What change do you expect if the value of the reaction quotient is greater than the value of the equilibrium constant?

Question

A- Rate of the forward reaction is greater than the rate of the reaction
B- The rate of the forward reaction is less than the rate of the reverse reaction. More reactant forms.
C- The rate of the forward reaction is less than the rate of the reverse reaction. More product forms.
D- The rate of the forward reaction is greater than the rate of the reverse reaction. More reactant forms.

I'm pretty sure A and B are wrong. I think the answer is C, just making sure.

Answer 1

I disagree with your answer.
You know the correct answer from the second sentence of C or D. Here is how you know. Consider the reaction at equilibrium.
A + B ==> C + D
So Keq = equilibrium constant = (C)(D)/(A)(B)
Q = reaction quotient = (C)(D)/(A)(B) and Q > K.
If Q > K you know that (C) and (D) are too large and (A)(B) are too small. That's how you get Q greater than K. So the products are too high and reactants too low which means the reaction will shift to the left to reach equilibrium (to reach K) and that means more reactant forms (I might add at the expense of the products). Hope this helps.

Answer 2

Answer B is correct.

Answer 3

B- The rate of the forward reaction is less than the rate of the reverse reaction. More reactant forms.