Page 1 of 1

### What does K say about stability?

Posted: Sun Mar 10, 2019 7:44 pm
For Q1 part c on the midterm, the question asked us to comment on the relative stability of the reactants and products, and the equilibrium constant (Kp) given was 3 x 10^4.

The solutions said the formation of SO3- was more stable because Kp is a large number, but I donâ€™t really understand why. Could someone explain why please?

Also, if K was a small number, would that mean the dissociation of reactants would be more stable?

### Re: What does K say about stability?

Posted: Sun Mar 10, 2019 8:20 pm
For this question, since K >10^3, at equilibrium and K = concentration of products / concentration of reactants, then there are more products than reactants and the reaction lies to the right. Therefore, there is more SO3 (the product) at equilibrium, which then refers to its stability. Since there is more SO3 at equilibrium, the formation of SO3 (the forward reaction) is considered more stable than the reverse reaction, which would mean the formation of SO2 and O2.

### Re: What does K say about stability?

Posted: Sun Mar 10, 2019 8:22 pm
To my understanding, a large K tells us that the forward reaction is favored. This means that the reactants have a tendency to split and/or change, making them unstable. Therefore, the products are relatively stable when compared to the reactants.

### Re: What does K say about stability?

Posted: Sun Mar 10, 2019 8:42 pm
The higher the K the lower the stability of the reactants and higher the stability of the products

### Re: What does K say about stability?

Posted: Sun Mar 10, 2019 8:47 pm
A large K means that there are more products than reactants at equilibrium. This means the products are more stable than the reactants, because reactions generally proceed to form the most stable things possible. Therefore, forming SO3 is more stable than forming SO2 and O2.

### Re: What does K say about stability?

Posted: Sun Mar 10, 2019 9:20 pm
I was wondering about this too! I got this wrong on the test, and I have still been confused about it. Thank you for all of the explanations, they were all really helpful!