Equilibrium Calculations Lesson 7. How can we describe an equilibrium system mathematically? Keq = =2.5 The Keq is the equilibrium constant- a number.

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Equilibrium Calculations Lesson 7

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

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!

1.At equilibrium at 25 o C, [SO 3 ] = M. [H 2 O] = 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!

mole PCl 5, 0.40 mole H 2 O, mole HCl, and 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] = moles= 0.10 M 2.0 L [POCl] = 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)

3. If mole of SO 3 and 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 ] = mole/2.00 L = M [SO 2 ] = mole/2.00 L = 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