KEY2CHEM

Comparing $$Q$$ and $$K$$

A process that is thermodynamically favored is one that is approaching equilibrium. As such, the magnitude of the reaction quotient ($$Q$$) in comparison to equilibrium constant ($$K$$) provides insight as to the net direction of the reaction. The format of $$Q$$ and $$K$$ expressions are the same, and at equilibrium $$Q = K$$. If $$Q < K$$, the reaction will proceed toward products (increase numerator, which contains products, and decrease denominator, which contains reactant values). If $$Q > K$$, the reaction will proceed toward reactants (increase reactant values, in the denominator, and decrease product values, in the numerator).

Example 1.

If $$Q > K$$, the reaction will proceed toward the ______.

A. products

B. right

C. reactants

Solution

C. reactants.

If $$Q > K$$, the reaction will consume products and generate reactants (will proceed toward reactants, or to the left).

Example 2.

A reaction is found to have $$Q = 25$$. If $$K = 50$$ for the reaction, which statement is true?

A. The reaction will proceed to the left.

B. The reaction will proceed to the right.

C. The reaction is at equilibrium.

Solution

B. The reaction will proceed to the right.

Since $$Q < K$$ (since $$25 < 50$$), the reaction will proceed to the right, consuming reactants and generating products.

Example 3.

For the reaction below, the initial concentrations are $$[H_2] = 0.040 \text{ M}$$, $$[I_2] = 0.016 \text{ M}$$, and $$[HI] = 0.010 \text{ M}$$. If $$K = 0.350$$, which statement is correct?

$$\require{mhchem}\ce{H2(g) + I2(g) <=> 2 HI(g)}$$

A. The reaction will proceed toward reactants.

B. The reaction will proceed toward products.

C. The reaction is at equilibrium.

Solution

B. The reaction will proceed to the right.

$$Q = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.020)^2}{(0.040)(0.016)} = 0.156$$

$$Q < K$$ (since $$0.156 < 0.350$$), so the reaction proceeds to the right (toward products).