To solve this problem, we need to set up an ICE (initial, change, equilibrium) table and use the equilibrium constant to determine the equilibrium molarity of N2.
Given:
- Volume of flask = 250 mL = 0.250 L
- Initial moles of N2 = 0.90 mol
- Initial moles of NO = 1.3 mol
- Equilibrium constant (Kc) = 0.168
Let x be the change in moles for the reactants (N2 and O2) and 2x for the product NO.
Initial:
N2: 0.90 mol
O2: 0 mol
NO: 1.3 mol
Change:
N2: -x
O2: -x
NO: +2x
Equilibrium:
N2: 0.90 - x
O2: x
NO: 1.3 + 2x
Using the equilibrium constant expression and the values from the ICE table:
Kc = [NO]^2 / ([N2][O2])
0.168 = (1.3 + 2x)^2 / ((0.90 - x)(x))
0.168 = (1.69 + 5.2x + 4x^2) / (0.90x - x^2)
Solving for x:
0.168(0.90x - x^2) = 1.69 + 5.2x + 4x^2
0.1512x - 0.168x^2 = 1.69 + 5.2x + 4x^2
0.168x^2 + 5.3512x - 1.69 = 0
Solving this quadratic equation gives x = 0.288 and x = -9.95. Since x must be positive, x = 0.288 mol.
Now we can find the equilibrium molarity of N2:
[N2] = (0.90 - 0.288) mol / 0.250 L
[N2] = 0.612 mol / 0.250 L
[N2] = 2.448 M
Therefore, the equilibrium molarity of N2 is 2.45 M.
Suppose a 250. mL flask is filled with 0.90 mol of N2 and 1.3 mol of NO. The following reaction becomes possible:
N2(g)+O2(g) <-> 2NO(g)
The equilibrium constant for this reaction is 0.168 at the temperature of the flask.
Calculate the equilibrium molarity of N2. Round your answer to two decimal places.
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