A 100L flask was filled with 200 mol each of gaseous SO2and

A 1.00-L flask was filled with 2.00 mol each of gaseous SO2and NO2and heated. It was found at equilibrium that 1.30 molof gaseous NO was present at equilibrium. Calculated the value of the equilibrium constant, Kc, for this reaction: SO2(g) + NO2(g) SO3(g) + NO(g

Solution

callculate equilibrium concentration of all species

initial conc of SO_2(g) = 2 mol / 1L = 2 M

initial conc of NO_2(g)  = 2 mol / 1L = 2 M

initial conc of SO_3(g)  = 0 M

initial conc of NO_(g)  = 0 M

At equilibrium change in concentration

equilibrium concentration of NO_(g) = 1.30/1 = 1.30 M

equilibrium concentration of SO_2(g) = according to reaction when 1 mole NO formed there is 1 mole SO₂ reacted therefore mole of SO₂  remain at equilibrium = 2 - 1.30 = 0.70 mole

equilibrium concentration of SO_2(g) = 0.70 / 1 = 0.70 M

equilibrium conc of SO_3(g) = According to reaction when 1 mole NO formed there is 1 mole SO₃ form therefore  when 1.30 mole NO form 1.30 mole SO₃ formed. Mole of SO₃ at equilibrium = 1.30 mole

equilibrium conc of SO_3(g) = 1.30/1 = 1.30 M

equilibrium concentration of NO_2(g) = according to reaction when 1 mole NO formed there is 1 mole NO₂ reacted therefore mole of NO₂  remain at equilibrium =  2 - 1.30 = 0.70 mole

equilibrium concentration of NO_2(g) = 0.70 / 1 = 0.70 M

K_c = [SO₃][NO] / [SO₂] [NO₂]

substitute value at equilibrium concentration

K_c  = (1.30) (1.30) / (0.70) (0.70) = 3.45

K_c = 3.45