A certain gas mixture is held at $395^{\circ} C$ $395^@ "C"$ has the following initial partial pressures: $P_{C l_{2}} = 351.4$ $P_(Cl_2) = 351.4$ , $P_{C O} = 342.0$ $P_(CO) = 342.0$ , $P_{C O C l_{2}} = 0$ $P_(COCl_2) = 0$ , all in $torr$ $"torr"$ . Find $K_{P}^{\circ}$ $K_P^@$ ?

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A certain gas mixture is held at $395^{\circ} C$ $395^@ "C"$ has the following initial partial pressures: $P_{C l_{2}} = 351.4$ $P_(Cl_2) = 351.4$ , $P_{C O} = 342.0$ $P_(CO) = 342.0$ , $P_{C O C l_{2}} = 0$ $P_(COCl_2) = 0$ , all in $torr$ $"torr"$ . Find $K_{P}^{\circ}$ $K_P^@$ ?

At equilibrium, the total pressure is $439.5 torr$ $"439.5 torr"$ . $V$ $V$ is held constant. Find $K_{P}^{\circ} = \frac{\frac{P_{C O C l_{2}}}{P^{\circ}}}{(\frac{P_{C l_{2}}}{P^{\circ}}) (\frac{P_{C O}}{P^{\circ}})}$ $K_P^@ = (P_(COCl_2)/P^@)/((P_(Cl_2)/P^@)(P_(CO)/P^@))$ for

$CO (g) + {Cl}_{2} (g) ⇌ {COCl}_{2} (g)$ $"CO"(g) + "Cl"_2(g) rightleftharpoons "COCl"_2(g)$

at $395^{\circ} C$ $395^@ "C"$ , where $P^{\circ} = 750.062 torr$ $P^@ = "750.062 torr"$ (and $χ_{A} P_{tot} = \frac{n_{A}}{n_{tot}} P_{tot} = P_{A}$ $chi_AP_"tot" = n_A/(n_"tot")P_"tot" = P_A$ , the partial pressure of $A$ $A$ ).

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.

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Answer 1 · 2016-10-10T09:47:50

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

We got $K_{P}^{\circ} = 22.17$ $K_P^@ = 22.17$ . The back of my book gives $K_{P}^{\circ} = 22.2$ $K_P^@ = 22.2$ .

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

$N \sum i P_{i} = P_{1} + P_{2} + . . . + P_{N} = P_{tot}$ $sum_i^N P_i = P_1 + P_2 + . . . + P_N = P_"tot"$

which we will use for $P_{i, e q}$ $P_(i,eq)$ and $P_{tot, e q}$ $P_("tot",eq)$ .

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

$CO (g) + {Cl}_{2} (g) ⇌ {COCl}_{2} (g)$ $"CO"(g) " "+" " "Cl"_2(g) " "rightleftharpoons" " "COCl"_2(g)$

$I 342.0 351.4 0$ $"I"" "342.0" "" "" "351.4" "" "" "" "" "0$
$C - x - x + x$ $"C"" "-x" "" "" "-x" "" "" "" "" "+x$
$E 342.0 - x 351.4 - x x$ $"E"" "342.0-x" "351.4-x" "" "" "x$

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

$P_{CO} + P_{{Cl}_{2}} + P_{{COCl}_{2}} = P_{tot, e q} = 439.5 torr$ $P_"CO" + P_("Cl"_2) + P_("COCl"_2) = P_("tot",eq) = "439.5 torr"$

$342.0 - x + 351.4 cancel(- x + x) = 439.5$

$693.4 - x = 439.5$

$color(green)(x) = 693.4 - 439.5 = color(green)("253.9 torr")$

From this, we can construct the expression to calculate $K_P^@$ (where $P^@ = "750.062 torr"$ is the standard pressure, which is also equal to $"1 bar"$ ):

$color(blue)(K_P^@) = (P_(COCl_2)/P^@)/((P_(Cl_2)/P^@)(P_(CO)/P^@))$

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

$= color(blue)(22.17)$

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A certain gas mixture is held at $395^{\circ} C$ $395^@ "C"$ has the following initial partial pressures: $P_{C l_{2}} = 351.4$ $P_(Cl_2) = 351.4$ , $P_{C O} = 342.0$ $P_(CO) = 342.0$ , $P_{C O C l_{2}} = 0$ $P_(COCl_2) = 0$ , all in $torr$ $"torr"$ . Find $K_{P}^{\circ}$ $K_P^@$ ?

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A certain gas mixture is held at $395^{\circ} C$ $395^@ "C"$ has the following initial partial pressures: $P_{C l_{2}} = 351.4$ $P_(Cl_2) = 351.4$ , $P_{C O} = 342.0$ $P_(CO) = 342.0$ , $P_{C O C l_{2}} = 0$ $P_(COCl_2) = 0$ , all in $torr$ $"torr"$ . Find $K_{P}^{\circ}$ $K_P^@$ ?