Sapling Week 1&2 Homework Question

[Mrudula Akkinepally]

What happens to a system at equilibrium when there is an increase in temperature? Will it favor products or reactants? Will Q be greater than, less than, or equal to K?

I don't remember this from lecture. I don't think we covered it yet lol.

But if you know, please let me know!

Thank you :)

[JaylinWangDis1L]

I believe that when there is an increase in temperature, K changes, not Q. If the reaction is endothermic and heat is required for the reaction, then adding heat will cause the reaction to favor product formation. If the reaction is exothermic and heat is given off (released), then adding heat will will cause the reaction to favor reactants. So (if I'm not mixing this up) when Q>K the system favors reactants, when Q<K the system favors products, and when Q=K the system is at equilibrium.

[Aaina 2D]

Hi! As far as I know, temperature is one of the only things that affect the equilibrium constant (K). The way I look at how temperature affects a reaction is by whether it's exothermic or endothermic. If the reaction is endothermic, heat would technically be on the reactants side, and if you increase temperature (add heat), your adding heat to the reactants side, and so by Le chatlier's principle, the direction would move to the right and favor the products. Similarly, if the reaction was exothermic, heat would be on the products side (heat is released), and if temperature increases, you are adding heat to the products side, so by Le chatlier's principle, the reaction would move to the left (reverse direction) and favor the reactants.

Hopefully this makes sense!

[Lucy_Balish_3G]

If you want a full explanation for this question, the chemical equilibrium part 4 video on Lavelle's website was helpful and describes the relationship between temperature increase, product and reactant increase, and the value of Q. It starts at 37:50 and the link is
https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Videos/Chemical_Equilibrium_Part4.wmv.html

Hope this helps1

[alebenavides]

there's just an increase in K and not Q

[Sophia Spungin 2E]

What happens to a system at equilibrium when there is an increase in temperature? Will it favor products or reactants? Will Q be greater than, less than, or equal to K?

I don't remember this from lecture. I don't think we covered it yet lol.

But if you know, please let me know!

Thank you :)

I had a similar question, as I was under the impression that we would have to know whether the reaction was exothermic or endothermic in order to predict how it would be effected by an increase in temperature?

[Ryan_Kien_1L]

The equilibrium would shift in a direction based on whether the reaction was endothermic or exothermic.

[Sandra Kim 2B]

based on what was said in the module 4 video, exothermic reactions (releases heat) shift to the right/products and K increases while endothermic reactions (uses up heat) shift to the left/reactants and K decreases. so, you would have to determine the type of reaction it is first if you wanted to see what side is favored i think.

[Tracey Huynh 2F]

Hi!

When there is an increase in temperature, the direction that the equilibrium will shift toward depends on whether the reaction is endothermic or exothermic.

When a reaction is endothermic, it indicates that the reactants have lower energy than the products since it required heat in order for the products to form from the reactants. Therefore, with an increase in heat due to an increase in temperature, the equilibrium constant, K, will favor the products since the reactants will utilize the heat to collide and form more products. This will cause K to increase since more products are being formed.

When a reaction is exothermic, the products not have a lower energy than the reactants since the reaction releases heat when the reactants form the products. Therefore, with more heat introduced into the system through the increase in temperature, the reverse reaction is favored as it goes from a lower to higher heat (from products to reactants). This will cause K to decrease since more reactants are being formed.

Normally, we're given ΔH to see whether or not a reaction is endothermic (positive ΔH) or exothermic (negative ΔH). However, there are some occasions when ΔH is omitted, such as in the textbook problem 5J #11 part b. In these cases, we would have to see whether bonds are being broken (endothermic) or formed (exothermic).