Equilibrium · Lesson 3 of 14

The Equilibrium Law and Concentration Constants

“Build an equilibrium expression from experimental patterns and learn how it follows the balanced equation.”

Learning Objectives

• Construct concentration equilibrium expressions from balanced equations. • Explain how hydrogen–iodine data support the equilibrium law. • Calculate a constant from equilibrium concentrations. • Transform a constant correctly when a reaction is reversed or scaled.

Different starting mixtures reveal one relationship

The same reaction can reach equilibrium from many starting compositions. The final concentrations need not match between experiments with different atom inventories. Yet a particular combination of these concentrations has a reproducible value at a fixed temperature. This relationship is the law of chemical equilibrium, developed from experimental work by Guldberg and Waage.

For $H_{2} (g) + I_{2} (g) ⇌ 2HI (g)$ at 731 K, the chapter reports the following measurements. Square brackets mean amount of a species per unit solution or gas volume, usually in moles per litre. One mole per litre is abbreviated M. In an equilibrium-constant expression, the concentrations are equilibrium values even when a subscript is not written.

Preparation	Initial H2 / M	Initial I2 / M	Initial HI / M	Equilibrium H2 / M	Equilibrium I2 / M	Equilibrium HI / M
1	0.0240	0.0138	0	0.0114	0.00120	0.0252
2	0.0240	0.0168	0	0.00920	0.00200	0.0296
3	0.0244	0.0198	0	0.00770	0.00310	0.0334
4	0.0246	0.0176	0	0.00920	0.00220	0.0308
5	0	0	0.0304	0.00345	0.00345	0.0235
6	0	0	0.0758	0.00860	0.00860	0.0586

Stoichiometry already constrains these data: consuming one mole of H2 consumes one mole of I2 and forms two moles of HI. But stoichiometry alone does not tell us how far the reaction goes. Try the simple concentration ratio $[HI] / ([H_{2}] [I_{2}])$ : it changes markedly between preparations. Squaring the HI concentration instead gives nearly the same value in all six experiments. Small differences reflect the rounded measurements.

Preparation	Ratio with HI to first power / numerical value	Ratio with HI squared / numerical value
1	1842	46.4
2	1609	47.6
3	1399	46.7
4	1522	46.9
5	1974	46.4
6	792	46.4

Hydrogen iodide equilibrium constantLaTeX

K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]}

The exponent 2 matches the coefficient of HI in the balanced reaction. At 731 K these rounded measurements give a value close to 47.

Coefficients become powers in the equilibrium expression

For a balanced reaction, multiply the product concentrations, raising each to its stoichiometric coefficient. Divide by the corresponding product of reactant concentrations. This is a statement about equilibrium. It is not permission to use the balanced equation as a reaction-rate law for every multistep reaction.

General equilibrium lawLaTeX

a A + b B ⇌ c C + d D, K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

The letters a, b, c and d are the coefficients in the balanced equation. The concentrations inserted here are the equilibrium concentrations at the stated temperature.

A homogeneous equilibrium has all reacting species in the same phase. Hydrogen, iodine vapour and hydrogen iodide form one gas phase. Dissolved iron(III), thiocyanate and their complex occupy one aqueous phase. A homogeneous liquid solution is a mixture: the amounts and activities of its components can vary, even if every component is labelled liquid. The next lesson distinguishes these from separate pure solid or liquid phases.

Example — Ammonia formation

Problem
At 500 K, the equilibrium concentrations are $[N_{2}] = 0.015 M$ , $[H_{2}] = 0.030 M$ and $[N H_{3}] = 0.012 M$ . Find the concentration-form constant for ammonia synthesis.

1.Balance the equation: $N_{2} + 3 H_{2} ⇌ 2N H_{3}$ . The hydrogen term must be cubed.
2.Write $K_{c} = [N H_{3}]^{2} / ([N_{2}] [H_{2}]^{3})$ .
3.Substitute $K_{c} = (0.012)^{2} / [0.015 (0.030)^{3}] = 355.6$ , or $3.56 \times 1 0^{2}$ with the stated numerical concentration convention.

Example — Nitric oxide formation

Problem
For $N_{2} + O_{2} ⇌ 2NO$ , equilibrium concentrations at 800 K are 0.0030 M, 0.0042 M and 0.0028 M respectively. Calculate the constant.

1.The expression is $K_{c} = [NO]^{2} / ([N_{2}] [O_{2}])$ .
2.The numerator is $(0.0028)^{2} = 7.84 \times 1 0^{- 6}$ and the denominator is $0.0030 \times 0.0042 = 1.26 \times 1 0^{- 5}$ .
3.Their ratio is $K_{c} = 0.622$ . All concentrations were supplied at equilibrium, so no change table is needed.

The constant belongs to a particular written reaction

An equilibrium constant must always be quoted with its balanced equation and temperature. Reversing an equation swaps the numerator and denominator, giving the reciprocal constant. Multiplying every coefficient by a number raises the entire expression to that power. A different written equation therefore has a different numerical constant even when it describes the same physical mixture.

Change in equation	Change in constant	Example if original K = 49
Reverse the reaction	$K_{reverse} = 1/ K$	$1/49$
Multiply all coefficients by 2	$K_{new} = K^{2}$	2401
Divide all coefficients by 2	$K_{new} = K^{1/2}$	7

Example — Combining reversal with scaling

Problem
At one temperature, $K = 49$ for $H_{2} + I_{2} ⇌ 2HI$ . Find the constant for $HI ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2}$ .

1.Reverse the original equation: the constant becomes $1/49$ .
2.Divide all coefficients of the reversed equation by two: take the square root of its constant.
3.The required constant is $(1/49)^{1/2} = 1/7$ . Reversing without scaling would give the wrong answer.

Changing the initial amounts at the same temperature changes the equilibrium composition, but it does not produce a new value of K for the same reaction and standard-state convention. Changing temperature generally does change K. This distinction lets us predict or calculate new equilibrium compositions while keeping one consistent constant.

Common mistake

Do not leave out coefficients when writing powers, and do not use starting concentrations as though they were equilibrium concentrations. The same algebraic ratio can be evaluated away from equilibrium, but it is then a reaction quotient rather than the equilibrium constant.

Quick check

For N2 + 3H2 forming 2NH3, which term appears in the denominator of Kc?

Quick check

What is the reverse constant if the forward constant is 20?

Quick check

If every coefficient is doubled, the new constant is

Quick check

Why is HI squared in its formation equilibrium expression?

Quick check

At a fixed temperature, changing starting amounts can change

Quick check

Which statement about equilibrium and rate laws is correct?

Practice Problems

Write Kc for 4NH3(g) + 5O2(g) forming 4NO(g) + 6H2O(g).
For A + B forming C + D, the equilibrium concentrations are 0.20, 0.50, 0.40 and 0.30 M. Find Kc.
Use preparation 5 in the hydrogen–iodine table to calculate the squared-HI ratio.
For a reaction with K = 16, find constants for its reverse and its half-sized equation.
Explain why two starting mixtures can have different equilibrium concentrations but the same K.
Distinguish a homogeneous gas equilibrium from an equilibrium containing a separate solid phase.

Key Takeaways

Equilibrium concentrations obey a characteristic product-to-reactant ratio at fixed temperature. Balanced coefficients determine the powers in the equilibrium expression. Reversal takes a reciprocal, and scaling an equation raises its constant to the same power.

Previous · Lesson 2

How Chemical Equilibrium Is Established

Next · Lesson 4

Gas Pressures and Heterogeneous Equilibria

Different starting mixtures reveal one relationship

Coefficients become powers in the equilibrium expression

The constant belongs to a particular written reaction

Quiz

Practice Problems

Key Takeaways