1. Equilibrium Calculations Lesson 7

2. How can we describe an equilibrium system mathematically? reactants products ⇌ reactants The Keq is the equilibrium constant- a number that does not change. Providing the temperature is kept constant. products =3.0 Keq =

3. Equilibrium Calculations An equilibrium system, at any given temperature, can be described by an equilibrium expression and equilibrium constant. aA+bB ⇌ cC+dD (aq) and (g) are included! (l) and (s) are not-constant concentration! [A] a [B] b [C] c [D] d Keq= Equilibrium Constant- a numberExpression- mathematical equation Keq= Products Reactants

4. SO 3(g) + H 2 O (g) ⇌ H 2 SO 4(l) 1.at 25 o C, [SO 3 ] = 0.200 M. [H 2 O] = 0.480 M, and [H 2 SO 4 ] = 24 M. Calculate the Keq. The Keq has no units but concentration units that go in the expression must be M! =10.4 (0.200)(0.480) 1 [SO 3 ] [H 2 O] 1 don’t count (l)! Use 1Keq= = At equilibrium No ICE

5. 2.0.500 mole PCl 5, 0.40 mole H 2 O, 0.200 mole HCl, and 0.400 mole POCl 3 are found in a 2.0 L container at 125 o C. Calculate the Keq. PCl 5(s) + H 2 O (g) ⇌ 2HCl (g) + POCl 3(g) [HCl] = 0.200 moles = 0.10 M 2.0 L [POCl 3 ] = 0.400 moles = 0.20 M 2.0 L = 0.20 M 0.40 moles 2.0 L [H 2 O] = Keq = [HCl] 2 [POCl 3 ] [H 2 O] Keq = [0.10] 2 [0.20] [0.20] Keq = 0.010 No ICE at equilibrium

6. 3. If 0.600 mole of SO 3 and 0.0200 mole of SO 2 are found in a 2.00 L container at equilibrium at 25 o C. Calculate the [O 2 ]. [SO 3 ] = 0.600 mole/2.00 L = 0.300 M [SO 2 ] = 0.0200 mole/2.00 L = 0.0100 M Keq = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] 798=(0.300) 2 (0.0100) 2 [O 2 ] (0.3) 2 = 798(0.01) 2 [O 2 ] [O 2 ] = (0.3) 2 798(0.01) 2 =1.13 M 2SO 2(g) + O 2(g) ⇌ 2SO 3(g) Keq = 798 1

7. 4.When 0.800 moles of SO 2 and 0.800 moles of O 2 a 2.00 L container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) Keq= [SO 3 ] 2 (0.3) 2 =36.0 [SO 2 ] 2 [O 2 ] (0.1) 2 (0.25) Equilibrium concentrations go in the equilibrium equation! Implies initial and not equilibrium concentrations 0.300 M0.250 M0.100 M 0 0.400M I +0.300 M-0.150 M -0.300 MC E - ICE = are placed into x1/2x1/2 x 2 / 2