Question

II Review | Constants 1 Periodic Table 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 1.6x10-Kp. Part A If a 1.55-L reaction vessel initially contains 110 torr 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. V ACO O 2 ? m2 ma, tmco+ma1,0 = mH2 +mco+mh,0 olo Submit Previous Answers Request Answer

Answer 1

volume = 1.55L Pressure = 110 torr Temperature = T= o.iuu fath 700k Tiror = atp x Pthe 1.600 2 x r A Cas) + Hogs Coco + this) Initial 00144 atm o o conversion an - ta ta Equilibrium (0.144-x) u Ike; pth x pco 1.60 x10-3 2x2 - (0.144-x) 1.60 x103 x 0.144.2) = 2² solving we get . x = o.oluy partial pressure of Co and H₂ =n= 0.0144 ath partial pressure of ho= 0.144 -0.0144 - = 0.1296 atm Total pressure of gay = photph to co -= 0.1296 to 0144to.oluy o.1584 atm According to Roult's law mole fraction a partial pressure tofal presund th= 0.0144 - 0.090 -0.1584 Xno - 0.1296 - 0.818 and Xico = 0.090

thoa 0.818 So 1 total Imole of mixure mole of th = mole of co = o.ogo рос \e 4 H 0 2 0 1 Now mass of th = 0.09 x 2 = 0.180 g mass of co= 0.098 28 = 2.52 mass ef 0.818 X 18 = 14.829 Totaal mass = 0.18 +2.52 +14.72 mass = 17.42 g percent of th = 0. 180x100 - 17.42 = 0.01038100 - 1.03 % Aug