Magnitude of the Equilbrium Constant | College Board AP Chemistry Revision Notes 2022

Magnitude of the Equilibrium Constant

The equilibrium constant (K) is the ratio of products and reactants in a system at equilibrium

$K_{C} = \frac{[products]}{[reactants]}$ $K_{P} = \frac{P_{products}}{P_{reactants}}$

The magnitude of the equilibrium constant indicates the extent to which the reaction proceeds
When K > 1, the products are favored at equilibrium
- If K is very large, the forward reaction essentially proceeds to completion
When K < 1, the reactants are favored at equilibrium
- If K is very small, the forward reaction barely proceeds. In other words, the reverse reaction essentially proceeds to completion
When comparing the equilibrium constants of two or more reactions at the same temperature, the reaction with the greatest value for K will have the greatest proportion of products at equilibrium

Worked example

C₂H₄(g) ⇌ C₂H₂(g) + H₂(g) K_P = 1.54 x10^-7

Pure C₂H₄(g) is placed in a 1.0 L rigid, sealed flask at 700 K and allowed to react according to the equation shown above. Which of the following statements correctly compares the partial pressures of C₂H₄(g) and H₂(g) after equilibrium is established?

$P_{C_{2} H_{4} (eq)} > P_{H_{2} (eq)}$ , because the smaller the K_P value, the greater the partial pressure of the reactants at equilibrium
$P_{C_{2} H_{4} (eq)} = P_{H_{2} (eq)}$ , because at equilibrium the partial pressures of the products and reactants must be constant and equal
$P_{H_{2} (eq)} > P_{C_{2} H_{4} (eq)}$ , because the initial rate of the forward reaction is greater than that of the reverse reaction leading to an increase in the partial pressure of the products at equilibrium
$P_{H_{2} (eq)} > P_{C_{2} H_{4} (eq)}$ , because the smaller the K_P value, the faster the rate of reaction and the greater the partial pressure of the products at equilibrium

Answer:

At equilibrium, the partial pressures of the products and reactants must remain constant, but they do not necessarily have to be equal
A very small K_P value indicates that the forward reaction barely proceeds and that the reactants are favored at equilibrium
So at equilibrium, the flask will container a much greater proportion of C₂H₄(g) than H₂(g)
As the gases occupy the same volume and are held at the same temperature, the partial pressure of C₂H₄(g) at equilibrium will be greater than that of H₂(g)

Worked example

Into two separate flasks, a solution containing 1.0 M OH^-(aq) is combined with a solution containing 1.0 M of the appropriate metal ion as shown in the table below. The ions react via an equilibrium reaction to form a complex ion.

Flask	Chemical Reaction	*K_c* at 25°C
A	Al³⁺ (aq) + 4OH^- (aq) ⇌ Al(OH)₄^- (aq)	1.1 x 10³³
B	Zn²⁺ (aq) + 4OH^-(aq) ⇌ Zn(OH)₄^2-(aq)	4.6 x 10¹⁷

After reaching equilibrium at 25°C, which flask contains a higher concentration of its respective complex ion?

Answer:

A very large K_c value indicates that the forward reaction essentially proceeds to completion
The larger the K_c value, the greater the proportion of products in the equilibrium mixture
As both flasks initially contain equimolar amounts of the reactants and the reaction in flask A has a larger K_c value, the concentration of the complex ion in flask A at equilibrium will be greater than that of flask B

Exam Tip

If you forget which side of a reaction is favored based on the value of K you can perform a quick test using simple concentrations. This is summarized in the table below.

….. Are favored

Concentrations at equilibrium

Value of K_c

Result

Products

2M = [products]

1M = [reactants]

K_{c} = \frac{2}{1}

K_c > 1

Reactants

1M = [products]

2M = [reactants]

K_{c} = \frac{1}{2}

K_c < 1

Magnitude of the Equilbrium Constant (College Board AP Chemistry)

Revision Note

Magnitude of the Equilibrium Constant

Worked example

Worked example

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