Tang 03 equilibrium law - keq 2

EQUILIBRIUM LAW - K eq A simple mathematical relationship defines each reaction at chemical equilibrium. H 2(g) + I 2(g) <===> 2 HI (g) K eq = [HI (g) ] 2 [H 2(g) ][I 2(g) ]

EQUILIBRIUM LAW - K eq ,[object Object],[object Object],[object Object],[object Object]

EQUILIBRIUM LAW - K eq ,[object Object],[object Object],[object Object],k eq = [NH 3(g) ] 2 [N 2(g) ][H 2(g) ] 3 k eq = [H 2 O (g) ] 2 [H 2(g) ] 2 [O 2(g) ] k eq = [N 2 O 4(g) ] [NO 2(g) ] 2 Example #1

EQUILIBRIUM LAW - K eq ,[object Object],[object Object],[object Object],ii) k eq = [N 2(g) ][H 2(g) ] 3 [NH 3(g) ] 2 i) k eq = [NH 3(g) ] 2 [N 2(g) ][H 2(g) ] 3

EQUILIBRIUM LAW - K eq If the direction of an equation is reversed, the equilibrium constant is the reciprocal of the original equilibrium constant. K eq forward = 1 / K eq reverse

EQUILIBRIUM LAW - K eq ,[object Object],[object Object],[object Object],i) k eq = [PCl 5(g) ] [PCl 3(g) ][Cl 2(g) ] ii) k eq = [PCl 5(g) ] 2 [PCl 3(g) ] 2 [Cl 2(g) ] 2

EQUILIBRIUM LAW - K eq When the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the exponent of the same factor. K eq rxn ii) = [K eq rxn i) ] x

EQUILIBRIUM LAW - K eq ,[object Object],[object Object],[object Object],k eq = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] k eq = [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 k eq = [NO 2(g) ] 4 [N 2(g) ] 2 [O 2(g) ] 4 K eq final = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 X

EQUILIBRIUM LAW - K eq When chemical equilibria are added together, the equilibrium constants are multiplied together. K eq final rxn = K eq rxn 1 x K eq rxn 2

EQUILIBRIUM LAW - K eq Example #2 At 25°C, K eq = 7.0 x 10 25 for: 2 SO 2(g) + O 2(g) <===> 2 SO 3(g) What is the value of K eq for: SO 3(g) <===> SO 2(g) + ½ O 2(g) K eq = 7.0 x 10 25 inversed, to the power of 0.5 = 1.195 x 10 -13 = 1.2 x 10 -13

EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq The value of K eq (large or small) can provide a hint to the ratio of reactants to products at equilibrium.

EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq Example #3 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

EQUILIBRIUM LAW - K eq EQUILIBRIUM INVOLVING PURE SOLIDS AND LIQUIDS

EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS What is the concentration of a solid or liquid? (i.e. H 2 O) Does the concentration of these pure compounds change? ex. 1 mol NaHCO 3 occupies 38.9 cm 3 2 mol NaHCO 3 occupies 77.8 cm 3 Molar concentration remains the same. Solids and liquids are unaffected by concentration

EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS In the equilibrium law, solids and liquids do not need to be included as it becomes part of the equilibrium constant. NH 3(aq) + H 2 O (l) <===> NH 4 + (aq) + OH - (aq) k eq = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ]

EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS Example #4 ,[object Object],[object Object],[object Object],[object Object],k eq = [CO 2(g) ] k eq = [H 2 O (g) ][CO 2(g) ] k eq = ___ 1__ [SO 2(g) ]

EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS Example #4 ,[object Object],[object Object],k eq = __ 1__ [Cl 2(g) ] k eq = _____ 1______ [NH 3(g) ][HCl (g) ]

EQUILIBRIUM LAW - K eq EFFECT OF TEMPERATURE ON K eq

EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE Will temperature change the value of K eq ? Why or why not? Reactions are endothermic or exothermic and therefore will be affected by the addition or removal of heat.

EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE ,[object Object],[object Object],[object Object],[object Object],k eq = ___ [NH 3(g) ] 2 ___ [H 2(aq) ] 3 [N 2(g) ] Shifts to the left. Since the denominator increases, K eq decreases (is smaller in value) Shifts to the right. Since the numerator increases, K eq increases (is larger in value)

EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE ,[object Object],[object Object],[object Object],[object Object],k eq = [PCl 3(g) ][Cl 2(g) ] [PCl 5(g) ] Shifts to the right. Since the numerator increases, K eq increases (is larger in value) Shifts to the left. Since the denominator increases, K eq decreases (is smaller in value)

Tang 03 equilibrium law - keq 2

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Tang 03 equilibrium law - keq 2