Some reactions can be represented as a sequence of two or more relations. For example, let’s consider the following reaction between NOBr and Cl_{2} gases:

2NOBr(*g*) + Cl_{2}(*g*) ⇆ 2NO(*g*) + 2 BrCl(*g*)

We can represent this net transformation as a sum of two reactions:

1) 2NOBr(*g*) ⇆ 2NO(*g*) + Br_{2}(*g*)

2) Br_{2}(*g*) + Cl_{2}(*g*) ⇆ 2BrCl(*g*)

To confirm that these two give the target equation, simply add them up and cancel out the identical components on both sides of the equation:

1) 2NOBr(*g*) ⇆ 2NO(*g*) + Br_{2}(*g*)

+

2) Br_{2}(*g*) + Cl_{2}(*g*) ⇆ 2BrCl(*g*)

2NOBr(*g*) + Br_{2}(*g*) + Cl_{2}(*g*) ⇆ 2NO(*g*) + Br_{2}(*g*) + 2 BrCl(*g*)

_______________________________________________

3) 2NOBr(*g*) + Cl_{2}(*g*) ⇆ 2NO(*g*) + 2 BrCl(*g*)

So, Br_{2} appeared on both sides of the equation and by canceling it, we obtained the final equation (3).

**Now, how is this related to the equilibrium constant?**

It turns out that we can **calculate the equilibrium constant** for a reaction if we know the equilibrium constants for **other reactions** that add up to this reaction. For this, you simply need to multiply the equilibrium constants of the two reactions.

**For example**, suppose the equilibrium constants of reactions (1) and (2) are 0.012 and 7.2 respectively:

1) 2NOBr(*g*) ⇆ 2NO(*g*) + Br_{2}(*g*)

2) Br_{2}(*g*) + Cl_{2}(*g*) ⇆ 2BrCl(*g*)

To obtain the equilibrium constant for reaction (3), we multiply the ones for reaction (1) and (2):

2NOBr(*g*) + Cl_{2}(*g*) ⇆ 2NO(*g*) + 2BrCl(*g*) (3)

Notice that we got an expression for *K*_{3} that matches what we’d write according to the chemical equation (3).

Therefore, the value of *K*_{3} is the product of *K*_{1} and *K*_{2}:

*K*_{3} = *K*_{1} x *K*_{2} = 0.012 x 7.3 = 0.088

If this is confusing, you can review the Hess’s law which uses a similar strategy to calculate the enthalpy of a reaction based on the enthalpy of other reactions that add up to it.

#### Practice

The equilibrium constant values for the reactions below were determined at a certain temperature:

S(*s*) + O_{2}(*g*) ⇆ SO_{2}(*g*) *K _{a}* = 3.2 x 10

^{45}

2S(*s*) + 3O_{2}(*g*) ⇆ 2SO_{3}(*g*) *K _{b}* = 3.2 x 10

^{124}

Using these data, determine the equilibrium constant *K*c for the following reaction:

2SO_{2}(*g*) + O_{2}(*g*) ⇆ 2SO_{3}(*g*) * K_{c}*

**= ?**

*K _{c}* = 3.1 x 10

^{33}

There are two SO_{2} reactant molecules in the target reaction, so reverse the first reaction and multiply by 2:

S(*s*) + O_{2}(*g*) ⇆ SO_{2}(*g*) ⇒ 2SO_{2}(*g*) ⇆ 2S(*s*) + 2O_{2}(*g*)

The equilibrium constant for this reaction is:

*K ^{r}_{a}* = (1/k

_{a})

^{2}= (1/3.2 x 10

^{45})

^{2 }= 9.77 x 10

^{-92}

Now, add the two reactions to make sure it matches the target equation:

2S(*s*) + 3O_{2}(*g*) ⇆ 2SO_{3}(*g*)

2SO_{2}(*g*) ⇆ 2S(*s*) + 2O_{2}(*g*)

__________________

2S(*s*) + 3O_{2}(*g*) + 2SO_{2}(*g*) ⇆ 2SO_{3}(*g*) + 2S(*s*) + 2O_{2}(*g*)

O_{2}(*g*) + 2SO_{2}(*g*) ⇆ 2SO_{3}(*g*)

The equations add up, so the last part is to calculate the equilibrium constant.

*K _{c}* =

*K*9.77 x 10

^{r}_{a }x K_{b }=^{-92 }x 3.2 x 10

^{124 }= 3.1 x 10

^{33}

The equilibrium constant values for the reactions below were determined at a certain temperature:

2NO(g) ⇆ N_{2}(g) + O_{2}(g) *K*_{a} = 3.8 x 10^{32}

NO(*g*) + 1/2Cl_{2}(*g*) ⇆ NOCl(*g*) *K*_{b} = 6.4

Using these data, determine the equilibrium constant *K*c for the following reaction:

1/2N_{2}(g) + 1/2O_{2}(g) + 1/2Cl_{2}(*g*) ⇆ NOCl(*g*) *Kc* = ?

*K _{c}* = 3.28 x 10

^{-16}

There are 1/2 moles of N_{2} reactant molecules in the target reaction, so reverse the first reaction and multiply by 1/2:

2NO(g) ⇆ N_{2}(g) + O_{2}(g) ⇒ 1/2N_{2}(g) + 1/2O_{2}(g) ⇆ NO(g)

The equilibrium constant for this reaction is:

*K ^{r}_{a}* = (1/k

_{a})

^{1/2}= (1/3.8 x 10

^{32})

^{1/2}

^{ }= 5.13 x 10

^{-17}

Now, add the two reactions to make sure it matches the target equation:

1/2N_{2}(g) + 1/2O_{2}(g) ⇆ NO(g)

NO(*g*) + 1/2Cl_{2}(*g*) ⇆ NOCl(*g*)

_____________________

1/2N_{2}(g) + 1/2O_{2}(g) + NO(*g*) + 1/2Cl_{2}(*g*) ⇆ NO(g) + NOCl(*g*)

1/2N_{2}(g) + 1/2O_{2}(g) + 1/2Cl_{2}(*g*) ⇆ NOCl(*g*)

The equations add up, so the last part is to calculate the equilibrium constant.

*K _{c}* =

*K*5.13 x 10

^{r}_{a }x K_{b }=^{-17}

^{ }x 6.4

^{ }= 3.28 x 10

^{-16}

**Check Also**

- Chemical Equilibrium
- Equilibrium Constant
*K*_{p}and Partial Pressure*K*and_{p }*K*Relationship_{c}*K*Changes with Chemical Equation- Equilibrium Constant K from Two Reactions
- Reaction Quotient –
*Q* - ICE Table – Calculating Equilibrium Concentrations
- ICE Table Practice Problems
- Le Châtelier’s principle
- Le Châtelier’s principle Practice Problems
**Chemical Equilibrium Practice Problems**