A mixture of 0.100 mol NO, 0.200 mol H2 and 0.0800 mol N2 were placed in a 2.00L reaction vessel, heated, and allowed to come to equilibrium conditions. At equilibrium, the molar concentration of N2 was 0.0500 mol/L. Calculate Kc for this reaction.

2NO + 2H2 <---> N2 + 2H2O

I HAVE begun this question by doing the ICE TABLE. but the problem is that at equilibrium, for No it is 0.0500! And the initial is 0.05 so that means it had 0 change in concentration??? And then 2H2O would be 0M at equilibrium. Is that correct? I know how to solve Kc its at the equilibrium concentrations I'm stuck. Thanks.

I assume all are gases. I assume initial H2O is 0.

Initial (NO) = 0.100/2 = 0.050M
Initial (H2) = 0.200/2 = 0.100M
Initial (N2) = 0.0800/2 = 0.0400M

........2NO + 2H2 ==> N2 + 2H2O
I......0.05...0.10....0.04..0
C......-2x....-2x....+x....+2x
E...0.05-2x.0.10-2x.0.04+x..2x

At equilibrium (N2) = 0.05; therefore, x ust be 0.01 since 0.04+x = 0.05
Then NO = 0.05-2x
(H2) = 0.10-2x
Kc goes from there.
I think you confused NO with N2.

To solve for the equilibrium concentrations and calculate Kc, you can follow these steps:

1. Start by writing the balanced equation for the reaction:
2NO + 2H2 ↔ N2 + 2H2O

2. Set up an ICE (Initial, Change, Equilibrium) table. Since the initial concentrations are given, fill in the 'Initial' row with the given values.

| NO | H2 | N2 | H2O
--------------------------------------------------
Initial | 0.100 | 0.200 | 0 | 0
Change | -x | -2x | +x | +2x
Equilibrium | 0.100-x | 0.200-2x | x | 2x

3. At equilibrium, the molar concentration of N2 is given as 0.0500 mol/L. So, you can write the equilibrium expression for Kc using the concentrations at equilibrium:

Kc = [N2][H2O]^2 / [NO]^2[H2]^2

Substituting the given values:
Kc = (0.0500)(2x)^2 / (0.100 - x)^2(0.200 - 2x)^2

4. Next, solve for x by plugging in the values into the equilibrium expression:

0.0500 = (0.0500)(2x)^2 / (0.100 - x)^2(0.200 - 2x)^2

This equation may look complex, but you can simplify it by assuming that x is small relative to 0.100 and 0.200. Therefore, you can neglect x in the denominators, resulting in a simpler equation:

0.0500 = (0.0500)(2x)^2 / (0.100)^2(0.200)^2

0.0500 = 4x^2 / (0.100)^2(0.200)^2

5. Solve for x by rearranging the equation:

x^2 = 0.0500 * (0.100)^2(0.200)^2 / 4

x^2 = 0.001000

x = √0.001000 ≈ 0.0316

6. Now that you have the value of x, you can substitute it back into the equilibrium concentration expressions to find the equilibrium concentrations:

[N2] = x = 0.0316 mol/L
[H2O] = 2x = 2 * 0.0316 = 0.0632 mol/L
[NO] = 0.100 - x = 0.100 - 0.0316 = 0.0684 mol/L
[H2] = 0.200 - 2x = 0.200 - 2 * 0.0316 = 0.1368 mol/L

7. Finally, substitute the equilibrium concentrations into the equilibrium expression to calculate Kc:

Kc = (0.0316)(0.0632)^2 / (0.0684)^2(0.1368)^2

Simplify the expression and calculate the value of Kc.

Please note that rounding the values during calculations will affect the final value of Kc.