# Suppose "2.83 g" of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is K_c for this reaction?

Nov 4, 2016

The equilibrium constant, ${K}_{c} = 9.05 \times {10}^{-} 5$ is calculated from the equilibrium concentrations of gases. If more $N {H}_{4} C l \left(s\right)$ is added, nothing happens because the gases are already at their equilibrium concentrations.

#### Explanation:

The first step in any equilibrium problem like this is to write the complete balanced equation:

$N {H}_{4} C l \left(s\right) \leftrightarrow N {H}_{3} \left(g\right) + H C l \left(g\right)$

The form of ${K}_{c}$ for this reaction (the subscript c means concentration, and the standard concentration is in units of $\text{mol/L}$) is:

${K}_{c} = \left[N {H}_{3}\right] \left[H C l\right]$

The $N {H}_{4} C l$ does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.

The starting amount of reactant is $\frac{2.83 g}{59.49 \frac{g}{m o l}} = 0.0476 m o l$

Because 40% of the reactant decomposed, the equilibrium concentrations are:
${\left[N {H}_{3}\right]}_{e} = {\left[H C l\right]}_{e} = \frac{0.0476 m o l \times 0.40}{2 L} = 0.00951 M$

The value of ${K}_{c}$ is therefore ${K}_{c} = {\left(0.00951\right)}^{2} = 9.05 \times {10}^{-} 5$