Question

Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine, CO(g) Cl2(g) <-------> COCl2(g) The value of Kc for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which PCO?

Answer 1

The question is incomplete, here is the complete question:

Phosgene, $COCl_2$, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

The value of $K_c$ for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which $p_{CO}=p_{Cl_2}=0.265atm$ and $p_{COCl_2}=0.000atm$ ?

Answer: The equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

Explanation:

The relation of $K_c\text{ and }K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure

$K_c$ = Equilibrium constant in terms of concentration = 5.79

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side = $n_{g,p}-n_{g,r}=1-2=-1$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = Temperature = 570 K

Putting values in above equation, we get:

$K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124$

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      $CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial:            0.265      0.265

At eqllm:        0.265-x    0.265-x        x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}$

Putting values in above equation, we get:

$0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59$

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = $(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $Cl_2=(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $COCl_2=x=0.008atm$

Hence, the equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.