Question

At a certain temperature, a 20.5-L contains holds four gases in equilibrium. Their masses are: 3.5 g SO3, 4.6 g SO2, 21.5 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

SO2 + N2O -> SO3 + N2

moles of SO2 = 4.6/ 64 = 0.071875

moles of SO3 =3.5/80 =0.04375

moles of N20=0.98/44 =0.02227

moles of N2= 21.5/28 =0.76786

equilibrium constant = [SO3][N2]/[SO2][N2O] = ( 0.04375*0.76786) / (0.071875*0.02227)

=20.987

Answer 2

SO2 + N2O <----> SO3 + N2 (balanced)
molar mass SO2 = 32+16 = 48 gm/mol; moles = 4.6/48=0.0958 mol; M= 0.0958/22L = 0.0044M
molar mass N2O = 28+16 = 44 gm/mol; moles = 0.98/44=0.022 mol; M = 0.022/22L = 0.001 M
molar mass SO3 = 32+48 = 80 gm/mol; moles = 3.5/80 = 0.0438 mol; M = 0.0438/22L = 0.002M
molar mass N2 = 14+14 = 28 gm/mol; moles = 11.1/28=0.396 mol; M =0.396/22L = 0.018M

Kc= [SO3][N2]/[SO2][N2O] = (0.002)(0.018)/(0.0044)(0.001) = 8.18

Answer 3

SO2 + N2O <----> SO3 + N2 (balanced)
molar mass SO2 = 32+16 = 48 gm/mol; moles = 4.6/48=0.0958 mol; M= 0.0958/20.5L = 0.00467M
molar mass N2O = 28+16 = 44 gm/mol; moles = 0.98/44=0.022 mol; M = 0.022/20.5L = 0.001073M
molar mass SO3 = 32+48 = 80 gm/mol; moles = 3.5/80 = 0.0438 mol; M = 0.0438/20.5L = 0.00213M
molar mass N2 = 14+14 = 28 gm/mol; moles = 21.5/28=0.767 mol; M =0.767/20.5L = 0.03741M

Kc= [SO3][N2]/[SO2][N2O] = (0.00213)(0.0374)/(0.00467)(0.001073) = 15.89