A 125.4 g quantity of water and an equal molar amount of carbon monoxide were pl

Question

A 125.4 g quantity of water and an equal molar amount of carbon monoxide were place in an empty 10.0L vessel, and the mixture was heated to 700 K. At equilibrium, the partial pressure of CO was 9.8 atm. The reaction is

CO(g) + H2O(g) === CO2(g) + H2(g)
I calculated Kp at 700 K to be 9.50.

b) An additional 31.4 g of water was added to the reaction vessel, and a new state of equilibrium was achieved. What are the equilibrium partial pressures of each gas in the mixture?
i got 6.86,16.9,33.1,33.1

c. What is the concentration of H2 in molecules/cm^3

step by step please.

Explanation / Answer

moles H2O = 125.4 g / 18.02 g/mol=6.96 pressure H2O = 6.96 mol x 0.08206 x 700/10.0 L=40 atm because moles CO = 6.96 pressure H2O = 40 atm CO + H2O CO2 + H2 initial pressure 40. .. . 40 change -x. . . . -x. . . . .+x .. . .+x at equilibrium 40-x. . .40-x. . . x. . . . .x 40-x = 9.80 x = 30.2 atm = partial pressure CO2 = partial pressure H2 Kp = (30.2)(30.2/ (9.80)(9.80)= 9.50

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