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Quick Equilibrium review

Quick Equilibrium review. The Concept of Equilibrium. As the substance warms it begins to decompose: N 2 O 4 ( g )  2NO 2 ( g ) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 ( g )  N 2 O 4 ( g ).

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Quick Equilibrium review

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  1. Quick Equilibrium review

  2. The Concept of Equilibrium • As the substance warms it begins to decompose: • N2O4(g)  2NO2(g) • When enough NO2 is formed, it can react to form N2O4: • 2NO2(g)  N2O4(g). • At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 • The double arrow implies the process is dynamic.

  3. The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate.

  4. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

  5. aA + bB cC + dD [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant To generalize this expression, consider the reaction • The equilibrium expression for this reaction would be

  6. (PC)c (PD)d (PA)a (PB)b Kp = The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

  7. Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp= Kc(RT)n Where n = (moles of gaseous product) − (moles of gaseous reactant)

  8. What Does the Value of K Mean? • If K >> 1, the reaction is product-favored; product predominates at equilibrium. • If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

  9. Kc = = 0.212 at 100C Kc = = (0.212)2 at 100C N2O4(g) 2 NO2(g) 2 N2O4(g) 4 NO2(g) [NO2]4 [N2O4]2 [NO2]2 [N2O4] Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

  10. Kc = = 4.72 at 100C Kc = = 0.212 at 100C N2O4(g) 2 NO2(g) 2 NO2(g) N2O4(g) [NO2]2 [N2O4] [N2O4] [NO2]2 1 0.212 = Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

  11. Applications of Equilibrium Constants • Predicting the Direction of Reaction • We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium • as • where [A], [B], [P], and [Q] are molarities at any time. • Q = K only at equilibrium.

  12. The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

  13. Applications of Equilibrium Constants • Predicting the Direction of Reaction • If Q > K then the reverse reaction must occur to reach equilibrium (go left) • If Q < K then the forward reaction must occur to reach equilibrium (go right)

  14. Applications of Equilibrium Constants • Predicting the Direction of Reaction • If Q > K then the reverse reaction must occur to reach equilibrium (go left) • If Q < K then the forward reaction must occur to reach equilibrium (go right)

  15. Le Châtelier’s Principle • Change in Reactant or Product Concentrations • Adding a reactant or product shifts the equilibrium away from the increase. • Removing a reactant or product shifts the equilibrium towards the decrease. • To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). • We illustrate the concept with the industrial preparation of ammonia

  16. Le Châtelier’s Principle • Effects of Volume and Pressure • The system shifts to remove gases and decrease pressure. • An increase in pressure favors the direction that has fewer moles of gas. • In a reaction with the same number of product and reactant moles of gas, pressure has no effect. • Consider

  17. Le Châtelier’s Principle • Effect of Temperature Changes • Removing heat (i.e. cooling the vessel), favors towards the decrease: • if H > 0, cooling favors the reverse reaction, • if H < 0, cooling favors the forward reaction. • Consider • for which DH > 0. • Co(H2O)62+ is pale pink and CoCl42- is blue.

  18. CATALYST—EQUILIBRIUM is achieved faster, but the equilibrium composition remains unaltered.

  19. Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

  20. Equilibrium and Solubility Equilibrium involving the solubility and precipitation of compounds

  21. Saturated solutions A saturated solution is a solution that is in equilibrium with undissolved solute Example: BaSO4 (s)D Ba+2 (aq)+ SO4-2 (aq)

  22. Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42-] where the equilibrium constant, Ksp, is called the solubility product.

  23. Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42-] where the equilibrium constant, Ksp, is called the solubility product.

  24. Solubility Products • Ksp is not the same as solubility. • Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

  25. Will a Precipitate Form? • In a solution, • If Q = Ksp, the system is at equilibrium and the solution is saturated. • If Q < Ksp, more solid will dissolve until Q = Ksp. • If Q > Ksp, the salt will precipitate until Q = Ksp.

  26. Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

  27. Common Ion Effect If a solution containing two dissolved substances share a common ion, then the solubility of the salt is more difficult to determine Adding “common ion” will cause the solubility to be less in the presence of the common ion Causes less of the substance with the smaller Ksp will dissolve.

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