Manipulating the Equilibrium Constant

An equilibrium constant applies to a specific reaction with specific stoichiometric coefficients at a particular temperature
Equilibrium constants and expressions can be manipulated algebraically to give new constants and expressions that can be applied to new reactions
All of the algebraic manipulations discussed below can be applied to K_c and K_P as well as Q
Consider the generic equation:

2A (g) + B (g) ⇌ C (g)

The equilibrium expression for this reaction is:

$K_{c} = \frac{[C]}{{[A]}^{2} [B]}$

If the reaction is reversed to:

C (g) ⇌ 2A (g) + B (g)

Then the equilibrium expression becomes

$new K_{c} = \frac{{[A]}^{2} [B]}{[C]}$

So, when a reaction is reversed the new K is the inverse of the original

$new K_{c} = \frac{1}{Original K_{c}}$

If the stoichiometric coefficients in the reaction are multiplied by a factor, c:

2cA (g) + cB (g) ⇌ cC (g)

The new K is equal to the original raised to the power of c

$new K_{c} = \frac{{[C]}^{c}}{{[A]}^{2 c} {[B]}^{c}}$ $new K_{c} = {(original K_{c})}^{c}$

When two or more chemical equations are combined, the new K_c is the product of the K_c’s of all of the summed equations

$new K_{c} = original K_{c 1} \times original K_{c 2}$

Worked example

2BrCl (g) ⇌ Br₂ (g) + Cl₂ (g) K_{c(500 K)}= 32

Based on the information above, determine the value of K_c for the following reactions at 500 K.

Br₂ (g) + Cl₂ (g) ⇌ 2BrCl (g)
6BrCl (g) ⇌ 3Br₂ (g) + 3Cl₂ (g)
½Br₂ (g) + ½Cl₂ (g) ⇌ BrCl (g)

Answers:

Answer 1:

The new equation is the original equation in reverse
The new K_c will be the inverse of the original

$new K_{c} = \frac{1}{original K_{c}}$

$new K_{c} = \frac{1}{32}$

$new K_{c} = 0.31$

Answer 2:

The stoichiometric coefficients in the new equation are three times larger than those in the original
The new K_c will be equal to the original K_c raised to the power of 3

$new K_{c} = {(original K_{c})}^{3}$

$new K_{c} = {(32)}^{3}$

$new K_{c} = 3.3 \times 10^{5}$

Answer 3:

The new equation is the original equation in reverse but the stoichiometric coefficients in the new equation are one-half those in the original
The new K_c will be the inverse of the original K_c raise to the power of ½

$new K_{c} = \frac{1}{{(original K_{c})}^{\frac{1}{2}}}$

$new K_{c} = \frac{1}{{(32)}^{\frac{1}{2}}}$

$new K_{c} = 0.18$

Worked example

Chemical reaction	K_P at 1300 K
C₂H₄ (g) ⇌ C₂H₂ (g) + H₂ (g)	0.333
3C₂H₄ (g) ⇌ C₆H₁₂ (g)	13.3

Based on the information above, calculate the equilibrium constant, K_P, at 1300 K for the reaction below.

C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

Answer:

Step 1: Manipulating the first equation

The desired equation contains three moles of C₂H₂ and three moles of H₂ as products
To achieve this, the stoichiometric coefficients in the first equation must be multiplied by three

3C₂H₄ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

The new K_c for this equation will be equal to the original K_c raised to the power of 3
- $new K_{c} = {(origianl K_{c})}^{3}$
- $new K_{c} = {(0.333)}^{3}$
- $new K_{c} = 3.683 \times 10^{2}$

Step 2: Manipulating the second equation

The desired equation contains one mole of C₆H₁₂ as a reactant
To achieve this, the second equation must be reversed

C₆H₁₂ (g) ⇌ 3C₂H₄ (g)

The new K_c will be the inverse of the original
- $new K_{c} = \frac{1}{original K_{c}}$
- $new K_{c} = \frac{1}{13.3}$
- $new K_{c} = 7.519 \times 10^{- 2}$

Step 3: Combining the equations

Combining the new equations gives the desired equation

3C₂H₄ (g) + C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g) + 3C₂H₄ (g)

C₆H₁₂ (g) ⇌ 3C₂H₂ (g) + 3H₂ (g)

The K_c of the desired equation is the product of the K_c’s of the summed equations
- $desired K_{c} = new K_{c 1} \times new K_{c 2}$
- $desired K_{c} = 3.693 \times 10^{2} \times 7.519 \times 10^{- 2}$
- $desired K_{c} = 2.78 \times 10^{- 3}$

Exam Tip

Be careful not to confuse the rules of manipulating equilibrium constants with those of changes in enthalpy. When the stoichiometric coefficients of a reaction are multiplied by a factor, c, the change in enthalpy of the reaction is multiplied by c but the equilibrium constant is raised to the power of c. This is shown in the table below.

Chemical equation	ΔH° (kJ/mol)	*K_c* at 375°C
½N₂ (g) + $\frac{3}{2}$ H₂ (g) ⇌ NH₃ (g)	-46	1.20
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)	2 x -46 = -92	(1.20)² = 1.44

Manipulating the Equilibrium Constant (College Board AP Chemistry)

Revision Note