Calculate each of these equilibrium concentrations based on the reaction below.
2 NO(g) + O2(g) 2 NO2(g) K = 1.71 1012
(a) [NO] = 0.0048 M; [O2] = 0.000057 M; [NO2] = ?
(b) [NO] = 0.0026 M; [O2] = 0.000023 M; [NO2] = ?
Are these equilibrium concns or their initial concns. The problem is clearly stated.
To calculate the equilibrium concentrations of [NO2] for the given reactions, we can use the equilibrium expression and the given initial concentrations of [NO] and [O2]. The equilibrium expression for the given reaction is:
K = [NO2]^2 / ([NO]^2 * [O2])
Let's solve for [NO2] for each case:
(a) [NO] = 0.0048 M; [O2] = 0.000057 M; [NO2] = ?
Using the equilibrium expression:
1.71 × 10^12 = [NO2]^2 / (0.0048^2 * 0.000057)
Rearranging the equation:
[NO2]^2 = 1.71 × 10^12 * (0.0048^2 * 0.000057)
Taking the square root of both sides:
[NO2] = √(1.71 × 10^12 * (0.0048^2 * 0.000057))
Calculating the result:
[NO2] ≈ 0.18 M
Therefore, the equilibrium concentration of [NO2] is approximately 0.18 M.
(b) [NO] = 0.0026 M; [O2] = 0.000023 M; [NO2] = ?
Using the same process as above:
1.71 × 10^12 = [NO2]^2 / (0.0026^2 * 0.000023)
Rearranging the equation:
[NO2]^2 = 1.71 × 10^12 * (0.0026^2 * 0.000023)
Taking the square root of both sides:
[NO2] = √(1.71 × 10^12 * (0.0026^2 * 0.000023))
Calculating the result:
[NO2] ≈ 0.028 M
Therefore, the equilibrium concentration of [NO2] is approximately 0.028 M.