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Equilibrium Law. Introduction to the Equilibrium law Read 14.3 to PE1 2H 2 (g) + O 2 (g) 2H 2 O (g) Step 1:Set up the equilibrium law equation Kc = Step.

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Presentation on theme: "Equilibrium Law. Introduction to the Equilibrium law Read 14.3 to PE1 2H 2 (g) + O 2 (g) 2H 2 O (g) Step 1:Set up the equilibrium law equation Kc = Step."— Presentation transcript:

1 Equilibrium Law

2 Introduction to the Equilibrium law Read 14.3 to PE1 2H 2 (g) + O 2 (g) 2H 2 O (g) Step 1:Set up the equilibrium law equation Kc = Step 2:Product concentrations go in numerator [H 2 O] 2H 2 (g) + O 2 (g) 2H 2 O (g) [H 2 O] 2 2H 2 (g) + O 2 (g) 2H 2 O (g) Step 3:Concentration in mass action expression is raised to the coefficient of the product 2H 2 (g) + O 2 (g) 2H 2 O (g) [H 2 ] 2 [O 2 ] Step 4:Reactant concentrations go in denominator Step 5:Concentrations in mass action expression are raised to the coefficients of reactants 2H 2 (g) + O 2 (g) 2H 2 O (g) [H 2 ] 2 [O 2 ] Mass action expression Equilibrium constant

3 Equilibrium law: important points State (g, l, s, aq) may or may not be added at this point since we will only be dealing with gasses for this section. Later it will matter. The equilibrium law includes concentrations of products and reactants in mol/L (M) The value of Kc will depend on temperature, thus this is listed along with the Kc value Tabulated values of Kc are unitless By substituting equilibrium concentrations into equilibrium law, we can calculate Kc … Do RE 14.31, 35, 36, 37 (pg. 589)

4 CO(g) + 2H 2 (g) CH 3 OH(g) [CH 3 OH] Equilibrium law: RE 14.31, Kc = [CO] [H 2 ] 2 [ ], Kc = [0.105] [0.250] 2 =0.398 C 2 H 4 (g) + H 2 O(g) C 2 H 5 OH(g) [C 2 H 5 OH] [C 2 H 4 ] [H 2 O] [0.150], Kc = [0.0222] [0.0225] Kc = = 300

5 [CH 3 OH] Equilibrium law: RE 14.36, CO(g) + 2H 2 (g) CH 3 OH(g) Kc = [CO] [H 2 ] 2 [x], Kc = [0.210] [0.100] 2 = N 2 (g) + 3H 2 (g) 2NH 3 (g) [NH 3 ] 2 [N 2 ] [H 2 ] 3 [0.280] 2, Kc = [ ] [x] 3 Kc = = 64 x = [0.210] [0.100] 2 (0.500) = M x 3 = [0.280] 2 / [ ] (64) =0.146 x 3 = 0.146x = 0.53 M

6 When Kc mass action expression We can use the equilibrium law to determine if an equation is at equilibrium or not If mass action expression equals equilibrium constant then equilibrium exists Q - consider: C 2 H 4 (g) + H 2 O(g) C 2 H 5 OH(g) If Kc = 300, []s = M, M, M which direction will the reaction need to shift? [C 2 H 5 OH] [C 2 H 4 ] [H 2 O] [0.175], Kc = [0.0197] [0.0200] Kc = = must be reduced to 300. Thus, the top must decrease and the bottom must increase. A shift to left is required to establish equilibrium.

7 More equilibrium law problems Do RE 14.32, 33 (pg. 589). For each, state in which direction the reaction needs to shift [PCl 5 ] [PCl 3 ] [Cl 2 ] [ ], Kc = [0.0520] [0.0140] Kc = = Top must, bottom must - shift to left is needed SO 2 (g) + NO 2 (g) NO(g) +SO 3 (g) [NO] [SO 3 ] [SO 2 ] [NO 2 ] [0.0100][0.0400], Kc = [ ] [ ] Kc = = Top must, bottom must - shift to left is needed PCl 3 (g) + Cl 2 (g) PCl 5 (g) For more lessons, visit


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