Calculating Equilibrium Constant (ICE Table)

This is part of the HSC Chemistry course under the topic Calculating Equilibrium Constant.

HSC Chemistry Syllabus

Deduce the equilibrium expression (in terms of K_eq) for homogeneous reactions occurring in solution.

Perform calculations to find the value of K_eq and concentrations of substances within an equilibrium system, and use these values to make predictions on the direction in which a reaction may proceed.

Qualitatively analyse the effect of temperature on the value of K_eq

Conduct an investigation to determine K_eq of a chemical equilibrium system, for example:

– K_eq of the iron (III) thiocyanate equilibrium

How to Calculate Equilibrium Constant

In this video, we will be exploring how to calculate the equilibrium constant and look at what the formula for providing the equilibrium expression is.

What is Equilibrium Constant?

The equilibrium constant (K_eq) represents the nature of a reaction at dynamic equilibrium. The value describes the relative quantity or concentration of reactants and products when the forward and backward reaction rates equal. In other words, it describes the relative quantity of reactant and product at equilibrium.

For a general equation involving reactants A, B and products C, D in aqueous form:

$$aA + bB \rightleftharpoons cC + dD$$

The equilibrium constant for this generic equation is:

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

Note:

K_eq must always be positive (K_eq > 0).
A large K_eqmeans there are more products present compared to reactants at equilibrium. This does not mean the forward reaction rate is greater than the backward reaction rate because the two are equal at equilibrium.

A small K_eq means there are more reactants present compared to products at chemical equilibrium.
K_eqremains constant at constant temperature. Therefore, changes in concentration, pressure and addition of a catalyst do not change the value of K_eq if temperature remains unchanged.

Only concentration/partial pressure of chemical species in aqueous and gas forms can be written in the equilibrium expression. This is because solids and pure liquids have constant and non-variable concentrations in a chemical reaction. For a solid, the concentration is essentially its density divided by molar mass — and that doesn’t change as long as some of the solid is present. For a pure liquid, like water, it’s the same idea: its molarity in a reaction is constant at a given temperature.

Manipulating Equilibrium Constant Expression

The expression and value of equilibrium constant change when the direction or stoichiometry of the corresponding chemical equation changes.

For example, the dynamic equilibrium involving iron (III) thiocyanate’s equilibrium:

$${Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)} \rightleftharpoons \; {FeSCN^{2+}}_{(aq)}$$

$$K_{eq}=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^–]}$$

Equilibrium constant of a reverse reaction is the reciprocal of that of the forward reaction:

$${FeSCN^{2+}}_{(aq)} \rightleftharpoons {Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)}$$

$$K_{eq(reverse)}=\frac{[Fe^{3+}][SCN^–]}{[FeSCN^{2+}]}=\frac{1}{K_{eq(forward)}}$$

Equilibrium constant changes when the stoichiometric ratio of a reaction changes. For example:

$${2Fe^{3+}}_{(aq)} + \; {2SCN^-}_{(aq)} \rightleftharpoons \; {2FeSCN^{2+}}_{(aq)}$$

$$K_{eq(new)}=\frac{[FeSCN^{2+}]^2}{[Fe^{3+}]^2[SCN^–]^2}={(K_{eq})}^2$$

Equilibrium Constant Calculation

Equilibrium constant calculation questions can be approached using the ‘ICE table’ method.

I: initial concentration

C: change concentration

E: equilibrium concentration

For example, for the reversible reaction between dinitrogen tetroxide and nitrogen dioxide, the ICE table is as follows:

$$N_2O_4(g) \rightleftharpoons 2NO_2(g)$$

There are two main categories of equilibrium constant calculations:

Calculation of concentrations when given the value of equilibrium constant
Calculation of equilibrium constant when given the concentrations of reactant(s) and/or product(s)

Example 1

A 0.0375 mole sample of dinitrogen tetroxide is allowed to come to equilibrium with nitrogen dioxide in a 1.0 L flask at constant temperature. This reaction is represented by the following equation.

$$N_2O_4(g) \rightleftharpoons 2NO_2(g) \hspace{1cm} K_{eq} = 4.6 \times 10^{-5}$$

Calculate the concentration of nitrogen dioxide at equilibrium.

Sample answer:

Set up an ICE table:

	`[N_2O_4(g)]`	`[NO_2(g)]`
Initial	`0.0375/1.0 = 0.0375 M`	0
Change	`-x`	`+2x`
Equilibrium	`0.0375 - x`	`2x`

Equilibrium constant expression:

$$K_{eq} = \frac{[NO_2]^2}{[N_2O_4]}$$

$$4.6 \times 10^{-5} = \frac{(2x)^2}{0.0375 - x}$$

Here, x can be solved by using the quadratic equation. However, we can apply an assumption to simplify the the algebra. Since the value of the equilibrium constant is small, it is valid to assume the change in concentration, x, is negligible compared to the initial concentration of 0.0375.

Therefore, we can simplify the equation to:

$$4.6 \times 10^{-5} = \frac{4x^2}{0.0375}$$

$$x = \sqrt{\frac{(4.6 \times 10^{-5})(0.0375)}{4}}$$

$$x = 6.6 \times 10^{-4}$$

$$[NO_2] = 2x = 2 \times 6.6 \times 10^{-4} = 1.3 \times 10^{-3} \text{ mol/L (2 s.f.)}$$

Example 2

A mixture of 5.00 mole H₂ and 10.0 mole I₂ are placed in a 5.00 L container at 450ºC and allowed to come to equilibrium. At equilibrium, the concentration of HI is 1.87 mol/L.

$$H_2(g)+I_2(g) \rightleftharpoons 2HI(g)$$

(a) Calculate the value for the equilibrium constant, K_eq, for this reaction under these conditions.

(b) What does the equilibrium constant inform us about the relative concentration of reactants and products at equilibrium?

$$\frac{1}{2}H_2(g)+\frac{1}{2}I_2(g) \rightleftharpoons HI(g)$$

Sample answers:

Part (a):

Set up an ICE table:

	`[H_2(g)]`	`[I_2(g)]`	`[HI(g)]`
Initial	`5.00/5.00=1.00 M`	`10.0/5.00=2.00 M`	0
Change	`-x`	`-x`	`+2x`
Equilibrium	`1.00-x`	`2.00-x`	`1.87 M`

$$0+2x=1.87 \text{ M}$$

$$x=0.935 \text{ M}$$

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

$$K_{eq}=\frac{{[HI]}^2}{[H_{2}][I_{2}]}$$

$$K_{eq}=\frac{{(1.87)}^2}{(1.0-0.935)(2.0-0.935)}$$

$$K_{eq}=50.5 \text{ (3 s.f.)}$$

Part (b):

The equilibrium constant is large (50.5) which means there are much more products than reactants at equilibrium.

Part (c):

Given the slightly different chemical equation, we should derive the new expression for equilibrium constant:

$$K_{eq} = \frac{[HI]}{[H_2]^{\frac{1}{2}}[I_2]^{\frac{1}{2}}}$$

Next, manipulate the new expression in terms of the old expression for the original equation:

$$K_{eq} = {(\frac{[HI]^2}{[H_2][I_2]})}^{\frac{1}{2}}$$

$$K_{eq} = (50.5)^{\frac{1}{2}} = 7.11 \text{ (3 s.f.)}$$

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Next section: Calculating Reaction Quotient, Effect of Temperature on Equilibrium