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Published bySheena Barber Modified over 8 years ago
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Equilibrium
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Ammonia N 2 + 3H 2 2NH 3
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Ammonia 3H 2 (g) + N 2 (g) 2NH 3 (g)
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N 2 O 4 (g) 2NO 2 (g)
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Chemical Equilibrium reaction that proceeds forward (reactants become products) and reverse (“products” become “reactants”) at the same rate does NOT mean that amounts of reactants and products are equal
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Equilibrium Constant an expression and resulting numerical value that describes the position of equilibrium for a given reaction K eq K sp KcKc KpKp KaKa KbKb
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Equilibrium Constant K c = 1.9 x 10 19
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Equilibrium Constant K c = 4.1 x 10 -31
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Equilibrium Constant K >> 1 K << 1 K ≈ 1 equilib favors products equilib favors reactants equilib midway btw
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Equilibrium Constant aW + bX cY + dZ K c = [Y] c [Z] d [W] a [X] b only solutions and gases appear in the equilibrium constant expression, NOT solids and pure liquids.
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Equilibrium Constant ClNO 2 (g) + NO(g) NO 2 (g) + ClNO(g) rate forward = rate reverse k f [ClNO 2 ][NO] = k r [NO 2 ][ClNO] k f [NO 2 ][ClNO] k r [ClNO 2 ][NO] = = K eq
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Equilibrium Constant Write the equilibrium constant expression for the following reactions: 3H 2 (g) + N 2 (g) 2NH 3 (g) CaCO 3 (s) CaO(s) + CO 2 (g)
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H 2 (g) + I 2 (g) 2HI(g) Calculate a numerical value for K eq if the equilibrium concentrations are [H 2 ] = 0.11 M [I 2 ] = 0.11 M [HI] = 0.78 M
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CH 3 CO 2 H (aq) + C 2 H 5 O (aq) CH 3 CO 2 C 2 H 5(aq) + OH - (aq) At 25 C, K c = 4.10. In an equilibrium mixture, the following concentrations were measured: [CH 3 CO 2 H] = 0.210 M [CH 3 CO 2 C 2 H 5 ] = 0.910 M [OH - ] = 0.00850 M What was the equilibrium concentration of ethanol?
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Comparing Multiple Equilibria reversible reaction written in opposite direction ½ N 2 O 4 NO 2 K forward = [NO 2 ] [N 2 O 4 ] 1/2 K forward = 0.11 NO 2 ½ N 2 O 4 K reverse = K reverse =
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Comparing Multiple Equilibria reversible reaction written in opposite direction = equilibrium constant is inverted
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Comparing Multiple Equilibria eqn that is a multiple of another equilib rxn ½ N 2 O 4 NO 2 K c = [NO 2 ] [N 2 O 4 ] 1/2 K c = 0.11 N 2 O 4 2NO 2 K c ’= K c ’=
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Comparing Multiple Equilibria eqn that is a multiple of another equilib rxn = equilib constant raised to that factor
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Comparing Multiple Equilibria equilib rxns added together to get an overall reaction A 2B 2B 3C KcKc KcKc
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Comparing Multiple Equilibria equilib rxns added together to get an overall reaction = equilib constant of overall reaction is the product of the individual equilib constants
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Do this without your calculator! Given the following equilibrium reactions: C 2 H 2 O 4 ↔ H + + C 2 HO 4 - K c = 6.5 x 10 -2 C 2 HO 4 - ↔ H + + C 2 O 4 2- K c = 6.1 x 10 -5 Calculate K c for the following reaction: C 2 H 2 O 4 ↔ 2H + + C 2 O 4 2-
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Reaction Quotient Q written the same as an equilib constant, but NOT at equilib conditions
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Reaction Quotient H 2 (g) + I 2 (g) 2HI(g)K c = 60. (at 350 o C) [H 2 ] = [I 2 ] = 0.010M [HI] = 0.050M
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Comparing Q and K Q = K Q < K K > Q Q > K K < Q rxn at equilb proceeds toward products proceeds toward reactants
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PCl 3 (g) + Cl 2 (g) PCl 5 (g) At a given temperature, the following equilibrium concentrations were observed: [PCl 3 ] eq = 0.00300 M [Cl 2 ] eq = 0.00200 M [PCl 5 ] eq = 0.00670 M Determine the value of the equilibrium constant at this temperature.
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At the same temperature, some Cl 2 (g), PCl 3 (g), and PCl 5 (g) are mixed at the following concentrations: [Cl 2 ] i = 0.0050 M [PCl 3 ] i = 0.063 M [PCl 5 ] i = 0.0038 M In which direction will the reaction proceed to achieve equilibrium?
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ICE Tables I nitial (Concentrations) C hange E quilibrium (Concentrations)
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