If kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4(g) ⇄ 2NO2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm).  If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

(a) Q = K;   The reaction is at equilibrium.
(b) Q < K;   The reaction will proceed to the right.
(c) Q > K;   The reaction will proceed to the left.

The answer is the option (c) Q > K; The reaction will proceed to the left, since Qp = 1, and 1 > 0.15.
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.

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The correct answer is:   bri hope this answer was ! : )