What change do you expect if the value of the reaction quotient is greater than the value of the equilibrium constant?
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 B is correct.
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.
B- The rate of the forward reaction is less than the rate of the reverse reaction. More reactant forms.
To determine the change expected when the value of the reaction quotient (Q) is greater than the value of the equilibrium constant (K), you need to consider the direction of the reaction and the concentrations of reactants and products involved.
The reaction quotient (Q) is calculated in the same way as the equilibrium constant (K), but it is determined using the concentrations of reactants and products at any given time during a reaction, not just at equilibrium. The equilibrium constant (K), on the other hand, is determined using the concentrations of reactants and products at equilibrium.
If Q is greater than K, it means that the concentration of products is relatively higher compared to the concentration of reactants, which indicates that the reaction has not reached equilibrium.
In this scenario, the reaction will tend to shift in the reverse direction to reduce the excess of products and increase the concentration of reactants until Q equals K and equilibrium is reached. As a result, the rate of the forward reaction will be less than the rate of the reverse reaction, and more reactant will form. Therefore, the correct answer is C: "The rate of the forward reaction is less than the rate of the reverse reaction. More product forms."