Finding Reactant Quotients

Question

1. A 5.0-L flask containing 17 g of NH₃, 14 g of N₂, and 12 g of H₂:

N2(g)+3H2(g)⇌2NH3(g)Kc=0.060N2(g)+3H2(g)⇌2NH3(g)

Kc=0.060

1. A 2.00-L flask containing 230 g of SO₃(g):

2SO3(g)⇌2SO2(g)+O2(g)Kc=0.2302SO3(g)⇌2SO2(g)+O2(g)

Kc=0.230

Answer 1

First we calculate all the concentrations, then we use the concentrations and the stoichiometric coefficients to construct the quotient as follows

17 g of NH₃ = 1 mol NH₃

[ NH₃ ] = 1 mol NH₃ / 5 L = 0.2 M

14 g of N₂ = 0.5 mol N₂

[ N₂ ] = 0.5 mol N₂ / 5 L = 0.1 M

12 g of H₂ = 6 mol H₂

[ H₂ ] = 6 mol H₂ / 5 L = 1.2 M

Hence the reaction quotient will be

0.2² / (0.1 x 1.2³) = 0.23 > 0.06 (some NH₃ will decompose as the system is moving leftwards towards the equilibrium)

In the second reaction mixture there are no products (but all the reagents are there). It means that the concentrations of the products are 0, and the reaction quotient is also 0.