Question

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0160 mol gaseous NH3 and 0.0220 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.70×10-3 M. Calculate Keq for the reaction at this temperature.

Answer 1

since volume is 1 L, number of moles will be same as concentration

4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g)
0.0160 0.0220 0 0 (initial)
0.0160-4x 0.0220-3x 2x 6x (at equilibrium)

[N2] = 2x = 2.70*10^-3 M
x= 1.35*10^-3 M = 0.00135 M

Keq = [H2O]^6 [N2]^2 / {[NH3]^4 [O2]^3}
= (6x)^6 (2x)^6 / {(0.0160-4x)^4 (0.0220-3x)^3}
= (6*0.00135)^6 (2*0.00135)^6 / {(0.0160-4*0.00135)^4 (0.0220-3*0.00135)^3}
= (2.824*10^-13)*(3.847*10^-16) / (1.262*10^-8 * 5.784*10^-6)
= 1.488*10^-15

Answer: 1.488*10^-15