Equilibrium constant calculations

In a nutshell

The equilibrium constant, $K_c$ , is the ratio of concentrations of the reactants and products. Unknown concentrations can be calculated when a known value of $K_c$ is given. The units of $K_c$ varies depending on the reaction.

Equations

word equation	symbol equation
$Equilibrium\>constant=\dfrac{(product\>concentration)^{product\>mole\>ratio}}{(reactant\>concentration)^{reactant\>mole\>ratio}}$	$K_c=\dfrac{[Product]^n}{[Reactant]^m}$
$Equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration$	$[]_{eqn}=[]_{Initial}\>-\>[]_{Reacted}$

Variable units

quantity name	symbol	unit name	unit
$concentration$	$c$	$mole\>per\>decimetre\>cubed$	$mol\>dm^{-3}$
$equilibrium\>constant$	$K_c$	$dependent\>on\>reaction$	$dependent\>on\>reaction$

K_cand concentration

The equilibrium constant, $K_c$ , is the ratio of concentrations of the reactants and products. The equilibrium constant can be calculated with known equilibrium concentrations. Sometimes, the concentration needs to be calculated with a known $K_c$ value and other known equilibrium concentrations.

When given a known value of $K_c$ , you can use the equation to solve for concentration.

$Equilibrium\>constant=\dfrac{(product\>concentration)^{product\>mole\>ratio}}{(reactant\>concentration)^{reactant\>mole\>ratio}} \\ \ \\ K_c=\dfrac{[Product]^n}{[Reactant]^m}$

procedure

1.	Write the $K_c$ expression.
2.	Insert the values you are given.
3.	Rearrange and solve for the unknown concentration.

Example

At $300\>K$ , a reaction between propanoic acid and ethanol reached equilibrium. The equilibrium mixture had $3.3\>mol\>dm^{-3}$ of propanoic acid and $4.1\>mol\>dm^{-3}$ of ethanol. The $K_c$ value is $0.5$ . What is the concentration of propyl ethanoate?

$C_3H_7OH(aq) + CH_3CO_2H(aq) \rightleftharpoons CH_3CO_2C_3H_7(aq) + H_2O(l)$

Write the $K_c$ expression:

$K_c=\dfrac{[Product]^n}{[Reactant]^m}$ 

$K_c=\dfrac{[CH_3CO_2C_3H_7][H_2O]}{[CH_3CO_2H][C_3H_7OH]}$

Insert the known values:

$5.0=\dfrac{[CH_3CO_2C_3H_7]}{(3.3)(4.1)}$

Rearrange for propyl ethanoate concentration:

$[CH_3CO_2C_3H_7]=(0.5)\times (3.3\>mol\>dm^{-3})\times (4.1\>mol\>dm^{-3})=6.765\>mol\>dm^{-3}$

The concentrations for propyl ethanoate and water are $\underline{6.765\>mol\>dm^{-3}}$ .

Reaction equation and concentration

The equilibrium concentrations can be calculated when known initial concentrations are given. Then $K_c$ can be calculated with these equilibrium concentrations.

procedure

1.	Work out the initial concentrations.
2.	Work out the equilibrium concentrations using: $Equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration$
3.	Write the $K_c$ expression.
4.	Insert known values to calculate $K_c$ .

Example

$600\>cm^3$ of $0.30\>mol\>dm^{-3}$ iron sulfate solution and $600\>cm^3$ of $0.30\>mol\>dm^{-3}$ silver nitrate solution were left together in a flask at $300\>K$ until equilibrium was reached. A sample was titrated and it was found the equilibrium concentration of $Fe^{2+}$ was $0.0421\>mol\>dm^{-3}$ . Calculate the value of $K_c$ . Give your answer to three significant figures.

$Fe^{2+}(aq)+Ag^+(aq) \rightleftharpoons Fe^{3+}(aq)+Ag(s)$

Ignore the silver solid as this isn't included in the $K_c$ expression.

Work out the initial concentration for $Fe^{2+}$ and $Ag^+$ . You half the concentration because the reactants have the same number of moles but in double the volume since they're mixed together:

$Initial\>concentration\>of\>Fe^{2+}\>and\>Ag^+\>=\dfrac{0.30\>mol\>dm^{-3}}{2}=0.15\>mol\>dm^{-3}$

Write the concentrations of $Fe^{2+}$ and $Ag^+$ at equilibrium:

 $[Fe^{2+}]=[Ag^+]=0.0421\>mol\>dm^{-3}$

Calculate the concentrations of $Fe^{3+}$ at equilibrium:

$Equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration$ 

$Fe^{3+}\>equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration \newline Fe^{3+}\>equilibrium\>concentration=0.15\>mol\>dm^{-3}\>-0.0421\>mol\>dm^{-3}=0.1079\>mol\>dm^{-3}$

Write the $K_c$ expression:

$K_c=\dfrac{[Product]^n}{[Reactant]^m}$ 

$K_c=\dfrac{[Fe^{3+}]}{[Fe^{2+}][Ag^+]}$

Insert the equilibrium concentrations to calculate $K_c$ :

$K_c=\dfrac{[0.1079\>mol\>dm^{-3}]}{[0.0421\>mol\>dm^{-3}][0.0421\>mol\>dm^{-3}]}=60.87756\>mol^{-1}\>dm^{3}$

Round to three significant figures:

$K_c=60.9\>mol^{-1}\>dm^{3}$

The value of $K_c$ is $\underline{60.9\>mol^{-1}\>dm^{3}}$ .

Unknown reactant equilibrium concentration

The equilibrium concentrations of reactants can be found if the equilibrium concentration of products and $K_c$ are given.

procedure

1.	Write the known initial and equilibrium concentrations.
2.	Work out the equilibrium concentrations using: $Equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration$
3.	Write the $K_c$ expression.
4.	Rearrange and solve for the unknown concentration for the reactant.

Example

$0.210\>mol\>dm^{-3}$ $Fe^{2+}$ ions are mixed with $0.410\>mol\>dm^{-3}$ $HBr$ and formed the equilibrium shown below. At equilibrium, the concentration of $[FeBr_4]^{2-}$ is $0.0511\>mol\>dm^{-3}$ . Calculate the value of $K_c$ . Round your answer to three significant figures.

$Fe^{2+}(aq)+4Br^- \rightleftharpoons [FeBr_4]^{2-}(aq)$

Write the known initial and equilibrium concentrations:

	$Fe^{2+}$	$Br^-$	$[FeBr_4]^{2-}$
$Initial\>concentration\>(mol\>dm^{-3})$	$0.210$	$0.410$	$n/a$
$Equilibrium\>concentration\>(mol\>dm^{-3})$			$0.0511$

The chemical equation shows that for every mole of $[FeBr_4]^{2-}$ formed, one mole of $Fe^{2+}$ reacted. For every mole of $[FeBr_4]^{2-}$ formed, four moles of $Br^-$ reacted. Work out the equilibrium concentrations:

$Equilibrium\>concentration=Initial\>concentration\>-\>Reacted\>concentration$ 

$[Fe^{2+}]=Initial\>concentration\>-\>Reacted\>concentration \newline [Fe^{2+}]=0.210\>mol\>dm^{-3}-0.0511\>mol\>dm^{-3}=0.1589\>mol\>dm^{-3}$

$[Br^-]=Initial\>concentration\>-\>Reacted\>concentration \newline [Br^-]=0.410\>mol\>dm^{-3}-(4\times 0.0511\>mol\>dm^{-3})=0.2056\>mol\>dm^{-3}$ 

Write the $K_c$ expression:

$K_c=\dfrac{[Product]^n}{[Reactant]^m}$ 

$K_c=\dfrac{[[FeBr_4] ^{2-}]}{[Fe^{2+}][Br^-]^4}$

Insert equilibrium concentrations to calculate $K_c$ :

$K_c=\dfrac{(0.0511\>mol\>dm^{-3})}{(0.1589\>mol\>dm^{-3})(0.2056\>mol\>dm^{-3})^4}=179.971842\>mol^{-4}\>dm^{12}$

Round to three significant figures:

$K_c=180\>mol^{-4}\>dm^{12}$

The value of $K_c$ is $\underline{180\>mol^{-4}\>dm^{12}}$ .