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### Equilibrium Constant

Posted: Mon Jan 06, 2020 5:38 pm
Dr. Lavelle mentioned today in lecture that a more stabilized molecule would have more concentration and a higher k value as a result. Can anyone please explain this to me? Thank you

### Re: Equilibrium Constant

Posted: Mon Jan 06, 2020 5:42 pm
Michael Du 1E wrote:Dr. Lavelle mentioned today in lecture that a more stabilized molecule would have more concentration and a higher k value as a result. Can anyone please explain this to me? Thank you

Hi! The value of K is equal to the concentration of the products (raised to the stoichiometric coefficient) divided by the concentration of reactants (raised to the stoichiometric coefficient) or [P]/[R]. If a reaction produces a product that is extremely stable, there will be a large concentration of the product at equilibrium. Specifically in class today, Lavelle said that if the K value is large, that means that the concentration of the products will be very high (because the numerator is larger) meaning that the products will be more stable.
I hope that helps

### Re: Equilibrium Constant

Posted: Mon Jan 06, 2020 5:50 pm
According to the constant formula (K = [P]/[R]) the value of K will be higher, meaning there is more product than reactant left at the end. This being said, a more stable molecule results in more product and less reverse reaction happening.

### Re: Equilibrium Constant

Posted: Mon Jan 06, 2020 7:01 pm
If a molecule that is more stable than the previous molecule can form, then it will readily form this new molecule, since nature favors stability. Therefore, if a product is more stable than the reactants, as the reaction proceeds, more product will form, resulting in a higher equilibrium constant since products > reactants.

You can think back to a similar concept from the strength of acids from 14A, where an acid is stronger if its anion is more stable. A more stable conjugate base would result in an overall stronger acid since more dissociation occurs.