Equilibrium Constants

Overview

Chemical equilibrium occurs when the rates of forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The equilibrium constant (K) quantifies the position of equilibrium.

Dynamic Equilibrium

At equilibrium:

• Forward rate = Reverse rate
• Concentrations are constant (not equal)
• Reactions continue at molecular level
• No net change in concentrations

Equilibrium Constant Expression

For a general reaction:

$$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$$

Concentration Constant ($K_c$)

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

Pressure Constant ($K_p$) - for gases

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

Rules for Writing K Expressions

1. Products over reactants
2. Coefficients become exponents
3. Only include gases and aqueous species
4. Omit pure solids and liquids (concentration = 1)

Examples

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$

$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$

$$K_c = [\text{CO}_2] \quad \text{(solids omitted)}$$

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

$$K_{sp} = [\text{Ag}^+][\text{Cl}^-] \quad \text{(solid omitted)}$$

Relationship Between $K_c$ and $K_p$

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

Where:

• $R$ = 0.0821 L·atm/(mol·K)
• $T$ = temperature (K)
• $\Delta n$ = (moles gas products) - (moles gas reactants)

Example

For $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, $\Delta n = 2 - 4 = -2$

$$K_p = K_c(RT)^{-2} = \frac{K_c}{(RT)^2}$$

Magnitude of K

K Value	Position	Meaning
$K \gg 1$	Right	Products favored
$K \approx 1$	Middle	Significant amounts of both
$K \ll 1$	Left	Reactants favored

Manipulating Equilibrium Constants

Reversing Reaction

$$K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}$$

Multiplying Coefficients

$$K_{\text{new}} = K^n$$

Adding Reactions

$$K_{\text{overall}} = K_1 \times K_2 \times K_3 \cdots$$

Example

If: $\text{A} \rightleftharpoons \text{B}$, $K_1 = 2.0$ $\text{B} \rightleftharpoons \text{C}$, $K_2 = 3.0$ Then: $\text{A} \rightleftharpoons \text{C}$, $K = K_1 \times K_2 = 6.0$

Reaction Quotient (Q)

Q has the same form as K but uses current (not equilibrium) concentrations.

$$Q = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \quad \text{(at any time)}$$

Comparing Q to K

Condition	Direction	Action
$Q < K$	Right (→)	More products form
$Q = K$	Equilibrium	No net change
$Q > K$	Left (←)	More reactants form

Equilibrium Calculations

ICE Table Method

I = Initial concentration C = Change in concentration E = Equilibrium concentration

Example Problem

For: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, $K_c = 54.3$ at 430°C Initial: $[\text{H}_2] = 0.50$ M, $[\text{I}_2] = 0.50$ M, $[\text{HI}] = 0$

	H₂	I₂	2HI
I	0.50	0.50	0
C	$-x$	$-x$	$+2x$
E	$0.50-x$	$0.50-x$	$2x$

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$ $$54.3 = \frac{(2x)^2}{(0.50-x)(0.50-x)} = \frac{4x^2}{(0.50-x)^2}$$ $$\sqrt{54.3} = \frac{2x}{0.50-x}$$ $$7.37(0.50-x) = 2x$$ $$3.69 - 7.37x = 2x$$ $$x = 0.39 \text{ M}$$ $$[\text{H}_2] = [\text{I}_2] = 0.11 \text{ M}, \quad [\text{HI}] = 0.78 \text{ M}$$

Special Equilibrium Constants

$K_w$ (Water)

$$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$$ $$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at 25°C}$$

$K_a$ (Acid Dissociation)

$$\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)$$ $$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$

$K_b$ (Base Dissociation)

$$\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)$$ $$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$

$K_{sp}$ (Solubility Product)

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$ $$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$

Temperature Dependence

K changes with temperature:

• Exothermic: K decreases as T increases
• Endothermic: K increases as T increases

Van't Hoff Equation

$$\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$