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 given chemical equation, we first need to understand the concept of equilibrium and the equilibrium constant.
At equilibrium, the rate of the forward reaction (CO(g) + H2O(g) → CO2(g) + H2(g)) is equal to the rate of the reverse reaction (CO2(g) + H2(g) → CO(g) + H2O(g)). The equilibrium constant, K_eq, is a value that represents the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
In this case, we are given the initial and final concentrations of the gases involved. Here are the given values:
Initial:
CO = 0.5 mol/L
H2O = 0.5 mol/L
CO2 = 0.5 mol/L
H2 = 0.5 mol/L
Final (at equilibrium):
CO = 0.40 mol/L
H2O = 0.40 mol/L
CO2 = 0.6 mol/L
H2 = 0.6 mol/L
To calculate K_eq, we need to use the formula:
K_eq = (CO2)(H2) / (CO)(H2O)
Substituting the given values:
K_eq = (0.6)(0.6) / (0.4)(0.4) = 2.25
Therefore, the value of K_eq for the system is 2.25.