If 0.8 mole of Sulfur dioxide and 0.4 mole of nitrogen dioxide are placed into a 2-liter container and allowed to reach equilibrium,

Question

what will be the molar concentration of sulfur dioxide at equilibrium (Products of the reaction are sulfur trioxide and nitrogen monoxide)? The equilibrium constant, Kc, is 64.00 for the reaction involved.

Answer 1

Write the correctly balanced equation:

SO₂(g) + NO₂(g) ==> SO₃(g) + NO(g) ... balanced equation

Write the equilibrium expression:

K = [SO₃][NO] / [SO₂][NO₂]

Set up an ICE table:

SO₂(g) + NO₂(g) ==> SO₃(g) + NO(g)

0.4 M........0.2 M............0................0..........Initial concentration

-x...............-x................+x..............+x.........Change in concentration

0.4-x.........0.2-x.............x................x...........Equilibrium concentrations

Plug values into equilibrium expression:

64.00 = (x)(x) / (0.4-x)(0.2-x)

64.00 = x² / x² - 0.6x + 0.08

x = 0.197 (use the quadratic formula to solve for x)

Molar concentration of SO2 at equilibrium will be 0.4 - 0.197 = 0.2 M

(be sure to check my math)