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Chemistry · Ch 6 — Equilibrium

Law of Chemical Equilibrium and Equilibrium Constant

6.3

Law of Chemical Equilibrium and Equilibrium Constant

The Law of Chemical Equilibrium

When a reversible reaction reaches equilibrium, the mixture of reactants and products that remains is called an equilibrium mixture. The central question is: what fixed relationship exists between the concentrations of the substances in that mixture?

Consider a general reversible reaction written in the form:

aA+bB⇌cC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

where A and B are reactants, C and D are products, and the lowercase letters are their stoichiometric coefficients from the balanced equation.

In 1864, the Norwegian chemists Cato Maximillian Guldberg and Peter Waage proposed, based on extensive experimental work, that at equilibrium the concentrations obey a simple mathematical relation. They called it the law of mass action — because in those days concentration was referred to as "active mass". Today we call it the law of chemical equilibrium.

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

The quantity KcK_c is the equilibrium constant. The expression on the right is the equilibrium constant expression. The subscript cc indicates that concentrations are measured in moles per litre (mol L−1^{-1}). It is understood that every concentration in the expression is an equilibrium concentration, so the subscript "eq" is usually omitted.


How the Law Was Discovered: The H₂–I₂–HI System

To see how Guldberg and Waage arrived at this form, examine the reaction between hydrogen and iodine gases in a sealed vessel at 731 K:

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

Six experiments were performed. In experiments 1–4, the vessel was charged with only H₂ and I₂ in various amounts. In experiments 5 and 6, the vessel was charged with only HI. In every case the system was allowed to reach equilibrium, and the equilibrium concentrations of all three gases were measured.

The data obtained from all six experiments is given in Table 6.2:

Table 6.2Initial and Equilibrium Concentrations of H2, I2 and HI -- six real experiments that Guldberg and Waage's data was drawn from, at 731 K.
Experiment numberInitial [H2_2(g)]Initial [I2_2(g)]Initial [HI(g)]Equilibrium [H2_2(g)]Equilibrium [I2_2(g)]Equilibrium [HI(g)]
12.4×10−22.4 \times 10^{-2}1.38×10−21.38 \times 10^{-2}01.14×10−21.14 \times 10^{-2}0.12×10−20.12 \times 10^{-2}2.52×10−22.52 \times 10^{-2}
22.4×10−22.4 \times 10^{-2}1.68×10−21.68 \times 10^{-2}00.92×10−20.92 \times 10^{-2}0.20×10−20.20 \times 10^{-2}2.96×10−22.96 \times 10^{-2}
32.44×10−22.44 \times 10^{-2}1.98×10−21.98 \times 10^{-2}00.77×10−20.77 \times 10^{-2}0.31×10−20.31 \times 10^{-2}3.34×10−23.34 \times 10^{-2}
42.46×10−22.46 \times 10^{-2}1.76×10−21.76 \times 10^{-2}00.92×10−20.92 \times 10^{-2}0.22×10−20.22 \times 10^{-2}3.08×10−23.08 \times 10^{-2}
5003.04×10−23.04 \times 10^{-2}0.345×10−20.345 \times 10^{-2}0.345×10−20.345 \times 10^{-2}2.35×10−22.35 \times 10^{-2}
6007.58×10−27.58 \times 10^{-2}0.86×10−20.86 \times 10^{-2}0.86×10−20.86 \times 10^{-2}5.86×10−25.86 \times 10^{-2}

The data revealed two important patterns:

  1. In experiments 1–4, the number of moles of H₂ that reacted equalled the number of moles of I₂ that reacted, and this was exactly half the number of moles of HI formed.
  2. In experiments 5 and 6 (starting from pure HI), at equilibrium [H2]eq=[I2]eq[\text{H}_2]_{\text{eq}} = [\text{I}_2]_{\text{eq}}.

Now, to find a constant relationship, several trial expressions were tested. The simple expression [HI][H2][I2]\frac{[\text{HI}]}{[\text{H}_2][\text{I}_2]} gave values that varied wildly from one experiment to the next — it was not constant at all.

But the expression [HI]2[H2][I2]\frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} gave the same numerical value for all six experiments, regardless of the starting composition. Notice that the exponent on HI (2) is exactly its stoichiometric coefficient in the balanced equation. This is the key: the power to which each concentration is raised in the equilibrium expression is the stoichiometric coefficient of that substance in the balanced chemical equation.

Both ratios, side by side for all six experiments, are summarised in Table 6.3:

Table 6.3Expression Involving the Equilibrium Concentration of Reactants for H2(g) + I2(g) ⇌ 2HI(g) -- shows [HI]^2/([H2][I2]) is the constant ratio, not the simpler [HI]/([H2][I2]).
Experiment number[HI(g)]eq[H2(g)]eq[I2(g)]eq\dfrac{[\text{HI}(g)]_{eq}}{[\text{H}_2(g)]_{eq}[\text{I}_2(g)]_{eq}}[HI(g)]eq2[H2(g)]eq[I2(g)]eq\dfrac{[\text{HI}(g)]^2_{eq}}{[\text{H}_2(g)]_{eq}[\text{I}_2(g)]_{eq}}
1184046.4
2161047.6
3140046.7
4152046.9
5197046.4
679046.4

Thus for this reaction:

Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}


Writing Equilibrium Constant Expressions

For any balanced chemical equation, the equilibrium constant expression is written by placing the product concentrations in the numerator, each raised to its stoichiometric coefficient, and the reactant concentrations in the denominator, each raised to its stoichiometric coefficient.

Watch out

The exponents in the equilibrium expression come from the stoichiometric coefficients in the balanced chemical equation as written. They are NOT the same as the orders in a rate law — those are determined experimentally and can be different. Do not confuse the two.

Example: For the reaction

4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g)

the equilibrium constant is:

Kc=[NO]4[H2O]6[NH3]4[O2]5K_c = \frac{[\text{NO}]^4 [\text{H}_2\text{O}]^6}{[\text{NH}_3]^4 [\text{O}_2]^5}

The square brackets denote molar concentrations at equilibrium. Phase symbols (s, l, g) are generally omitted from the expression — they are not needed because the concentrations themselves already imply the physical state.


Properties of the Equilibrium Constant

The equilibrium constant has several important properties that follow directly from its definition. Each one is derived below.

Property 1: The Equilibrium Constant for the Reverse Reaction

If a reaction is written in the reverse direction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

Proof: Take the forward reaction

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

for which

Kc=[HI]2[H2][I2]=xK_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = x

Now write the reverse reaction:

2HI(g)⇌H2(g)+I2(g)2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)

Its equilibrium constant, call it Kc′K'_c, is:

Kc′=[H2][I2][HI]2K'_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

But this is exactly 1/x1/x, which is 1/Kc1/K_c. Therefore:

Kc′=1KcK'_c = \frac{1}{K_c}

Important

The equilibrium constant for the reverse reaction is always the reciprocal of the equilibrium constant for the forward reaction at the same temperature.

Property 2: Multiplying the Equation by a Factor

If the stoichiometric coefficients in a balanced equation are multiplied by a factor nn, the new equilibrium constant is the original constant raised to the power nn.

Proof: Start again with

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

and Kc=[HI]2[H2][I2]=xK_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = x.

First, multiply the equation by 12\frac{1}{2}:

12H2(g)+12I2(g)⇌HI(g)\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{I}_2(g) \rightleftharpoons \text{HI}(g)

The equilibrium constant for this version, call it Kc′′K''_c, is:

Kc′′=[HI][H2]1/2[I2]1/2K''_c = \frac{[\text{HI}]}{[\text{H}_2]^{1/2}[\text{I}_2]^{1/2}}

Now square both sides:

(Kc′′)2=[HI]2[H2][I2]=x(K''_c)^2 = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = x

Therefore Kc′′=x1/2=Kc1/2K''_c = x^{1/2} = K_c^{1/2}.

Now multiply the original equation by a general factor nn:

nH2(g)+nI2(g)⇌2nHI(g)n\text{H}_2(g) + n\text{I}_2(g) \rightleftharpoons 2n\text{HI}(g)

The equilibrium constant for this multiplied equation, Kc′′′K'''_c, is: …

Table 6.4Relations between Equilibrium Constants for a General Reaction and its Multiples.
Operation on the equationNew equilibrium constant
Original: aA+bB⇌cC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}KcK_c
Reverse: cC+dD⇌aA+bBc\text{C} + d\text{D} \rightleftharpoons a\text{A} + b\text{B}Kc′=1/KcK'_c = 1/K_c