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Equilibrium Law

Equilibrium Law. Mass action expression. Equilibrium constant. Kc =. Introduction to the Equilibrium law. [H 2 O]. [H 2 O] 2. [H 2 ] 2 [O 2 ]. [H 2 ] 2 [O 2 ]. Read 14.3 to PE1. 2H 2 (g) + O 2 (g)  2 H 2 O (g). 2 H 2 (g) + O 2 (g)  2H 2 O (g). 2H 2 (g) + O 2 (g)  2H 2 O (g).

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Equilibrium Law

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  1. Equilibrium Law

  2. Mass action expression Equilibrium constant Kc = Introduction to the Equilibrium law [H2O] [H2O]2 [H2]2[O2] [H2]2[O2] • Read 14.3 to PE1 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) Step 1: Set up the “equilibrium law” equation Step 2: Product concentrations go in numerator Step 3: Concentration in mass action expression is raised to the coefficient of the product Step 4: Reactant concentrations go in denominator Step 5: Concentrations in mass action expression are raised to the coefficients of reactants

  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. Kc = Kc = [0.00261] ,Kc= [0.105][0.250]2 [0.150] ,Kc= [0.0222][0.0225] Equilibrium law: RE 14.31, 14.35 [CH3OH] CO(g) + 2H2(g)  CH3OH(g) =0.398 [CO][H2]2 C2H4(g) + H2O(g)  C2H5OH(g) [C2H5OH] = 300 [C2H4][H2O]

  5. Kc = Kc = [x] ,Kc= [0.210][0.100]2 [0.280]2 ,Kc= [0.00840][x]3 Equilibrium law: RE 14.36, 14.37 [CH3OH] = 0.500 CO(g) + 2H2(g)  CH3OH(g) [CO][H2]2 x = [0.210][0.100]2 (0.500) = 0.00105 M N2(g) + 3H2(g)  2NH3(g) [NH3]2 = 64 [N2][H2]3 x3 = [0.280]2/ [0.00840] (64) = 0.146 x3 = 0.146 x = 0.53 M

  6. Kc = [0.175] ,Kc= [0.0197][0.0200] 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: C2H4(g) + H2O(g)  C2H5OH(g) If Kc = 300, []s = 0.0197 M, 0.0200 M, 0.175 M which direction will the reaction need to shift? [C2H5OH] = 300 [C2H4][H2O] 444  300 444 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. PCl3 (g) + Cl2(g)  PCl5(g) Kc = Kc = [0.00600] ,Kc= [0.0520][0.0140] [0.0100][0.0400] ,Kc= [0.00150][0.00300] More equilibrium law problems • Do RE 14.32, 33 (pg. 589). For each, state in which direction the reaction needs to shift [PCl5] = 0.18 [PCl3][Cl2] 8.24  0.18 Top must , bottom must -shift to left is needed SO2(g) + NO2(g)  NO(g) +SO3(g) [NO] [SO3] = 85 [SO2][NO2] 88.9  85 Top must , bottom must -shift to left is needed For more lessons, visit www.chalkbored.com

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