The equilibrium constant, KC, for the following reaction is 2.5 × 10-3N2(g) + O2(g) ⇌ 2NO(g)If [N2]eq = 0.75 M and [NO]eq = 0.090, then [O2]eq
Question
The equilibrium constant, KC, for the following reaction is 2.5 × 10-3N2(g) + O2(g) ⇌ 2NO(g)If [N2]eq = 0.75 M and [NO]eq = 0.090, then [O2]eq
Solution
The equilibrium constant (Kc) for a reaction is given by the formula:
Kc = [Products] / [Reactants]
In this case, the reaction is N2(g) + O2(g) ⇌ 2NO(g), so the equilibrium constant expression is:
Kc = [NO]^2 / ([N2] * [O2])
We are given that Kc = 2.5 × 10^-3, [N2]eq = 0.75 M, and [NO]eq = 0.090 M. We can substitute these values into the equilibrium constant expression to solve for [O2]eq:
2.5 × 10^-3 = (0.090)^2 / (0.75 * [O2]eq)
Solving this equation for [O2]eq gives:
[O2]eq = (0.090)^2 / (2.5 × 10^-3 * 0.75)
[O2]eq = 0.0081 / 0.001875
[O2]eq = 4.32 M
So, the equilibrium concentration of O2 is 4.32 M.
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