# M5-S6: Calculating Equilibrium Constant

• Deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring in solution

• Perform calculations to find the value of Keq 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 Keq

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

- Keq of the iron (III) thiocyanate equilibrium

### Equilibrium Constant

• The equilibrium constant (Keq) 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.
• For a general equation involving reactants A, B and products C, D in aqueous form:

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

• The equilibrium constant is:

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

• Keq must always be positive (Keq > 0).
• A large Keq means 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 Keq means there are more reactants present compared to products at chemical equilibrium.
• Only concentration/partial pressure of chemical species in aqueous and gas forms can be written in the equilibrium expression.
• For example, the dynamic equilibrium involving iron (III) thiocyanate’s equilibrium:

$${Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)} \rightleftharpoons \; {FeSCN^{2+}}_{(aq)} \;\;\; \Delta{H}<0$$

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

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

$${FeSCN^{2+}}_{(aq)} \rightleftharpoons {Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)} \;\;\; \Delta{H}>0$$

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

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

$${2Fe^{3+}}_{(aq)} + \; {2SCN^-}_{(aq)} \rightleftharpoons \; {2FeSCN^{2+}}_{(aq)} \;\;\; \Delta{H}<0$$

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

### Equilibrium Quotient (Q)

For chemical reactions that are not yet at equilibrium, the concentration/partial pressure of reactants and products can be expressed in the form of an equilibrium quotient (Q).

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

• If Q < Keq, the forward reaction rate of the chemical system will be greater than the reverse reaction rate. This is so that the concentration of product(s) increases while that of reactant(s) decreases. Eventually, Q will increase in value and become Keq.
• If Q > Keq, the reverse reaction rate of the chemical system will be greater than the forward reaction rate. This is so that the concentration of reactant(s) increases while that of product(s) decreases. Eventually, Q will decrease in value and become Keq.

Therefore, a chemical system will proceed in such a way that quotient Q will eventually equal to the equilibrium constant Keq.

### Equilibrium Calculation

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

I: initial reaction concentration

C: change in reaction concentration

E: equilibrium reaction concentration.

Example 1

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

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

$$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$$

 [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

$$0+2x=1.87 M$$

$$x=0.935 M$$

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

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

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

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

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

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

(c) Using your answer in part (a), calculate the equilibrium constant for the reaction:

$$\frac{1}{2}H_{2(g)}+\frac{1}{2}I_{2(g)} \rightleftharpoons HI_{(g)}$$

### Effect of Temperature on Equilibrium Constant

• Temperature is the only factor (change in condition) that affects the equilibrium constant of a reaction.

Endothermic reactions

• An increase in temperature increases the equilibrium constant of an endothermic reaction. Vice versa.

Consider a generic reversible reaction at equilibrium:

$$A_{(g)}+B_{(g)}\rightleftharpoons AB_{(g)} \;\;\; \Delta{H} >0$$

The equilibrium expression of this reaction is:

$$K_{eq}=\frac{[AB]}{[A][B]}$$

An increase in temperature favours the forward reaction because it has a higher activation energy compared to the reverse reaction. So, reactants A and B will be consumed to form AB.

 [A_{(g)}] [B_{(g)}] [AB_{(g)}] Initial [A] [B] [AB] Change -x -x +x Equilibrium [A]-x [B]-x [AB]+x

After heating, the new equilibrium constant becomes:

$$K_{eq}=\frac{[AB]+x}{([A]-x)([B]-x)}$$

As the numerator increases and denominator decreases, the equilibrium constant increases in value.

Exothermic reactions

• An increase in temperature decreases the equilibrium constant of an exothermic reaction. Vice versa.

Consider a generic reversible reaction at equilibrium:

$$A_{(g)}+B_{(g)}\rightleftharpoons AB_{(g)} \;\;\; \Delta{H} <0$$

The equilibrium expression of this reaction is:

$$K_{eq}=\frac{[AB]}{[A][B]}$$

An increase in temperature favours the reverse reaction because it has a higher activation energy compared to the forward reaction. So, product AB will be consumed to form A and B.

 [A_{(g)}] [B_{(g)}] [AB_{(g)}] Initial [A] [B] [AB] Change +x +x -x Equilibrium [A]+x [B]+x [AB-x]

After heating. the new equilibrium constant becomes:

$$K_{eq}=\frac{[AB]-x}{([A]+x)([B]+x)}$$

As the numerator decreases and denominator increases, the equilibrium constant decreases in value.