1.0 mol of each of the gases, CO, H20, CO2 and H2 are placed in a 2.00L container and allowed to reach equilibrium. It is found at equilibrium 0.40 mol of CO is present. Calculate the value of K eq for the system.
C0(g) + H20(g)--- CO2(g) + H2(g)
Intial
CO= 0.5 mol/L
H20= 0.5 mol/L
CO2= 0.5 mol/L
H2= 0.5 mol/L
change
CO= -0.1
H20= -0.1
C02= +0.1
H2= +0.1
Equilibrium
C0= 0.40 mol/L
H20=0.40 mol/L
CO2= 0.6 mol/L
H2=0.6 mol/L
k eq= (C02)(H2)/(C0)(H20)
= (0.6)(0.6)/(0.4)(0.4)= 2.25
To calculate the value of K eq for the system, we can use the formula:
K eq = (CO2)(H2) / (CO)(H2O)
Given the equilibrium concentrations:
CO = 0.40 mol/L
H2O = 0.40 mol/L
CO2 = 0.60 mol/L
H2 = 0.60 mol/L
Substituting the values into the formula, we get:
K eq = (0.60)(0.60) / (0.40)(0.40)
= 0.36 / 0.16
= 2.25
Therefore, the value of K eq for the system is 2.25.