# A certain gas mixture is held at 395^@ "C" has the following initial partial pressures: P_(Cl_2) = 351.4, P_(CO) = 342.0, P_(COCl_2) = 0, all in "torr". Find K_P^@?

## At equilibrium, the total pressure is $\text{439.5 torr}$. $V$ is held constant. Find ${K}_{P}^{\circ} = \frac{{P}_{C O C {l}_{2}} / {P}^{\circ}}{\left({P}_{C {l}_{2}} / {P}^{\circ}\right) \left({P}_{C O} / {P}^{\circ}\right)}$ for ${\text{CO"(g) + "Cl"_2(g) rightleftharpoons "COCl}}_{2} \left(g\right)$ at ${395}^{\circ} \text{C}$, where ${P}^{\circ} = \text{750.062 torr}$ (and ${\chi}_{A} {P}_{\text{tot" = n_A/(n_"tot")P_"tot}} = {P}_{A}$, the partial pressure of $A$). My guess is somehow I have to find the partial pressures at equilibrium, but there's not enough obvious information for me to immediately figure this out.

Oct 10, 2016

Since Michael and Stefan managed to help me to figure this out, I'll put an answer here.

We got ${K}_{P}^{\circ} = 22.17$. The back of my book gives ${K}_{P}^{\circ} = 22.2$.

The main idea is that for ideal gases, at fixed $V$ and at the same/fixed $T$ (the actual value doesn't matter), a change in $m o l s$ of gas influences the change in the total pressure, which follows Dalton's Law of partial pressures:

${\sum}_{i}^{N} {P}_{i} = {P}_{1} + {P}_{2} + . . . + {P}_{N} = {P}_{\text{tot}}$

which we will use for ${P}_{i , e q}$ and ${P}_{\text{tot} , e q}$.

First, we can construct an ICE table to determine the expression for each equilibrium partial pressure.

${\text{CO"(g) " "+" " "Cl"_2(g) " "rightleftharpoons" " "COCl}}_{2} \left(g\right)$

$\text{I"" "342.0" "" "" "351.4" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "-x" "" "" "" "" } + x$
$\text{E"" "342.0-x" "351.4-x" "" "" } x$

The equilibrium partial pressures can be expressed as Dalton's Law of partial pressures:

${P}_{\text{CO" + P_("Cl"_2) + P_("COCl"_2) = P_("tot",eq) = "439.5 torr}}$

$342.0 - x + 351.4 \cancel{- x + x} = 439.5$

$693.4 - x = 439.5$

$\textcolor{g r e e n}{x} = 693.4 - 439.5 = \textcolor{g r e e n}{\text{253.9 torr}}$

From this, we can construct the expression to calculate ${K}_{P}^{\circ}$ (where ${P}^{\circ} = \text{750.062 torr}$ is the standard pressure, which is also equal to $\text{1 bar}$):

$\textcolor{b l u e}{{K}_{P}^{\circ}} = \frac{{P}_{C O C {l}_{2}} / {P}^{\circ}}{\left({P}_{C {l}_{2}} / {P}^{\circ}\right) \left({P}_{C O} / {P}^{\circ}\right)}$

= ((253.9)/(750.062))/(((351.4 - 253.9)/(750.062))((342.0 - 253.9)/(750.062))

$= \textcolor{b l u e}{22.17}$