Question

URGENT!! 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=1.60×10−3. Part A If a 1.55-L reaction vessel initially contains 121 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.

I've tried 4 times, I keep getting it wrong. Please show me how to do it! It is due very soon.

Answer 1

1 bar = 0.9869 atm
0.121 bar = 0.1194 atm

	H2O	CO	H2
Initial partial pressure (atm)	0.1194	0	0
Change in partial pressure (atm)	-x	x	x
Equilibrium partial pressure (atm)	0.1194-x	x	x

The equilibrium constant

This is a quadratic equation of the type

It has solutions

The negative value is discarded as pressure cannot be negative.

Hence, the equilibrium partial pressures of H2 and CO gas are0.01304 atm each. The equilibrium partial pressure of gaseous water is atm

The number of moles are directly proportional to partial pressure. So mole percent is equal to the percent calculated from partial pressures.

Mole percent of H2 and CO are % The mole percent of gaseous water is %

Thus 100 moles of the mixture will have 80.3 moles of water, 9.85 moles of H2 and 9.85 moles of CO.

The molar masses of H2O, H2 and CO are 18 g/mol, 2 g/mol and 28 g/mol respectively.

Mass of gaseous water

The percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium %