Question

At a certain temperature, a 19.0-L contains holds four gases in equilibrium. Their masses are: 3.5 g SO3, 4.6 g SO2, 20.1 g N2, and 0.98 g N2O. What is the value of the equilibrium constant at this temperature for the reaction of SO2 with N2O to form SO3 and N2 (balanced with lowest whole-number coefficients)?

Answer 1

S02 + N2O --> SO3 + N2.

We know Kc = [SO3][N2] / [SO2][N2O]. We know concentration is proportional to pressure and pressure is proportional to number of mole (remember PV = nRT).

Therefore Kc = n(SO3).n(N2) / n(SO2).n(N2O)

To solve we just need to work out the number of mole of each gas.

MW(SO3) ~ 80 g/mol [1]
MW(N2) ~ 14 g/mol [2]
MW(N2O) ~ 44 g/mol [3]
MW(SO2) ~ 64 g/mol [4]

Therefore moles of the gases
n(SO3) = 3.5 (g) / 80 (g/mol) = 4.37 * 10^-2 mol
n(N2) = 20.1 (g) / 14 (g/mol) = 1.43 mol
n(SO2) = 4.6 (g) / 64 (g/mol) = 7.18 * 10^-2 mol
n(N2O) = 0.98 (g) / 44 (g/mol) = 2.22 * 10^-2 mol

Kc = n(SO3).n(N2) / n(SO2).n(N2O)

Therefore Kc = 4.37 * 10^-2 * 1.43 / (7.18 * 10^-2 * 2.22 * 10^-2) = 39.20

equilibrium constant Kc = 39.20