Question

The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K:

2CH2Cl2(g) CH4(g) + CCl4(g)

Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.856 atm at 350 K.

PCH2Cl2 = ________ atm

PCH4 = _________ atm

PCCl4 = _______ atm

Answer 1

balanced reaction is
4NH3 + 5O2 --> 4NO + 6 H2O

according to reaction
4 mol of NH3 required 5 mol of O2
so, number of mole of O2 required = 5 mol

according to reaction
reaction taking place
2CH2Cl2 <--> CH4 + CCl4
0.856 0 0 intial
0.856-2x x x at equlibrium

Kp = (pCH4*pCCl4)/(pCH2Cl2)^2
10.5 = x^2/(0.856-2x)^2
10.5 = {x/(0.856-2x)}^2
{x/(0.856-2x)} = 3.24
x = 3.24*(0.856 - 2x)
x = 2.77 - 6.48x
7.48x = 2.77
x = 0.371

pCH2Cl2 = 0.856-2x
= 0.856-(2*0.371)
= 0.856 - 0.742
= 0.114 atm

pCH4 = x
= 0.371 atm

pCCl4 = x
= 0.371 atm