The equilibrium constant  

In a nutshell

The equilibrium constant, $$K_c$$​, can be calculated from the ratio of the concentrations of the reactants and products. The value is given at a certain temperature. The units can vary depending on the equilibrium reaction. 

Equations

This is a general expression for $$K_c$$. You will need to be able to write $$K_c$$ expressions for any given reaction.

​

​$$K_c=\dfrac{[Products]^n}{[Reactants]^m}$$

$$n=Number\>of\>product\>moles$$

$$m=Number\>of\>reactant\>moles$$​​​​

There is a calculation to find the concentration. This involves knowing the volume and moles. 

Variable units    

Equilibrium constant, K_c

The equilibrium constant shows which reaction the equilibrium is closer to. It can be calculated at a given temperature with the known molar concentrations of the reactants and products at equilibrium. 

This definition is usually applied for homogeneous equilibria. All the reactants and products are in the same physical state.

Writing the $$K_c$$ expression:  

​$$K_c=\dfrac{[Products]^n}{[Reactants]^m}$$​​

Procedure

Example 

​$$2 NO(g)+O_2(g) ⇌ 2 NO_2(g)$$

Write the $$K_c$$​ expression for the above reaction.  

​

​$$\underline{K_c = \dfrac{[NO_2]^2}{[NO]^2 [O_2]}}$$

Note: The square brackets, [ ], indicate concentration.   

K_c units 

The units for the equilibrium constant, $$K_c$$​, can vary depending on the reaction. You need to figure out the units every single time you calculate $$K_c$$​. 

Procedure 

Example

​$$2NO(g)+O_2(g) ⇌ 2NO_2(g)$$​

The $$K_c$$​ expression is: 

$$K_c = \dfrac{[NO_2]^2}{[NO] ^2[O_2]}$$​​

Insert the concentration units into the equation:

​$$K_c = \dfrac{[mol\,dm^{-3}]^2}{[mol\,dm^{-3}]^2[mol\,dm^{-3}]}$$​​

The simplified version by canceling out any common units is:

​$$K_c = \dfrac{1}{mol\,dm^{-3}} = mol^{-1}\,dm^3$$

The units for $$K_c$$​ are $$\underline{mol^{-1}\,dm^3}$$. 

Calculating K_c

You can calculate the value of the equilibrium constant, $$K_c$$​, with known equilibrium concentrations. Sometimes you have to work out the equilibrium concentrations first.

procedure 

Example 

$$0.30 \>mole$$ of phosphorous (V) bromide​ reacts at $$550\> K$$​ in a reaction tube of $$6.00 \>dm^3$$. The equilibrium mixture has $$0.080 \>mole$$ of bromine. Calculate $$K_c$$​, including units.​

$$PBr_5 (g) ⇌ PBr_3 (g)+ Br_2(g)$$

The equation is balanced. Note down the moles before equilibrium. The moles before are mentioned in the question: 

​$$PBr_5\>moles\>before\>equilibrium=0.30\>mol$$

Note down the moles after equilibrium. The moles of bromine after equilibrium is mentioned in the question:

$$Br_2\>moles\>after\>equilibrium= 0.080\>mol$$

Calculate the moles after equilibrium for the reactants. Both products, $$Br_2$$​ and $$PBr_3$$​, have $$1\>mole$$​ in the balanced equation. This means they have a $$1:1$$​ ratio. Therefore, the moles produced after equilibrium will be the same:

$$Br_2 \>moles\>after\>equilibrium=PBr_3\>moles\>after\>equilibrium=0.080\>mol$$

Calculate the moles for $$PBr_5$$​ after equilibrium. Since $$0.080\>moles$$ of $$Br_2$$​ was produced, that means $$0.080\>moles$$ of $$PBr_5$$​ reacted:

$$PBr_5 \>moles\>after\>equilibrium=0.30-0.080=0.22\>mol$$

Calculate the concentration of each species. The concentration of each species can be calculated using the moles after equilibrium and the volume given in the question:

​$$[PBr_5]=\dfrac{0.22\>mol}{6\>dm^3} = 0.036 \>mol\>dm^{-3}$$

$$[Br_2] =\dfrac{ 0.080\>mol}{6 \>dm^3} = 0.013\>mol\>dm^{-3}$$

$$[PBr_3] =\dfrac{0.080 \>mol}{ 6\>dm^3} =0.013 \>mol\>dm^{-3}$$

Write down the $$K_c$$​ expression:

$$K_c = \dfrac{[PBr_3][Cl_2]}{[PBr_5]}$$

Insert the concentration values to calculate $$K_c$$:

$$K_c=\dfrac{(\dfrac{ 0.080\>mol}{6 \>dm^3}\>mol\>dm^{-3})\times(\dfrac{ 0.080\>mol}{6 \>dm^3}\>mol\>dm^{-3})}{(\dfrac{0.22\>mol}{6\>dm^3}\>mol\>dm^{-3})} =0.00485\>mol\>dm^{-3}$$

The $$K_c$$​ value is $$\underline{0.00485\>mol\>dm^{-3}}$$ .  

Finding concentration with known K_c

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

Procedure

Experimental data and K_c

The equilibrium constant, $$K_c$$​, can be calculated from experimental data made in the laboratory. An experiment between iron(II) sulfate solution and silver nitrate solution at $$298\>K$$ will reach equilibrium. 

You can titrate it to work out the equilibrium concentrations of the $$Fe^{2+}$$​ ions. Then you'll be able to work out the other equilibrium concentrations and $$K_c$$​.

Example 

$$400\>cm^3$$ of $$0.2\>mol\>dm^{-3}$$​ iron(III) sulfate solution and $$400\>cm^3$$ of $$0.2\>mol\>dm^{-3}$$​ silver nitrate solution were left together in a flask at $$298\> K$$ until equilibrium was reached. A sample was titrated and it was found that the equilibrium concentration for $$Fe^{2+}$$​ was $$0.0411\>mol\>dm^{-3}$$. What is the value of $$K_c$$?

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

The balanced equation is heterogeneous, you can ignore the solid silver. 

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:  

$$Initial\> concentration \>of \>Fe^{2+} and \>Ag^+= \dfrac{0.2\>mol\>dm^{-3}}{2}=0.1\>mol\>dm^{-3}$$

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

$$Fe^{2+}equilibrium\>concentration = \>Ag^+equilibrium\>concentration=0.0411\>mol\>dm^{-3}$$

Calculate the concentration for $$Fe^{3+}$$ at equilibrium:

$$Fe^{3+}equilibrium\>concentration=Initial \>concentration-Reactants\>equilibrium \>concentration$$

$$Fe^{3+}equilibrium\>concentration=0.1-0.0411=0.0589\>mol\>dm^{-3}$$​​​

Insert the equilibrium concentration values into the $$K_c$$ equation: 

$$K_c= \dfrac{[Fe^{3+}]}{[Fe^{2+}][Ag^+]}= \dfrac{0.0589\>mol\>dm^{-3}}{0.0411\>mol\>dm^{-3}\times0.0411\>mol\>dm^{-3}}=34.9\>mol^{-1}dm^3$$

$$Ag(s)$$​ is ignored in the $$K_c$$ expression because it's a solid. The value of $$K_c$$ is $$\underline{34.9\>mol^{-1} \, dm^3}$$.

K_c value 

You can only calculate one $$K_c$$​ value at a given temperature. The $$K_c$$​ value will change if you change the temperature. 

Catalysts don't affect the value of $$K_c$$​.