Can anyone explain how temperature changes can cause changes in K?

[305597516]

Hello, so in the lecture on how equilibria responds to change, it was mentioned that temperature changes can also lead to changes in the equilibrium constant. Furthermore, the professor talked about how an endothermic reaction will favor P formation and an exothermic reaction will favor R formation. Then, Lavelle drew a diagram relating enthalpy to R and P concentrations (I think?). What I don't understand is that on the diagram, R was higher than P, and moving from R to P led to an endothermic reaction. Wouldn't R be losing heat because it's going down to P? Or am I just completely misunderstanding it??? Thanks in advance.

[Sohan Talluri 1L]

In the January 7th lecture, I believe Professor Lavelle was referring to an exothermic reaction when he drew the diagram with reactants at a higher enthalpy than the products (timestamp 35:55). Also, the diagram shown doesn't reflect the concentrations of either the reactants or products, only their relative enthalpies.

Hopefully this helps and please correct me if there are any mistakes.

[Sydney Ngao 3I]

Hi, I'm not sure if this is completely correct but this is how I think of it. The reactions do not want to get too hot, so when you raise the temperature, it will shift in the direction that does not release more heat into the environment.
Hopefully, the enthalpy part will be clearer once we learn about enthalpy.

[Michelle_Duong_3H]

Hello,

Concentration changes in the reactants and products do not affect whether the reaction is endothermic or exothermic. Rather, the direction of heat transfer will affect how the reaction responds to temperature changes.

If a reaction is endothermic, then heating will favor product formation because you are adding more energy to fuel that reaction. For exothermic reactions, heating will favor reactant formation--we can think about the reverse reaction being endothermic and follow the same logic that we are adding energy to fuel that reaction.

Note that temperature changes affect the equilibrium constant K, but not the reaction quotient Q.

Hope this helps!

[Hannah_Pon_1F]

The equilibrium constant depends on the temperature of the reaction because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant.

[Vivek Chotai 2C]

I think because temperature affects the rate of reaction for both product and reactant formation. So with different rates, you would need a new ratio at the different temperature.

[Amanda Tran 1D]

When we change the temperature, the rates of the forward and reverse reactions are no longer the same. This is because adding heat to an endothermic reaction will favor products, and adding heat to an exothermic reaction will favor reactants. This means that the concentrations/proportions of products and reactions will change, causing a change in K.

[Quade Albert 2J]

Equilibrium is based off of the rates of forward and reverse reactions being equal. If the temperature changes, the resulting change in those reaction rates will change the equilibrium constant.

[Allison Peng 1D]

In an exothermic reaction K changes in to favor the reactants if temperature is increased, but in an endothermic reaction, it's the opposite. This is because the equilibrium constant K is dependent on the rate constants, Kf and Kr, which change differently with temperature change depending on if that specific direction of a reaction is endothermic or exothermic. In particular, if you look at the thermodynamics, the direction that is endothermic will have a much larger Ea, so by using the arrhenius equation, its rate constant will increase much more with an increase with temperature. Since equilibrium is when the rate of forward and reverse reaction are the same, or Keq = Kf/Kr, if the forward direction is exothermic and the reverse direction is endothermic, since Kr increases more than Kf, overall Keq will decrease and it will favor reactants more.

An easier way I like to think about it is, for an endothermic reaction
A + B --> C + D (delta H >0),
I rewrite it like
A + B + heat --> C + D, where "heat" is now a '"reactant". Therefore, by using the usual Le Chatelier's, adding heat will cause an endothermic reaction to shift forward anyway (without even considering change in K). And then, the K change will be associated with this concentration shift at equilibrium, so if it shifts right, K will increase along with it.

[Joellen 1B]

When temperature changes it can cause an increase in either the reactants or products depending on if the forward or reverse reaction is endothermic or exothermic. If the forward reaction is endothermic, the reverse reaction is exothermic. So if you are looking at a reaction that is endothermic, adding heat/ increasing temperature will cause the reaction to produce more products because it wants heat since it is endothermic. This can cause a change in K because to find k, it is the concentration of products divided by the reactants and after increasing the heat of an endothermic reaction, there is more products, leading to a greater K value.