Question

Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3If a 1.55-L reaction vessel initially contains 143 mbar of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Answer 1

Volume = 1.55 L,

Pressure = 143 mbar = 0.143 bar = 0.141 atm

Temperature = T =700 K

Given,

By solving it

partial pressure of CO & H₂ = x = 0.0142 atm

partial pressure of H₂O = 0.141-0.0142 = 0.1268 atm

Total pressure of gass = pH₂O + pH₂ + pCO = 0.1268 + 0.0142 + 0.0142 =0.1552 atm

According to Roult's law

mole fraction = partial pessure/total pressure

mole fraction of H₂ = 0.092

moles of H2 = 0.092 mol

mole fraction of H₂O =

moles of H2O = 0.816 mol

Mole fraction of CO = 0.092

moles of CO = 0.092 mol

Mass of H₂ =0.092 * 2 = 0.184 g

Mass of CO =0.092 * 28 = 2.576 g

Mass of H₂O = 0.1268 * 18 = 14.688 g

Total mass of gases =0.184 g + 2.576 g + 14.688 g = 17.45 g

Mass percent of H₂ =

Mass percent of H₂ = 1.05 %