# Consider the following equilibrium at 388.6 K: NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g). The partial pressure of each gas is 0.203 atm. How do you calculate K_P and K_C for the reaction?

Mar 13, 2017

${K}_{P} = 0.203$
($\text{atm}$)

${K}_{C} = 6.37 \times {10}^{- 3}$
($\text{M}$)

The first thing you can do is write out ${K}_{P}$.

${K}_{P} = \frac{{P}_{N {H}_{3}} {P}_{{H}_{2} S}}{{P}_{N {H}_{4} H S}}$

where ${P}_{i}$ is the partial pressure of gas $i$.

The stoichiometries are all 1:1:1, so all the exponents are $1$. Thus, as all partial pressures are given, we have a very simple equilibrium expression to evaluate:

$\textcolor{b l u e}{{K}_{P}} = \left(\left(\text{0.203 atm")("0.203 atm"))/(("0.203 atm}\right)\right)$

$=$ $\textcolor{b l u e}{\text{0.203}}$
$\text{ }$($\text{atm}$)

Converting to ${K}_{C}$, we assume all the gases dealt with are ideal gases, so that we can use the ideal gas law:

$P V = n R T$

Given that ${n}_{i} / {V}_{i} = \left[i\right]$ for a given gas $i$, we can substitute all the pressure terms in ${K}_{P}$ with concentration terms as follows. For a given gas reaction

$a A + b B \to c C + \mathrm{dD}$,

we would have written

${K}_{P} = \frac{{P}_{C}^{c} {P}_{D}^{d}}{{P}_{A}^{a} {P}_{B}^{b}}$.

Thus, by substituting

${P}_{i}^{{n}_{i}} = {\left(\frac{{n}_{i} R T}{V} _ i\right)}^{{n}_{i}} = {\left(\left[i\right] R T\right)}^{{n}_{i}}$,

we then have:

${K}_{P} = \frac{{\left(\left[C\right] R T\right)}^{c} {\left(\left[D\right] R T\right)}^{d}}{{\left(\left[A\right] R T\right)}^{a} {\left(\left[B\right] R T\right)}^{b}}$

$= \stackrel{{K}_{C}}{\overbrace{\frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}}} \frac{{\left(R T\right)}^{c} {\left(R T\right)}^{d}}{{\left(R T\right)}^{a} {\left(R T\right)}^{b}}$

At this point we've separated out the very definition of ${K}_{C}$. Now, just use the properties of exponents to condense the $R T$ terms together.

$= {K}_{C} \frac{{\left(R T\right)}^{c + d}}{{\left(R T\right)}^{a + b}}$

$= {K}_{C} {\left(R T\right)}^{\left(c + d\right) - \left(a + b\right)}$

But since the stoichiometries of a reaction can be related through the moles of reactants and products, we might as well call the exponent

$\Delta {n}_{g a s} = \left({n}_{c} + {n}_{d}\right) - \left({n}_{a} + {n}_{b}\right)$,

where $\Delta {n}_{g a s}$ is the mols of product gases minus the mols of reactant gases.

Therefore, to convert from ${K}_{P}$ to ${K}_{C}$, we simply have:

$\textcolor{g r e e n}{{K}_{P} = {K}_{C} {\left(R T\right)}^{\Delta {n}_{g a s}}}$

For this expression, be sure to use $\boldsymbol{R = \text{0.08206 L"cdot"atm/mol"cdot"K}}$. To get ${K}_{C}$ then, we simply have:

$\textcolor{b l u e}{{K}_{C}} = \frac{{K}_{P}}{{\left(R T\right)}^{\Delta {n}_{g a s}}}$

= (0.203 cancel"atm")/[("0.08206 L"cdotcancel"atm""/mol"cdotcancel"K")(388.6 cancel"K")]^((1 + 1) - (1))

$= \textcolor{b l u e}{6.37 \times {10}^{- 3}}$

Technically, the units are ("mol"/"L")^((1+1) - (1)) = "mol"/"L", but generally ${K}_{C}$ is reported without units.