# For the reaction NH_4HS_((s)) rightleftharpoons NH_(3(g)) + H_2S_((g)), K_c = 1.2*10^(-4) at 491K. What concentration of NH_3 will be present at equilibrium if a sample of NH_4HS is placed in a vessel at 491K?

May 24, 2015

The equilibrium concentration of ammonia will be equqal to $1.1 \cdot {10}^{- 2} \text{M}$.

Even before doing any calculations, you can look at the magnitude of the equilibrium constant, ${K}_{c}$, and predict what's going to happen when you place the ammonium hydrosulfide in the vessel.

The first important thing to notice is that ammonium hydrosulfide is actually a solid. This means that its concentration is assumed to be constant during the reaction.

This is why the problem doesn't provide an initial concentration for the ammonium hydrosulfide.

Moreover, since ${K}_{c}$ is smaller than 1, the equilibrium will lie to the left, favoring the reactant. This means that you can expect smaller equilibrium concentrations, compared with that of the reactant, which will remain unchanged, for both products.

Use an ICE table to help you determine what the equilibrium concentrations of the two products will be

$\text{ } N {H}_{4} S {H}_{\left(s\right)} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(g\right)} + {H}_{2} {S}_{\left(g\right)}$
I.........$-$........................0...............0
C.......$-$.......................(+x)............(+x)
E.......$-$.........................x................x

The expression of the eequilibrium constant will look like this

${K}_{c} = \left[N {H}_{3}\right] \cdot \left[{H}_{2} S\right] = x \cdot x = {x}^{2}$

This means that $x$ will be equal to

$x = \sqrt{{K}_{c}} = \sqrt{1.2 \cdot {10}^{- 4}} = 0.01095$

Rounded to two sig figs, the equilibrium concentration of ammonia will thus be

$\left[N {H}_{3}\right] = \textcolor{g r e e n}{1.1 \cdot {10}^{- 2} \text{M}}$