Try to solve this Chemistry Problem

[Xylitol]

It's not homework
It's a random textbook question I pulled up that I can't seem to solve

At 2000degreesC, the equilibrium constant fro the reaction
2NO(g) <-> N2(g) + O2(g)

is Keq = 2.4*10^3. If the initial partial pressure of NO is 37.3 atm, what are the equilibrium partial pressures of NO, N2, and O2?

Please solve it and then explain to me how you did it
Thanks

Edit: Got it thanks

 

[Oil]

I got:
NO = 0.0377 atm
N2 = 1.846 atm
O2 = 1.846 atm
The answers seem a little low so I may be wrong. Use an ICE table to solve

 

[cirthix]

dalton's law of partial pressures. you can find the # of moles of each substance by the equilibrium constant. you know the initial pressure, so work from there.

 

[Xylitol]

Originally posted by: OSx86
I got:
NO = 0.0377 atm
N2 = 1.846 atm
O2 = 1.846 atm
The answers seem a little low so I may be wrong. Use an ICE table to solve

Yea the answers are those multiplied by 10

 

[amol]

I got that both N2 and O2 are about 8.02 atm and NO is 21.258 atm.

Do you have the correct answers? Or is it no answer provided?

edit: oops, never mind. haha, it's been a while since i've done an Keq problem, but i have the AP in 2 months. better start brushing up!

 

[imported_quasi]

ICE tables work wonders

......................... 2NO <--> N2 + O2
-----------------------------------------------
Initial................. 37.3 ......... 0 ....... 0
Change.............. -2x ............x ....... x
Equlibrium.......... 37.3-2x ..... x........ x

2.4*10^3 = products/reactants = x^2 / (37.3-2x)
0 = x^2 + 4800x + 89520, so x = 18.6

NO = 37.3 - 2x = .1 atm
N2 = x = 18.6 atm
O2 = x = 18.6 atm

Little off from the answers you said were correct (after multipled by 10)

 

[Oil]

Originally posted by: RockSolid
ICE tables work wonders

......................... 2NO <--> N2 + O2
-----------------------------------------------
Initial................. 37.3 ......... 0 ....... 0
Change.............. -2x ............x ....... x
Equlibrium.......... 37.3-2x ..... x........ x

2.4*10^3 = products/reactants = x^2 / (37.3-2x)
0 = x^2 + 4800x + 89520, so x = 18.6

NO = 37.3 - 2x = .1 atm
N2 = x = 18.6 atm
O2 = x = 18.6 atm

Little off from the answers you said were correct (after multipled by 10)

The K equation should be: 2.4*10^3 = x^2/(37.3-2x)^2 (NO should be squared)

My mistake was that I wrote down 3.73 instead of 37.3

 

[imported_quasi]

Originally posted by: OSx86

Originally posted by: RockSolid
ICE tables work wonders

......................... 2NO <--> N2 + O2
-----------------------------------------------
Initial................. 37.3 ......... 0 ....... 0
Change.............. -2x ............x ....... x
Equlibrium.......... 37.3-2x ..... x........ x

2.4*10^3 = products/reactants = x^2 / (37.3-2x)
0 = x^2 + 4800x + 89520, so x = 18.6

NO = 37.3 - 2x = .1 atm
N2 = x = 18.6 atm
O2 = x = 18.6 atm

Little off from the answers you said were correct (after multipled by 10)

The K equation should be: 2.4*10^3 = x^2/(37.3-2x)^2 (NO should be squared)

My mistake was that I wrote down 3.73 instead of 37.3

Ah, you're right, I missed that (already did 2 hours of equilibrium homework tonight).