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Equilibrium Calculations Lesson 7. How can we describe an equilibrium system mathematically ?. products. reactants. products. = 3.0. Keq =. ⇌. reactants. The Keq is the equilibrium constant - a number that does not change. Providing the temperature is kept constant.
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Equilibrium Calculations Lesson 7
How can we describe an equilibrium system mathematically? products reactants products = 3.0 Keq = ⇌ reactants The Keq is the equilibrium constant- a number that does not change. Providing the temperature is kept constant.
Equilibrium Calculations An equilibrium system, at any given temperature, can be described by an equilibrium expressionand equilibrium constant. aA + bB ⇌cC + dD Products Keq= Reactants [C]c[D]d Keq = [A]a[B]b Equilibrium Constant- a number Expression- mathematical equation (aq) and (g) are included! (l) and (s) are not-constant concentration!
1.at 25oC, [SO3] = 0.200 M. [H2O] = 0.480 M, and [H2SO4] = 24 M. Calculate the Keq. At equilibrium No ICE SO3(g) + H2O(g) ⇌ H2SO4(l) 1 Keq = don’t count (l)! Use 1 [SO3] [H2O] 1 = (0.200)(0.480) = 10.4 The Keq has no units but concentration units that go in the expression must be M!
2. 0.500 mole PCl5, 0.40 mole H2O, 0.200 mole HCl, and 0.400 mole POCl3 are found in a 2.0 L container at 125 oC. Calculate the Keq. at equilibrium No ICE PCl5(s) + H2O(g)⇌ 2HCl(g) + POCl3(g) 0.200 moles [HCl] = = 0.10 M 2.0 L [HCl]2[POCl3] 0.400 moles Keq = = 0.20 M [POCl3] = 2.0 L [H2O] [H2O] = 0.40 moles = 0.20 M 2.0 L [0.10]2[0.20] Keq = [0.20] Keq = 0.010
3. If 0.600 mole of SO3 and 0.0200 mole of SO2 are found in a 2.00 L container at equilibrium at 25 oC. Calculate the [O2]. 2SO2(g) + O2(g)⇌ 2SO3(g) Keq = 798 [SO 3]2 Keq = [SO3] = 0.600 mole/2.00 L = 0.300 M [SO2]2[O2] [SO2] = 0.0200 mole/2.00 L = 0.0100 M 798 = (0.300)2 1 (0.0100)2[O2] (0.3)2 = 798(0.01)2[O2] [O2] = (0.3)2 798(0.01)2 =1.13 M
4. When 0.800 moles of SO2 and 0.800 moles of O2 a 2.00 L container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq value. are placed into x1/2 x 2/2 Implies initial and notequilibrium concentrations - ICE 2SO2 (g) + O2 (g)⇋ 2SO3 (g) I 0.400 M 0.400M 0 -0.150 M +0.300 M C -0.300 M E 0.100 M 0.250 M 0.300 M Equilibrium concentrations go in the equilibrium equation! (0.3)2 [SO3]2 = = 36.0 Keq = (0.1)2(0.25) [SO2]2[O2]