CHEMISTRY COMMUNITY

What are the rules for when Qc is greater than Kc and when Qc is less than Kc? And what happens when the two are exactly equal?

If they are equal then the reaction is at equilibrium. If Q>K then the forward reaction or products are favored. The opposite is true for Q<K the reactants are favored.

Q will tend to go towards Kc, so when Q >Kc it will shift towards the reactants and when Q<Kc it will shift towards the products.

Q > K, system shifts left
Q < K, system shifts right
Q = K, system is at equilibrium, no shift in either direction

When Q and Kc are equal then the reaction has reached equilibrium and Q will stay the same. This doesn't mean that the reaction is no longer occurring rather that both the forward reaction and the reverse reaction are happening at the same rate

If Q>K, then the products are greater and the reverse reaction is favored. If Q<K, then reactants are greater and the forward reaction is favored.

Since Q is calculated similarly, when Q is higher than K, we know that there are more products than reactants at Q. If there are more products than at equilibrium, then the reaction has to be proceeding in reverse to go to equilibrium. And the reverse is also true.

Q > K ----> Reactants are favored
Q < K -----> Products favored
Q = K -----> Equilibrium

Think about the concentration of reactants and products: When Q > K, the concentration (or activities) of products increase (numerator increases) causing the rxn to make more reactants

For Q=K, the reaction is at equilibrium
For Q>K, the forward reaction is favored (towards the products)
For Q<K, the reverse reaction is favored (towards the reactants)

When Qc is more than Kc, reactants would be formed to get that ratio back to equilibrium.
When Qc is less than Kc, products would be formed to get the ratio up to equilbrium.
When they are equal, it's at equilibrium.

If Q < K, the forward reaction is favored. If Q > K, the reverse reaction is favored.

Q is just the same calculation as K, just at a certain point in time and does not necessarily have to be at equilibrium, correct?

John Calonia 1D wrote:Q is just the same calculation as K, just at a certain point in time and does not necessarily have to be at equilibrium, correct?

Yes, that is correct. This makes it so the Q and K values are useful to compare like people explained in previous comments.

When Qc > Kc, the reverse reaction is favored (reactants are favored).
When Qc < Kc, the forward reaction is favored (products are favored).
When Qc = Kc, neither reactants nor products are favored.

When Q>K the system shifts left (the amount of products is greater (P/R) so to reach equilibrium it moves towards reactant)
When Q<K the system shifts right (amount of reactant is greater than product so to reach eq moves toward products)
When Q=K the neither are favored, Q is considered K (the equilibrium constant)
this applies for both Qc/Qp and Kc/Kp

Q<K = Reaction is going to the product side (right)- forward direction.
Q>K = Reaction is going in the reverse direction, when making too many products.