At 2273K, equilibrium constant for the reaction
2NO(g)<->N2(g)+O2(g)
Kc=2.4x10^3
What would be equilibrim concentration of N2, if initial concentration of NO was 0.1160M?
Have you tried setting up an ICE chart? That's the obvious way to go.
yes I did, but im still not getting the right answer.
I got .1136 which is not the correct answer.
what do you have for
initial:
change:
equilibrium
2NO <-> N2 O2
initial .116 0 0
change -x +x +x
equil .116-x x x
Did you look at your post from last night where we arrived at K = something x 10^-192? That problem didn't ask for K, it asked for pK. I'll get back on the current problem.
Make it
initial:
NO = 0.1160
N2 = 0
O2 = 0
change:
NO = -2x
N2 = x
O2 = x
equilibrium:
NO = 0.1160 - 2x
N2 = x
O2 = x
and solve for x.
yes, I figured that one out. Thanks so much for the help!
that works! thank you so much!!
I'm curious about the problem last night. Was the pK about 192?
To determine the equilibrium concentration of N2, we need to use the equilibrium constant, the initial concentration of NO, and apply the stoichiometry of the reaction.
The equilibrium constant expression for the given reaction is Kc = [N2][O2] / [NO]^2, where [N2], [O2], and [NO] represent the molar concentrations of N2, O2, and NO, respectively.
Given that the equilibrium constant is Kc = 2.4 x 10^3, and the initial concentration of NO is [NO] = 0.1160 M, we can set up the equation as follows:
2.4 x 10^3 = [N2][O2] / (0.1160^2)
Since the initial concentration of NO is squared in the denominator, the equilibrium concentrations of N2 and O2 will both be half the equilibrium constant.
Let x be the equilibrium concentration of N2. Therefore, the equilibrium concentration of O2 is also x.
Now, we can rewrite the equation as:
2.4 x 10^3 = (x)(x) / (0.1160^2)
Multiplying both sides by (0.1160^2) gives:
2.4 x 10^3 x (0.1160^2) = x^2
Simplifying:
x^2 = (2.4 x 10^3 x 0.1160^2)
x^2 = 0.0333
x = √0.0333
x ≈ 0.182 M
Therefore, the equilibrium concentration of N2 is approximately 0.182 M.