Question

1.00 g each of CO, H_2O, and H_2 are sealed in a 1.47 - L vessel and brought to equilibrium at 600 K. CO (g) + H_2O(s) CO_2(g) + H_2(g) KC = 23.2 How many grams of CO_2 will be present in the equilibrium mixture?

Answer 1

Molar masses : CO= 28, H2O= 18 and H2=2

Moles: mass/Molar mass

Moles : CO= 1/28=0.036, H2O= 1/18= 0.056, H2=1/ 2 =0.5, CO2=44

Concentration :Moles/Liter

Concentrations : CO= 0.036/1 = 0.036, H2O= 0.056/1= 0.056 and H2= 0.5/1

These are initial concentrations

Let x= drop in concentration of CO to reach equilibrium

At equilibrium : CO= 0.036-x, H2O= 0.056-x, H2= 0.5+x and CO2=x

For the reaction CO+H2O<->CO2 + H2

Kc = [CO2] [H2]/ [CO] [H2O] = x*(0.5+x)/ (0.036-x)*(0.056-x) = 23.2

The equation when solved using excel gives x= 0.0218 moles/L

Since the volume is one litres, moles of CO2= 0.0218

Mass of CO2= moles* Molar mass =0.0218*44=0.99592 gms