8.3 The Law of Mass Action | Chemical Equilibrium | Chemistry 11

The Law of Mass Action was proposed by chemists C.M. Guldberg and P. Waage in 1864 to describe the state of chemical equilibrium.

It states that:

"the rate at which a substance reacts is proportional to its active mass, and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances."

Active Mass: For dilute solutions, this is the molar concentration of the reactants and products, expressed in moles per decimeter cubed ( $m o l d m^{- 3}$ ).
The rate of a reaction is proportional to the product of the molar concentrations of each reactant, with each concentration raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

Deriving the Equilibrium Constant ( $K_{c}$ )

Consider a general, reversible gaseous reaction: $a A_{(g)} + b B_{(g)} ⇌ c C_{(g)} + d D_{(g)}$

Where:

$A, B, C, D$ are the chemical species.
$a, b, c, d$ are their stoichiometric coefficients in the balanced equation.

According to the Law of Mass Action:

Rate of the Forward Reaction ( $R_{f}$ ): The rate at which reactants A and B form products C and D. $R_{f} \propto [A]^{a} [B]^{b}$ $R_{f} = k_{f} [A]^{a} [B]^{b}$ where $k_{f}$ is the rate constant for the forward reaction.
Rate of the Reverse Reaction ( $R_{r}$ ): The rate at which products C and D revert to reactants A and B. $R_{r} \propto [C]^{c} [D]^{d}$ $R_{r} = k_{r} [C]^{c} [D]^{d}$ where $k_{r}$ is the rate constant for the reverse reaction.
At Equilibrium: The rate of the forward reaction equals the rate of the reverse reaction. $R_{f} = R_{r}$ $k_{f} [A]^{a} [B]^{b} = k_{r} [C]^{c} [D]^{d}$
The Equilibrium Constant ( $K_{c}$ ): Rearranging the equation gives the ratio of the rate constants, which is a new constant called the equilibrium constant, $K_{c}$ . $\frac{k _{f}}{k _{r}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ This is known as the equilibrium constant expression. The square brackets $[...]$ denote the molar concentration of the species at equilibrium.

Conditions for Equilibrium (Features of $K_{c}$ )

$K_{c}$ applies only at the equilibrium state. The subscript 'c' indicates that concentrations are in $m o l d m^{- 3}$ .
$K_{c}$ is dependent on temperature but independent of the initial concentrations of reactants or products.
The expression for $K_{c}$ is derived directly from the balanced chemical equation:
- Products are in the numerator.
- Reactants are in the denominator.
- Each concentration is raised to the power of its stoichiometric coefficient.
The magnitude of $K_{c}$ indicates the position of equilibrium:
- If $K_{c} > 1$ : The concentration of products is greater than reactants at equilibrium. The equilibrium lies to the right (favors products).
- If $K_{c} < 1$ : The concentration of reactants is greater than products at equilibrium. The equilibrium lies to the left (favors reactants).
- If $K_{c} ≫ 1$ : The reaction proceeds almost to completion.

Equilibrium Expressions in Different Units

For reactions involving gases, equilibrium can be expressed in terms of partial pressures ( $K_{p}$ ), number of moles ( $K_{n}$ ), or mole fractions ( $K_{x}$ ).

1. Partial Pressure ( $K_{p}$ )

Based on Henry's Law, at a constant temperature, the partial pressure of a gas is directly proportional to its molar concentration.

For the general reaction $a A_{(g)} + b B_{(g)} ⇌ c C_{(g)} + d D_{(g)}$ : $K_{p} = \frac{P _{C}^{c} \times P _{D}^{d}}{P _{A}^{a} \times P _{B}^{b}}$ Where $P_{A}, P_{B}, P_{C}, P_{D}$ are the partial pressures of the gases at equilibrium.

The relationship between $K_{p}$ and $K_{c}$ is: $K_{p} = K_{c} (R T)^{Δ n}$

$Δ n = (moles of gaseous products) - (moles of gaseous reactants) = (c + d) - (a + b)$
$R$ is the ideal gas constant ( $0.0821 d m^{3} a t m K^{- 1} m o l^{- 1}$ ).
$T$ is the absolute temperature in Kelvin (K).

Special cases:

If $Δ n = 0$ : $K_{p} = K_{c}$
If $Δ n > 0$ : $K_{p} > K_{c}$
If $Δ n < 0$ : $K_{p} < K_{c}$

2. Number of Moles ( $K_{n}$ )

$K_{n} = \frac{n _{C}^{c} \times n _{D}^{d}}{n _{A}^{a} \times n _{B}^{b}}$ Where $n_{A}, n_{B}, n_{C}, n_{D}$ are the number of moles of each species at equilibrium.

The relationship between $K_{p}$ and $K_{n}$ is: $K_{p} = K_{n} (\frac{P}{n})^{Δ n}$

$P$ is the total pressure of the mixture at equilibrium.
$n$ is the total number of moles of all species at equilibrium.

3. Mole Fraction ( $K_{x}$ )

$K_{x} = \frac{x _{C}^{c} \cdot x _{D}^{d}}{x _{A}^{a} \cdot x _{B}^{b}}$ Where $x_{A}, x_{B}, x_{C}, x_{D}$ are the mole fractions of each species.

The relationship between $K_{p}$ and $K_{x}$ is: $K_{p} = K_{x} (P)^{Δ n}$

Worked Examples

Writing Equilibrium Constant Expressions

Problem Solving Strategy:

Write the products in the numerator and reactants in the denominator within square brackets.
Raise each concentration to the power of its coefficient from the balanced chemical equation.

Example 8.1: Haber Process $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$ Expression: $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$

Example 8.2: Carbon Monoxide Oxidation $C O_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ C O_{2 (g)}$ Expression: $K_{c} = \frac{[ C O _{2} ]}{[ C O ] [ O _{2} ] ^{\frac{1}{2}}}$

Calculating $K_{c}$ from Equilibrium Concentrations

Example 8.3

The following equilibrium concentrations were observed for the reaction at $50 0^{\circ} C$ . Calculate $K_{c}$ . $A_{2 (g)} + B_{2 (g)} ⇌ 2 A B_{(g)}$

Species	Equilibrium Concentration ( $m o l d m^{- 3}$ )
$A_{2}$	0.50
$B_{2}$	0.50
$A B$	1.00

Solution: $K_{c} = \frac{[ A B ] ^{2}}{[ A _{2} ] [ B _{2} ]} = \frac{( 1.00 ) ^{2}}{( 0.50 ) ( 0.50 )} = \frac{1.00}{0.25} = 4.0$

Since $K_{c} = 4.0 > 1$ , the equilibrium favors the formation of products.

Calculating $K_{p}$ from $K_{c}$

Example 8.4

Calculate $K_{p}$ at $105 0^{\circ} C$ if $K_{c} = 2.3 \times 1 0^{22}$ for the reaction: $2 C O_{(g)} + O_{2 (g)} ⇌ 2 C O_{2 (g)}$

Solution:

Step 1: Calculate $Δ n$ : $Δ n = 2 - (2 + 1) = - 1$

Step 2: Convert temperature to Kelvin: $T = 1050 + 273 = 1323 K$

Step 3: Apply $K_{p} = K_{c} (R T)^{Δ n}$ : $K_{p} = (2.3 \times 1 0^{22}) (0.0821 \times 1323)^{- 1}$ $K_{p} = \frac{2.3 \times 1 0 ^{22}}{108.6}$ $K_{p} \approx 2.12 \times 1 0^{20}$

Note: $K_{p} < K_{c}$ because $Δ n < 0$ .

The Law of Mass Action was proposed by chemists C.M. Guldberg and P. Waage in 1864 to describe the state of chemical equilibrium.

It states that:

"the rate at which a substance reacts is proportional to its active mass, and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances."

Active Mass: For dilute solutions, this is the molar concentration of the reactants and products, expressed in moles per decimeter cubed ( $m o l d m^{- 3}$ ).
The rate of a reaction is proportional to the product of the molar concentrations of each reactant, with each concentration raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

Deriving the Equilibrium Constant ( $K_{c}$ )

Consider a general, reversible gaseous reaction: $a A_{(g)} + b B_{(g)} ⇌ c C_{(g)} + d D_{(g)}$

Where:

$A, B, C, D$ are the chemical species.
$a, b, c, d$ are their stoichiometric coefficients in the balanced equation.

According to the Law of Mass Action:

Rate of the Forward Reaction ( $R_{f}$ ): The rate at which reactants A and B form products C and D. $R_{f} \propto [A]^{a} [B]^{b}$ $R_{f} = k_{f} [A]^{a} [B]^{b}$ where $k_{f}$ is the rate constant for the forward reaction.
Rate of the Reverse Reaction ( $R_{r}$ ): The rate at which products C and D revert to reactants A and B. $R_{r} \propto [C]^{c} [D]^{d}$ $R_{r} = k_{r} [C]^{c} [D]^{d}$ where $k_{r}$ is the rate constant for the reverse reaction.
At Equilibrium: The rate of the forward reaction equals the rate of the reverse reaction. $R_{f} = R_{r}$ $k_{f} [A]^{a} [B]^{b} = k_{r} [C]^{c} [D]^{d}$
The Equilibrium Constant ( $K_{c}$ ): Rearranging the equation gives the ratio of the rate constants, which is a new constant called the equilibrium constant, $K_{c}$ . $\frac{k _{f}}{k _{r}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ $K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ This is known as the equilibrium constant expression. The square brackets $[...]$ denote the molar concentration of the species at equilibrium.

Conditions for Equilibrium (Features of $K_{c}$ )

$K_{c}$ applies only at the equilibrium state. The subscript 'c' indicates that concentrations are in $m o l d m^{- 3}$ .
$K_{c}$ is dependent on temperature but independent of the initial concentrations of reactants or products.
The expression for $K_{c}$ is derived directly from the balanced chemical equation:
- Products are in the numerator.
- Reactants are in the denominator.
- Each concentration is raised to the power of its stoichiometric coefficient.
The magnitude of $K_{c}$ indicates the position of equilibrium:
- If $K_{c} > 1$ : The concentration of products is greater than reactants at equilibrium. The equilibrium lies to the right (favors products).
- If $K_{c} < 1$ : The concentration of reactants is greater than products at equilibrium. The equilibrium lies to the left (favors reactants).
- If $K_{c} ≫ 1$ : The reaction proceeds almost to completion.

Equilibrium Expressions in Different Units

For reactions involving gases, equilibrium can be expressed in terms of partial pressures ( $K_{p}$ ), number of moles ( $K_{n}$ ), or mole fractions ( $K_{x}$ ).

1. Partial Pressure ( $K_{p}$ )

Based on Henry's Law, at a constant temperature, the partial pressure of a gas is directly proportional to its molar concentration.

The relationship between $K_{p}$ and $K_{c}$ is: $K_{p} = K_{c} (R T)^{Δ n}$

$Δ n = (moles of gaseous products) - (moles of gaseous reactants) = (c + d) - (a + b)$
$R$ is the ideal gas constant ( $0.0821 d m^{3} a t m K^{- 1} m o l^{- 1}$ ).
$T$ is the absolute temperature in Kelvin (K).

Special cases:

If $Δ n = 0$ : $K_{p} = K_{c}$
If $Δ n > 0$ : $K_{p} > K_{c}$
If $Δ n < 0$ : $K_{p} < K_{c}$

2. Number of Moles ( $K_{n}$ )

$K_{n} = \frac{n _{C}^{c} \times n _{D}^{d}}{n _{A}^{a} \times n _{B}^{b}}$ Where $n_{A}, n_{B}, n_{C}, n_{D}$ are the number of moles of each species at equilibrium.

The relationship between $K_{p}$ and $K_{n}$ is: $K_{p} = K_{n} (\frac{P}{n})^{Δ n}$

$P$ is the total pressure of the mixture at equilibrium.
$n$ is the total number of moles of all species at equilibrium.

3. Mole Fraction ( $K_{x}$ )

$K_{x} = \frac{x _{C}^{c} \cdot x _{D}^{d}}{x _{A}^{a} \cdot x _{B}^{b}}$ Where $x_{A}, x_{B}, x_{C}, x_{D}$ are the mole fractions of each species.

The relationship between $K_{p}$ and $K_{x}$ is: $K_{p} = K_{x} (P)^{Δ n}$

Worked Examples

Writing Equilibrium Constant Expressions

Problem Solving Strategy:

Write the products in the numerator and reactants in the denominator within square brackets.
Raise each concentration to the power of its coefficient from the balanced chemical equation.

Example 8.1: Haber Process $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}$ Expression: $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$

Example 8.2: Carbon Monoxide Oxidation $C O_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ C O_{2 (g)}$ Expression: $K_{c} = \frac{[ C O _{2} ]}{[ C O ] [ O _{2} ] ^{\frac{1}{2}}}$

Calculating $K_{c}$ from Equilibrium Concentrations

Example 8.3

The following equilibrium concentrations were observed for the reaction at $50 0^{\circ} C$ . Calculate $K_{c}$ . $A_{2 (g)} + B_{2 (g)} ⇌ 2 A B_{(g)}$

Species	Equilibrium Concentration ( $m o l d m^{- 3}$ )
$A_{2}$	0.50
$B_{2}$	0.50
$A B$	1.00

Solution: $K_{c} = \frac{[ A B ] ^{2}}{[ A _{2} ] [ B _{2} ]} = \frac{( 1.00 ) ^{2}}{( 0.50 ) ( 0.50 )} = \frac{1.00}{0.25} = 4.0$

Since $K_{c} = 4.0 > 1$ , the equilibrium favors the formation of products.

Calculating $K_{p}$ from $K_{c}$

Example 8.4

Calculate $K_{p}$ at $105 0^{\circ} C$ if $K_{c} = 2.3 \times 1 0^{22}$ for the reaction: $2 C O_{(g)} + O_{2 (g)} ⇌ 2 C O_{2 (g)}$

Solution:

Step 1: Calculate $Δ n$ : $Δ n = 2 - (2 + 1) = - 1$

Step 2: Convert temperature to Kelvin: $T = 1050 + 273 = 1323 K$

Step 3: Apply $K_{p} = K_{c} (R T)^{Δ n}$ : $K_{p} = (2.3 \times 1 0^{22}) (0.0821 \times 1323)^{- 1}$ $K_{p} = \frac{2.3 \times 1 0 ^{22}}{108.6}$ $K_{p} \approx 2.12 \times 1 0^{20}$

Note: $K_{p} < K_{c}$ because $Δ n < 0$ .