Equilibrium Constant

Equilibrium constant in chemical equilibrium

The equilibrium constant (K) is a value that expresses the ratio of the concentrations of products to reactants for a reversible chemical reaction at equilibrium, with each concentration raised to the power of its respective coefficient in the balanced chemical equation. The equilibrium constant is a crucial concept in understanding chemical equilibrium.

General expression for equilibrium constant

Consider a general reversible reaction:

$aA + bB \rightleftharpoons cC + dD$

The equilibrium constant expression is given by:

$K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

Where:

• $[C]$, $[D]$ are the molar concentrations of the products (C and D) at equilibrium.
• $[A]$, $[B]$ are the molar concentrations of the reactants (A and B) at equilibrium.
• a, b, c, d are the coefficients from the balanced chemical equation.

The equilibrium constant, K, is a dimensionless number that provides information about the relative concentrations of products and reactants at equilibrium. Its value depends on the reaction and temperature.

Types of equilibrium constants

There are different types of equilibrium constants, which are specific to the type of reaction:

• $K_c$ – The equilibrium constant in terms of concentrations (molarity) for reactions occurring in solution.
• $K_p$ – The equilibrium constant in terms of partial pressures for reactions involving gases.

For gaseous reactions, the equilibrium constant expression can also be written in terms of partial pressures:

$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

Where P represents the partial pressures of the gases involved in the reaction.

Relation between $K_c$ and $K_p$

For reactions involving gases, there is a relation between $K_c$ and $K_p$:

$K_p = K_c (RT)^{\Delta n}$

Where:

• R is the gas constant (0.0821 L·atm/mol·K).
• T is the temperature in Kelvin.
• $\Delta n$ is the change in the number of moles of gas, calculated as (moles of gaseous products) – (moles of gaseous reactants).

Interpreting the equilibrium constant

The value of the equilibrium constant (K) can be interpreted as follows:

1. If K > 1: The reaction favors the products at equilibrium (more products than reactants). The equilibrium lies to the right.
2. If K < 1: The reaction favors the reactants at equilibrium (more reactants than products). The equilibrium lies to the left.
3. If K ≈ 1: The concentrations of products and reactants are approximately equal at equilibrium.

Example 1: The formation of nitrogen dioxide

Consider the following reversible reaction:

$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$

The equilibrium constant expression for this reaction is:

$K_c = \frac{[NO_2]^2}{[NO]^2 [O_2]}$

If at equilibrium, the concentrations of $NO_2$, $NO$, and $O_2$ are known, you can substitute those values into the equilibrium constant expression to calculate the equilibrium constant, $K_c$.

Example 2: The dissociation of acetic acid

Consider the dissociation of acetic acid in water:

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

The equilibrium constant expression is:

$K_c = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

At equilibrium, the concentrations of acetic acid, acetate ions, and hydrogen ions will be known, and you can substitute these values into the equation to calculate the equilibrium constant, $K_c$.

Equilibrium constant calculations

Example 1: Calculating the equilibrium constant for a reaction

Consider the reversible reaction:

$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$

At equilibrium, the concentrations of the gases are as follows:

• $[NO]$ = 0.50 M
• $[O_2]$ = 0.25 M
• $[NO_2]$ = 1.0 M

The equilibrium constant expression is:

$K_c = \frac{[NO_2]^2}{[NO]^2 [O_2]}$

Substitute the values of the concentrations into the equilibrium constant expression:

$K_c = \frac{(1.0)^2}{(0.50)^2 (0.25)}$

$K_c = \frac{1.0}{(0.25)(0.25)}$

$K_c = \frac{1.0}{0.0625}$

$K_c = 16$

Thus, the equilibrium constant for this reaction is $K_c = 16$.

Example 2: Calculating equilibrium concentration using $K_c$

For the following reaction:

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

The equilibrium constant is given as $K_c = 1.8 \times 10^{-5}$, and the initial concentration of acetic acid ($CH_3COOH$) is 0.10 M. Calculate the equilibrium concentrations of $CH_3COO^-$ and $H^+$.

Let the concentration of $CH_3COOH$ decrease by x at equilibrium. Then, the concentration of $CH_3COO^-$ and $H^+$ will each increase by x at equilibrium.

At equilibrium:

• $[CH_3COOH]$ = 0.10 - x
• $[CH_3COO^-]$ = x
• $[H^+]$ = x

The equilibrium constant expression is:

$K_c = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Substitute the equilibrium concentrations into the equation:

$1.8 \times 10^{-5} = \frac{(x)(x)}{(0.10 - x)}$

$1.8 \times 10^{-5} = \frac{x^2}{0.10 - x}$

Assuming that x is very small compared to 0.10, we can approximate:

$1.8 \times 10^{-5} = \frac{x^2}{0.10}$

Now solve for x:

$x^2 = (1.8 \times 10^{-5}) \times 0.10$

$x^2 = 1.8 \times 10^{-6}$

$x = \sqrt{1.8 \times 10^{-6}}$

$x \approx 1.34 \times 10^{-3}$

The equilibrium concentrations are:

• $[CH_3COO^-]$ = $1.34 \times 10^{-3}$ M
• $[H^+]$ = $1.34 \times 10^{-3}$ M
• $[CH_3COOH]$ = $0.10 - 1.34 \times 10^{-3} \approx 0.0987$ M

Thus, the equilibrium concentrations are approximately:

• $[CH_3COO^-] \approx 1.34 \times 10^{-3}$ M
• $[H^+] \approx 1.34 \times 10^{-3}$ M
• $[CH_3COOH] \approx 0.0987$ M

Example 3: Calculating $K_p$ from $K_c$ and partial pressures

Consider the following reaction at 298 K:

$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$

The equilibrium constant in terms of concentration is given as:

$K_c = 16$

We need to calculate the equilibrium constant in terms of partial pressures ($K_p$). The change in the number of moles of gas, $\Delta n$, is:

$\Delta n = (\text{moles of products}) - (\text{moles of reactants}) = 2 - (2 + 1) = -1$

Using the relation between $K_c$ and $K_p$:

$K_p = K_c (RT)^{\Delta n}$

Substitute the known values:

$K_p = 16 \times (0.0821 \times 298)^{-1}$

$K_p = 16 \times (24.4758)^{-1}$

$K_p \approx \frac{16}{24.4758}$

$K_p \approx 0.654$

Thus, the equilibrium constant in terms of partial pressure is $K_p \approx 0.654$.

Importance of the equilibrium constant

The equilibrium constant provides valuable information about the extent of a reaction at equilibrium. A large K suggests that the reaction strongly favors the formation of products, whereas a small K suggests that the reaction favors the reactants.

The equilibrium constant also helps predict the effect of changes in concentration, pressure, and temperature on the equilibrium position. By applying Le Chatelier's Principle, one can understand how shifts in equilibrium are likely to occur in response to such changes.