Question

The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm

Answer 1

Answer: The partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

Explanation:

We are given:

$K_c=4.63\times 10^{-3}$

$p_{COCl_2}=0.760atm$

Relation of $K_p$ with $K_c$ is given by the formula:

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

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $4.63\times 10^{-3}$

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

T = temperature = $527^oC=527+273=800K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304$

The chemical reaction for the decomposition of phosgene follows the equation:

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

At t = 0          0.760            0     0  

At $t=t_{eq}$      0.760-x          x      x

The expression for $K_p$ for the given reaction follows:

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

We are given:

$K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x$

Putting values in above equation, we get:

$0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656$

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

$p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm$

Hence, the partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.