Question

Consider the following equilibrium.
CS2(g) + 3 O2(g) equilibrium reaction arrow CO2(g) + 2 SO2(g)
If the reaction is started in a container with 5.59 atm CS2 and 13.3 atm O2, what is Kp if the partial pressure of CO2 is 3.76 atm at equilibrium? (There is no change in temperature and the initial partial pressures of the products are equal to 0.)

Answer 1

Given:

Initial pressure of CS₂ = 5.59 atm

Initial pressure of O₂ = 13.3 atm

At equilibrium, Partial pressure of CO₂ = 3.76 atm

The equilibrium reaction is

CS₂ (g) + 3O₂ (g) CO₂ (g) + 2SO₂ (g)

The equilibrium constant of pressure is defined as the ratio of products of the pressure of products and products of the pressure of reactants each raised to the power of stoichiometric coefficients.

CS₂ (g)                  + 3O₂ (g)   CO₂ (g)          + 2SO₂ (g)

Initial pressure	5.59 atm	13.3 atm	0 atm	0 atm
Number of moles reacted	-x	-3x	+x	+2x
At equilibrium, Partial pressure	5.59 -x	13.3 - 3x	x	2x

At equilibrium:

Given that Partial pressure of CO₂ () = 3.76 atm = x

Partial pressure of SO₂ () = 2x = 2 * 3.76 atm = 7.52 atm

Partial pressure of CS₂ () = 5.59 – x

= 5.59 – 3.76 atm

Partial pressure of CS₂ ()= 1.83 atm

Partial pressure of O₂ ()= 13.3 - 3x

= 13.3 – 3(3.76) atm = 13.3 – 11.28 atm

= 2.02 atm.

Now, substitute the values in the equilibrium constant of pressure formula, and simplify.

  

The equilibrium constant of pressure is 14.1 atm^-1