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 on C please

Explanation / Answer

c) Kp = Kc(RT)^dn , where dn= gas product moles-gas reactnats moles ( stochiometrically in equation)   = 2-2=0, hence

Kp = Kc = 9.50 ,

Kc = [H2][CO2]/[H2O][CO]

water moles = (125.4+ 31.4)/18 = 8.711 ,

CO moles = water moles initially = (125.4/18) = 6.967

at equi [H2O] = (8.711-x)/10 , [CO] =(6.967-x)/10,

[CO2]=[H2] =x,

Kc = 9.5 = (x/10)^2 /(8.711-x)(6.967-x)/100

9.5 = x^2/(8.711-x)(6.967-x)

8.5x^2 -148.94x + 576.55 =0,

x =5.778 moles ,

[H2] =5.778/10 = 0.5778 moles/liter , 1 liter = 1000 cm3

hence [H2] = 0.5778/1000 = 5.778 x10^-4 moles/cm3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.