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

1 answer:

6 0

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<span>(g) ⇄ 2NO</span>2(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<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

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

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