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Q = reaction quotient Q = K eq only at equilibrium

Q = reaction quotient Q = K eq only at equilibrium If Q<K eq then the forward reaction is favored If Q>K eq then the reverse reaction is favored If K eq >1, the reaction is considered spontaneous. K p. Equilibrium constant when reactants and products are gases.

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Q = reaction quotient Q = K eq only at equilibrium

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  1. Q = reaction quotient Q = Keq only at equilibrium If Q<Keq then the forward reaction is favored If Q>Keq then the reverse reaction is favored If Keq >1, the reaction is considered spontaneous

  2. Kp • Equilibrium constant when reactants and products are gases. • Treated same as Keq, but using atm or kPa rather than molarities.

  3. 2HI(g) D H2(g) + I2(g) • Originally, a system contains only HI at a pressure of 1.00 atm at 520°C. The equilibrium partial pressure of H2 is found to be 0.10atm. Calculate the equilibrium Kp for iodine, hydrogen, and the system. -0.20 +0.10 +0.10 0.10 0.80 0.10

  4. Kp=(.10)(.10)= .0156 (.80)2

  5. Converting Kcto Kp PV=nRT or P=nRT P = [A]RT V Kp= Kc(RT)Δn Δn = (mol gaseous product) – (mol gas reactant) If Δn = 0, then Kp=Kc R is .0821 L·atm/mol·K

  6. 2NO(g) + O2(g) n 2NO2(g) The value for Kc for the above reaction is 5.6 x 1012 at 290K. What is the value of Kp? ∆n=-1 Kp = 5.6 x 1012 [(.0821)(290)]-1 = 5.6 x 1012=2.4 x 1011 (.0821)(290)

  7. Ksp -Solubility product constant- Like Keq but don’t include the solid

  8. NaCl(s) D Na+(aq) + Cl-(aq) Ksp = [Na+][Cl-] For CaCl2(s)D Ca2+(aq) + 2Cl-(aq) Ksp = [Ca2+][Cl-]2

  9. Solubility Product Constant AgBr(s) D Ag+(aq) + Br-(aq) Keq = [Ag+][Br-] [AgBr] Solids are omitted from equilibrium expressions, so. . . Ksp = [Ag+][Br-] For AgBr (at 25°C), Ksp = 5.01 x 10-13 Find [Br-] Hint: [Br-] = [Ag+] [Br-] = 7.08 x 10-7M

  10. What is the solubility expression for a solution of Cu3(PO4)2? Cu3(PO4)2 D 3Cu2+(aq) + 2PO43- Ksp = [Cu2+]3[PO43-]2 What is the [Be2+] in a saturated solution of Be(OH)2? Ksp = 1.58 x 10-22 For every Be2+ ion, there are 2OH- ions. 1.58 x 10-22 = x(2x)2 = 4x3 X = 3.41 x 10-8 M

  11. Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it. PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = [Pb2+][Cl-]2 Convert the amount of dissolved lead(II) chloride into moles per liter. 0.2207 g PbCl2 x(1 mol PbCl2/278.1 g PbCl2)= 7.94 x 10-4 mol 7.94 x 10-4 mol = 0.0159 M PbCl2 .0500L

  12. create an "ICE" table Substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp. Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

  13. Find [IO3-] in a saturated solution of copper (II) iodate. • Ksp of Cu(IO3)2 is 7.41 x 10-8 • Cu(IO3)2D Cu2+ + 2IO3- • If [Cu2+]=x then [IO3-]= • Ksp=[Cu2+][IO3-]2 = x(2x)2 = 4x3 7.41 x 10-8=4x3 X= 2.65 x 10-3M and [IO3-] = 5.30 x 10-3M 2x

  14. Common Ion Effect • Ksp is unchanged by the addition of a solute. • The solubility of a slightly soluble salt is reduced by the presence of a second solute that produces a common ion. • Addition of a common ion will decrease the concentration of the ion it bonds with, creating a precipitate. • If the ions’ concentrations > Ksp, ppt will form

  15. What is the [Tl+] when .050mol of NaBr is added to 500.0ml of a saturated solution of TlBr? • What should happen according to LeChatelier’s Principle? TlBr D Tl+ Br- • Ksp= 3.39 x 10-6

  16. Ksp = [Tl+][Br-] = 3.39 x 10-6 • 3.39 x 10-6 = x (.10 +x) • X ͌ 0.00184 so .10 + 0.00184 = .10 • 3.39 x 10-6 = x (.10) 3.39 x 10-5 = x

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