Answer to Question #90118 in General Chemistry for Michael Maruszewski

Question

Initially 0.500 mol/L NOCl(g) reaches equilibrium as shown in the equation below:

2 NOCl(g) ↔ Cl2(g) + 2 NO(g)

At equilibrium concentration of Cl2(g) is 0.15 mol/L.

The value of the equilibrium constant Kc is a.bc?

Write your answer for a.bcd? in the blank, including the decimal.

Your Answer:

Answer 1

Let V=1L.

  X mol      0.150 mol   Y mol

2NOCl(g) ↔ Cl₂(g) + 2NO(g)

  2 mol        1 mol        2 mol

[Cl₂] = 0.150 M.

n_react(NOCl) = X = 0.300 mol.

n_res(NOCl) = n_in(NOCl) –  n_react(NOCl) = 0.500 mol – 0.300 mol = 0.200 mol.

[NOCl] = 0.200 M.

n_form(NO) = Y = 0.300 mol.

[NO] = 0.300 M.

K_c = [Cl₂]×[NO]² / [NOCl]² = 0.150 × (0.300)² / (0.200)² = 0.3375 ≈ 0.34.