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.