equilibrium-law-handout

equilibrium-law-handout - OH(g Equilibrium law RE 14.31...

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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 2:Product concentrations go in numerator [H 2 O] 2 Step 3:Concentration in mass action expression is raised to the coefficient of the product [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 Mass action expression Equilibrium constant 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) CO(g) + 2H 2 (g) CH 3
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Unformatted text preview: OH(g) Equilibrium law: RE 14.31, 14.35 C 2 H 4 (g) + H 2 O(g) ↔ C 2 H 5 OH(g) Equilibrium law: RE 14.36, 14.37 CO(g) + 2H 2 (g) ↔ CH 3 OH(g) N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) 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 = 0.0197 M, 0.0200 M, 0.175 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 = = 300 444 ≠ 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. More equilibrium law problems • Do RE 14.32, 33 (pg. 589). For each, state in which direction the reaction needs to shift SO 2 (g) + NO 2 (g) ↔ NO(g) +SO 3 (g) PCl 3 (g) + Cl 2 (g) ↔ PCl 5 (g)...
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This note was uploaded on 04/12/2011 for the course CHEM 1010 taught by Professor Marshall during the Spring '11 term at North Texas.

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