1. Equilibrium Calculations Lesson 7

2. How can we describe an equilibrium system mathematically? Keq = =2.5 The Keq is the equilibrium constant- a number that does not change. Providing the temperature is kept constant. reactants products reactants

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

5. 1.At equilibrium 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. Do not use ice! SO 3(g) + H 2 O (g) ⇌ H 2 SO 4(l) Keq=1 [SO 3 ] [H 2 O] don’t count (l)! Use 1 =1=1 (0.200)(0.480) =10.4 The Keq has no units but concentration units that go in the expression must be M!

6. 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 equilibrium at 25 o C. Calculate the Keq. PCl 5(s) + H 2 O (g) ⇄ 2HCl (g) + POCl 3(g) Do not count the solid! [HCl] = 0.200 moles= 0.10 M 2.0 L [POCl] = 0.400 moles = 0.20 M 2.0 L [H 2 O] = 0.40 moles = 0.20 M 2.0 L Keq= [HCl] 2 [POCl 3 ] [H 2 O] =(0.10) 2 (0.20)=0.010 (0.20)

7. 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 ]. 2SO 2(g) + O 2(g) ⇄ 2SO 3(g) Keq = 798 [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 1[0.0100] 2 [O 2 ] (0.3) 2 =798(0.01) 2 [O 2 ] [O 2 ]= (0.3) 2 =1.13 M 1 798(0.01) 2