IB DP Chemistry: SL复习笔记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 Kexpression 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 Kexpression for this reaction is:

7.1.3-Kc-expression-for-NH3

  • If you double the stoichiometry the equation becomes

N2(g)    +    3H2(g)  ⇌ 2NH3(g)

  • The new Kexpression for this reaction is then:

7.1.3-Kc-expression-for-NH3-doubled

  • 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:

7.1.3-Kc-relationships-squared

  • If we reverse the equation:

2NH3(g) ⇌ N2(g)    +    3H2(g)

  • Kc  becomes the reciprocal of the original Kvalue:

7.1.3-Kc-expression-for-NH3-reversed

  • Test your understanding in the following example:

Worked Example

Kc for 2NO2 (g) + F2 (g) ⇌ 2NO2F (g) is 7.1 × 1032What is Kc for the following reaction, at the same temperature?

NO2F (g) ⇌  NO2 (g) + ½F2 (g)

7.1.3-Worked-Example-Question6

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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