Question

PS8.4. At 25 °C, 0.540 mol of O2 and 0.400 mol of N2O were placed in a 1.00 liter vessel and allowed to react according to the equation 2N2O(g) + 3O2(g) 4NO2(g) When the system reached equilibrium, the concentration of NO2 was found to be 0.02748 M. a) What are the equilibrium concentrations of N2O and O2? b) What is the value of KC for this reaction at 25 °C

Answer 1

2N2O + 3O2 <-------------------> 4NO2

initially [O2] = 0.54 / 1 = 0.54 M

[N2O] = 0.4/1 = 0.4 M

[NO2] = 0 M

at equilbrium [NO2] = 0.02748 M

0.02748 / 2 = 0.01374 M [N2O] consumed

so at equilibrium [N2O] = 0.4 - 0.01374 = 0.38626 M

(0.01374 / 2) x 3 = 0.02061 M O2 consumed

so at equilibrium [O2] = 0.54 - 0.02061 M = 0.51939

Kc = [NO2]⁴ / [N2O]² [O2]³

Kc = [0.02748]⁴  / [0.38626]² [0.51939]³

Kc = 5.70 x 10^-7  / 0.1492 x 0.140

Kc = 5.70 x 10^-7  / 0.0209

Kc = 2.73 x 10^-5