Having the toughest time with this chemistry question

Xylitol

Diamond Member
Aug 28, 2005
6,617
0
76
A mixture of .10 mol of NO, .050 mol of H2, and .10 mol of H2O is placed in a 1.0-L vessel at 300 K. The folloowing equilibrium is established:
2NO(g) + 2H2(g) <-> N2(g) + 2H2O (g)

At equilibrium PartialPressureNO = 1.53 atm

Calculate the equilibrium partial pressures of H2, N2, and H20

The answer is in the back of the book, but I dont know how they got that answer
Help is appreciated
 

atomicacid55

Member
Jan 10, 2006
112
0
71
Molar ratios... then you use equilibrium constant stuff. Sum of partial pressures also equals total pressure. Blah blah. You get it. Multiple equations, multiple variables. Voila!
 

Xylitol

Diamond Member
Aug 28, 2005
6,617
0
76
Originally posted by: atomicacid55
Molar ratios... then you use equilibrium constant stuff. Sum of partial pressures also equals total pressure. Blah blah. You get it. Multiple equations, multiple variables. Voila!

Whats the total moles
.3 moles right
So mole fraction for say... H2 would be .5/.3 right?
 

TheChort

Diamond Member
May 20, 2003
4,203
0
76
first figure out how many moles of NO produces 1.53 atm
once you have that, use the mole ratio of the equation to figure out how many moles of H2 you have as well. (that will give you the atm for H2)

but for N2 and H2O, you can't use the same work (well, not right away at least). Let me just say there is some adding/subtracting involved.
 

Oil

Diamond Member
Aug 31, 2005
3,552
5
81
Using an ICE table would be the easiest if you were taught that method

 
Mar 8, 2005
126
0
0
Taking AP Chem? We just started equilibrium a few days ago.

At equilibrium, the partial pressure of NO is 1.53 atm, so with PV=NRT, you can determine the number of moles. (.062)

Then set up an ICE table (Initial, Change, Equilibrium) for the reaction. Because the change in NO was -.038 mol, H2 should be -.038 mol change, N2 should be .019 mol change and H20 should be .038 change. Then just use PV=NRT to convert to pressure

Final equilibrium concentration and partial pressure:
H2 = .012 mol = .30 atm
N2 = .019 mol = .47 atm
H20 = .138 mol = 3.4 atm

It makes sense in my head, but I may have gone wrong in several places.