Answer:
The equilibrium molarity of O2 is 0.6 M
Explanation:
Equation of reaction:
1 mole of N2 reacts reversibly with 1 mole of O2 to form 2 moles of NO.
Volume of flask (V) = 250 mL = 250/1000 = 0.25 L
Initial concentration of N2 = number of moles of N2/V = 0.3/0.25 = 1.2 M
Initial concentration of NO = number of moles of NO/V = 0.7/0.25 = 2.8 M
Equilibrium constant (K) = 7.70
Let the equilibrium concentration of O2 be y
From the equation of reaction,
Mole ratio of N2 to O2 is 1:1, equilibrium concentration of N2 is (1.2 - y).
Mole ratio of O2 to NO is 1:2, equilibrium concentration of NO is (2.8 - 2y)
K = [NO]^2/[N2][O2]
7.7 = (2.8 - 2y)^2/(1.2 - y)y
7.84 - 11.2y + 4y^2 = 9.24y - 7.7y^2
4y^2 + 7.7y^2 - 11.2y - 9.24y + 7.84 = 0
11.7y^2 - 20.44y + 7.84 = 0
Divide each term by 11.7
y^2 - 1.75y + 0.67 = 0
The value of y is obtained using the quadratic formula
y = [1.75 - sqrt(1.75^2 - 4×1×0.67)] ÷ 2×1 = [1.75 - sqrt(0.3825)] ÷ 2 = (1.75 - 0.62) ÷ 2 = 1.13/2 = 0.565 = 0.6 (to 1 decimal place)
Equilibrium concentration of O2 is 0.6 M
