# Question #f5a83

Aug 24, 2017

$\text{0.303 atm}$.

Given the reaction, write the mass action expression using the partial pressures of each gas:

${K}_{P} = \frac{{P}_{{H}_{2} , e q} {P}_{S , e q}}{{P}_{{H}_{2} S , e q}} = 0.833$

All the coefficients were $1$ so there are no exponents to worry about. We know that ${P}_{{H}_{2} S , i} = \text{0.163 atm}$. An ICE table gives:

$\text{H"_2"S"(g) " "rightleftharpoons" " "H"_2(g) + "S} \left(g\right)$

$\text{I"" "0.163" "" "" "" "0" "" "" } 0$
$\text{C"" "-x" "" "" "" "+x" "" } + x$
$\text{E"" "0.163 - x" "" "x" "" "" } x$

We define $x$ as the partial pressure of ${\text{H}}_{2} \left(g\right)$ or $\text{S} \left(g\right)$ at equilibrium, i.e. ${P}_{{H}_{2} , e q} = {P}_{S , e q}$.

This becomes:

$0.833 = {x}^{2} / \left(0.163 - x\right)$

The ${K}_{P}$ is not small, so we'll have to solve this quadratic equation in full.

$0.833 \left(0.163 - x\right) - {x}^{2} = 0$

You should get:

$x \equiv {P}_{{H}_{2} , e q} = {P}_{S , e q} = \text{0.140 atm}$

This says that $\text{H"_2"S}$ decomposes almost completely, as ${P}_{{H}_{2} S , e q} = \text{0.023 atm}$.

$\textcolor{b l u e}{{P}_{\text{tot}}} = {P}_{1} + {P}_{2} + . . .$

$= \left(0.163 - x\right) + \left(x\right) + \left(x\right)$

$= 0.163 + x$

$= \text{0.163 atm" + "0.140 atm}$

$=$ $\textcolor{b l u e}{\text{0.303 atm}}$