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Equilibrium

Equilibrium. For equilibrium to occur:. System must be closed. Temperature must be constant. Reactions must be reversible (do not go to completion). H 2 (g) + Cl 2 (g) 2HCl(g) + energy No visible change…… A dynamic equilibrium exists.

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Equilibrium

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  1. Equilibrium

  2. For equilibrium to occur: • System must be closed. • Temperature must be constant. • Reactions must be reversible (do not go to completion). • H2(g) + Cl2(g) 2HCl(g) + energy • No visible change…… A dynamic equilibrium exists. • The rate of forward rx. = the rate of the reverse rx. • Homogeneous Equilibria: all gaseous or aqueous phases.

  3. Equilibrium in N2O4(g) + q 2 NO2 (g) • Initially, [N2O4] large. Rate f = Rate r • It decreases rapidly then more slowly. • The [NO2] starts at zero; increases rapidly. concentration N2O4 NO2 • Both eventually, plateau…no change • Equilibrium reached. Time Teq

  4. N2O4(g) + q 2 NO2 (g) What happened here? concentration Here? Time Teq

  5. Predicting Changes Le Chatelier’s Principle: when a stress is applied to a system at equilibrium…. The reaction will “shift” in a direction to minimize the stress. 2H2(g) + O2(g) 2H2O(g) + energy “shift” = rxn speeds up in a particular direction. E.g. shifts right…. Forward rxn speeds up. More products “appear”. We say the reaction is “favored” to the right or the “position of the equilibrium” shifts right. • Types of stress: change in concentrations; change in volume; in temperature…

  6. 2H2(g) + O2(g) 2H2O(g) + energy [H2] Make more [H2O] [O2] Use it up! [H2O] [H2] [O2] Temp Make more pressure Volume [H2O] [O2] [H2] What would a catalyst do to the equilibrium position? Nothing! It would simply be reached sooner. What would happen if we added He gas ? Nothing! It does not affect the partial press of other gases.

  7. N2O4(g) + q 2 NO2 (g) colorless orange Describe what happens when your instructor removes the tube from the freezer, containing the system described above. Explain your observation(s) using LeChatelier’s Principle, and all of the appropriate terminology.

  8. Quantitative Aspect of Equilibrium • Measurements in the N2O4 - NO2 system @ 100C If we divide Equilibrium product values by reactant, each factor raised to the correct power (explained in a moment)…. What do you notice????

  9. Quantitative Aspect of Equilibrium N2O4(g) + q 2 NO2(g)

  10. Mass Action Expression Given the general equation: aA(aq)* + bB(aq)cC(aq) + dD(aq) *also if g = gas [C]c [D]d [A]a [B]b K eq equilibrium constant expression @ constant Temp. ( Keq can also be Kp , Kc , Ksp , Ka , Kb , etc.) Product(s) Right K eq = 0r Reactant(s) Left

  11. What does the K value tell us? Prod. large • If K is > 1 At Equilibrium… there is more product than reactant! Reactant small • If K is < 1 Prod. At Equilibrium… there is more reactant than product! small Reactant Large • If K is = 1 [Products] = [Reactants] • If K is verysmall…..then practically no product • is formed.

  12. Eg. Weak Acid: HC2H3O2 (+ H2O) H+ (aq) + C2H3O2-(aq) [H+] [C2H3O2- ] = 1.8 x 10-5 Ka = [HC2H3O2] What is there more of……reactant or product? Reactant ! HC2H3O2 only 3% ionizes. There is almost no product present. This is why it is a weak acid. Strong acids 100% ionize….Ka is LARGE. Insoluble salts: PbI2 (s) (+ H2O) Pb+2 (aq) + 2 I-(aq) Ksp = [Pb+2][I-]2 = 8.4 x 10-9 What does this value tell us? This solid is very insoluble; mostly solid PbI2 present.

  13. Keqis used to calculate concentrations of species at equilibrium. Given: N2(g) + O2(g) = 2NO(g). At 25°C the Kc = 1.0 x 10-30. [N2] =0.040 & [O2] = 0.010. What is the concentration of NO? [NO]2 M.A.E. : Kc = = 1.0 x 10-30 [N2] [O2] Small K value…….very little product! [NO]2 = [N2] [O2] • 1.0 x 10-30 = (0.040) (0.010) (1.0 x 10-30) = 4.0 x 10-34 [NO] = 2.0 x 10-17mol/L

  14. For the system: CO2 + H2 = CO + H2O Kc = 0.64 @ 900°C. The initial concentrations of reactants are both 0.100mol/L. When the system reaches equilibrium what are the concentrations of reactants and products? Orig. Conc (mol/L) Change in Conc. Equilib. Conc. CO2 H2 CO H2O 0.100 0.100 0.000 0.000 -x 0.100-x -x 0.100-x +x +x x x [CO] [H2O] = x2 Kc = 0.64 = [CO2] [H2] (0.100-x)2 x = 0.044; [CO] = [H2O] = 0.044mol/L [CO2] = [H2] = 0.100 - 0.044 = 0.056 mol/L

  15. Heterogeneous equilibrium • Reactions in which one or more of the substances involved is a pure liquid or solid. CO2(g) + H2(g) CO(g) + H2O (l) • Experimentally, we find, that the position of the equilibrium is independent of the amount of solid or liquid. Adding or removing a liquid or a solid has no effect on the equilibrium. • We do not need to include terms for solids or liquids in the expression for Keq. They are totally ignored. [CO] Thus, Keq = [CO2] [H2]

  16. Solubility & K: (read pgs.597-612) • Calculating solubility from Ksp PbSO4(s) = Pb+2(aq) + SO4-2 (aq) Given, PbSO4 , Ksp= 1.6 x 10-8. Calculate its solubility, [Pb+2] & [SO4-2]. , Ksp = [Pb+2][SO4-2] = 1.6 x 10-8 x2 = 1.6 x 10-8 (since x = [Pb+2] & [SO4-2]) x = 1.3 x 10-4. [Pb+2] & [SO4-2] = 1.3 x 10-4M

  17. Try this problem: Mg(OH)2(s) = Mg+2(aq) + 2OH -1(aq) Ksp = 1.5 x 10 -11 . Calculate the [Mg+2], [OH-1], and the [Mg(OH)2] at equilibrium. If X = [Mg+2] [OH-1] = 2x So… how do we set up the mass action expression? (x) (2x)2 = 1.5 x 10-11 x = 1.6 x10 -4 = [Mg+2]; [OH-1] = 3.2 x10 -4 [Mg(OH)2] = 1.6 x10 -4

  18. More Problems: 1. Bromine(I) chloride gas is formed in an endothermic reaction. At 400°C, after the reaction reaches equilibrium, the mixture contained 0.82M BrCl, 0.20M Br2(g)and 0.48M Cl2(g). a. write the equation for this reaction: 1 Cl2 (g) + 1Br2(g) + q = 2 BrCl (g) b. Write the equilibrium Expression, the MAE [BrCl]2 = Keq [Cl2][Br2] [.82]2 = 7.0 c. Calculate the value for Keq: [.20] [.48] d. What direction is favored?

  19. 2. Will a precipitate form when the following two solutions are combined? (assume volumes are additive.) 0.0025 M Pb(NO3)2 + 0.000036 M NaI The Ksp for PbI2 = 8.4 x 10-9 [Pb+2] = 2.5 x 10-3 , [I-] = 3.6 x 10-5 Reaction Quotient: (2.5 x 10-3) (3.6 x 10-5)2 = 3.24 x 10-12 Since the rx. Quotient < the Ksp , No ppt. Forms!

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