# 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.

##### 1 Answer
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))

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