Question

Pure phosgene gas (COCl2), 0.0340 mol, was placed in a 1.50−L container. It was heated to 700.0 K, and at equilibrium, the pressure of CO was found to be 0.509 atm. Calculate the equilibrium constant KP for the reaction.CO(g) + Cl2(g) ⇌ COCl2(g)KP =

Answer 1

Answer: The value of $K_p$ for the given equation is 3.065

Explanation:

To calculate the partial pressure of phosgene we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = ?

V = Volume of the gas = 1.50 L

T = Temperature of the gas = 700 K

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

n = number of moles of phosgene = 0.0340 moles

Putting values in above equation, we get:

$p_{COCl_2}\times 1.50L=0.0340\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 700K\\\\p_{COCl_2}=\frac{0.0340\times 0.0821\times 700}{1.50}=1.303atm$

As, initially phosgene is present in the system. So, the reaction is going backwards.

We are given:

Equilibrium partial pressure of CO = 0.509 atm

For the given chemical equation:

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

Initial:                                          1.303

At eqllm:        x            x             1.303-x

Evaluating the value of 'x'

$\Rightarrow x=0.509$

So, equilibrium partial pressure of chlorine gas = x = 0.509 atm

Equilibrium partial pressure of phosgene = 1.303 - x  = [1.303 - 0.509] = 0.794 atm

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:

$K_p=\frac{0.794}{0.509\times 0.509}\\\\K_p=3.065$

Hence, the value of $K_p$ for the given equation is 3.065