Question

A 1.5 dm3
flask containing 0.03 mol of pure phosgene gas, COCl2 was heated
at 8000C. At equilibrium, the partial pressure of CO was found to be 0.5 atm.
a. What is the equilibrium constant Kp and Kc for the reaction?
b. What is the degree of dissociation of COCl2?

Answer 1

The reaction above proceeds as thus;

COCl2_(g) $\iff$ CO_(g) + Cl2_(g)

(a) Calculating Kp

$Kp =\dfrac{(PCO)^1(PCl2)^1}{(PCOCl2)^1}$

P_CO = 0.05 atm

P_COCl2 = ?

PV=nRT

$P =\dfrac{nRT}{V}$

$PCOCl2 =\dfrac{(0.03mol)(0.08314Jmol^-1K^-1)(1073K)}{1.5dm^3}$

P _COCl2 = 1.784 atm

P_T = P₁ + P₂

P_Cl2 = (P_COCL2) - (P_CO)

P_Cl2 = 1.784 - 0.5 = 1.284atm

$Kp =\dfrac{(0.5)^1(1.284)^1}{(1.784)^1}$

Kp = 0.35987

Calculating Kc

Kp = Kc(RT)$\Delta$ⁿ

$\Delta$n = Mol of product gas - Mol of reactant gas

= 2mol -1mol

= 1 mol

Therefore;

0.35987 = Kc (0.08314Jmol^-1K^-1 x 1073K)¹

0.35987 = Kc (89.20922Jmol-1)

Divide both sides by 89.20922

Kc = 4.034x10^-3

(b) Calculating the degree of dissociation

Let the initial amount of COCl2 be n and its degree of dissociation be $\alpha$. Then we will have

COCl2_(g)$\iff$ CO_(g) + Cl2_(g)

t=0 n

t_eq n(1- $\alpha$) n$\alpha$ n$\alpha$

Total amount of gases = n(1-$\alpha$) + n$\alpha$ + n$\alpha$

= n(1+$\alpha$)

Now, the volume of the flask would be

$V=\dfrac{n(1+\alpha)RT}{\rho}$

density of the mixture is therefore;

$\rho =$ $\dfrac{nMCOCl2}{V} = \dfrac{nMCOCl2\rho}{n(1+\alpha)RT}=\dfrac{MCOCl2\rho}{(1+\alpha)RT}$

V = 1.5dm3

$\rho$ = $\dfrac{Mass}{V}$

Mass = moles x molar mass = 0.03mol x 99g/mol =2.97g

$\rho = \dfrac{2.97g}{1.5dm3} = 1.98g/L$

$\alpha=\dfrac{PMCOCl2}{\rho RT} - 1$

Substituting the given values

$\alpha=\dfrac{(1.784atm) (99g/mol)}{(1.98g/L)(0.08314LatmK^-1mol^-1)(1073K)T} - 1$

= 0.999 - 1

= 0.001