G is the change in Gibb's Free Energy, H is the change in enthalpy, T is the temperature, and S is the entropy.

Before dissecting this equation and using it to explain why an increase in temperature will make the equilibrium shift towards the endothermic reaction, one important fact needs to be made clear: reactions always want to be at the lowest energy state, more specifically the lowest Gibb's Free Energy state (G in the equation).

In the equation, we see that:
As we increase temperature, entropy (S) becomes more important than enthalpy (H) in order to obtain a low G (notice the minus sign in front of the expression of T*S).

As we lower the temperature, enthalpy (H) becomes more important and so in order to obtain a low G, the reaction will shift towards the most stable (bond making) direction, which is the exothermic direction (as exothermic reactions have a negative enthalpy value). There will be a favoring of the products of the exothermic reaction. Hence, the K value will change accordingly.

I don't think you need to worry about this reasoning as in the syllabus, there is a section on thermodynamics, and Dr. Lavelle should explain this equation in detail then.

### Re: Le Chatelier's Principle: Change in temperature

Posted: Sat Jan 12, 2019 11:37 am
The equilibrium constant of a reaction is essentially a ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction (we have not discussed this in class). These rate constants are temperature dependent. This is why the K is dependent on temperature.