Question

Pure phosgene gas (COCl2,), 3.30 x 10-2 mol, was placed in a 1.50-L container. It was heated to 800K, and at equilibrium the pressure of CO was found to be 0.497 atm. Calculate the equilibrium constant, Kp, for the reaction: CO(g) Cl2(g) COCl2(g)

Answer 1

Answer : The value of equilibrium constant
`K_p` for the reaction is, 3.82

Explanation :

First we have to calculate the pressure of phosgene gas.

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of phosgene gas = ?

V = Volume of phosgene gas = 1.50 L

n = number of moles phosgene =
`3.30* 10^(-2)mol`

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of phosgene gas = 800 K

Putting values in above equation, we get:

`P* 1.50L=3.30* 10^(-2)mol* (0.0821L.atm/mol.K)* 800K`

`P=1.44atm`

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is:

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

Initial pressure 0 0 1.44

At eqm. P P (1.44-P)

The expression of equilibrium constant
`K_p` for the reaction will be:

`K_p=((p_(COCl_2)))/((p_(CO))* (p_(Cl_2)))`

As we are given that:

`p_(CO)=0.497atm`

That means, P = 0.497 atm

`K_p=((1.44-P))/((P)* (P))`

Now put all the values in this expression, we get :

`K_p=((1.44-0.497))/((0.497)* (0.497))`

`K_p=3.82`

Thus, the value of equilibrium constant
`K_p` for the reaction is, 3.82