# 7.1.3 Equilibrium Constant Relationships

### Equilibrium constant relationships

•  In the previous section we saw that the concentrations of the substances are raised to the power of the coefficients from the balanced equation
• This means the Kc expression is dependent on a specific equation
• For example, take the reaction between nitrogen and hydrogen to make ammonia

½N2(g)    +    1½H2(g)  ⇌     NH3(g)

• The Kc expression for this reaction is:

• If you double the stoichiometry the equation becomes

N2(g)    +    3H2(g)  ⇌ 2NH3(g)

• The new Kc expression for this reaction is then:

• What is the relationship between these two Kc values? You can probably see that when we double the coefficient the new Kc is the square of the original value:

• If we reverse the equation:

2NH3(g) ⇌ N2(g)    +    3H2(g)

• Kc  becomes the reciprocal of the original Kc value:

• Test your understanding in the following example:

#### Worked Example

Kc for 2NO2 (g) + F2 (g) ⇌ 2NO2F (g) is 7.1 × 1032

What is Kc for the following reaction, at the same temperature?

NO2 (g) + ½F2 (g) ⇌  NO2F (g) Answer

The correct option is B.

The original equation has been reversed and halved, so the Kvalue must be square rooted and inverted to obtain the reciprocal

#### Exam Tip

You must use square brackets in equilibrium constant expressions as they have a specific meaning, representing concentrations. In an exam answer you would lose the mark if you used round brackets

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