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

Variable units

K_c and 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 

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

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

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:  

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 $$\underline{180\>mol^{-4}\>dm^{12}}$$.