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# 17.1 Liquid – vapour equilibrium

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1. 17.1 Liquid – vapour equilibrium 17.1.1 Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature changes. 17.1.2 Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of the kinetic theory. 17.1.3 State and explain the relationship between the enthalpy of vaporization, boiling point and intermolecular forces.

2. Physical equilibrium • Rate of vapourisation = rate of condensation • Pressure exerted by the particles in the vapour phase is known as vapour pressure of the liquid. • Changing the liquid’s surface area affects both rates equally = no overall effect on VP. • It can affect how long it takes to reach equilibrium. More surface area = faster rate.

3. Enthalpy of vapourisation • Vapourisation is an endothermic reaction, so there is a minimum amount of energy that is required for the liquid molecules to escape as vapour molecules • They must overcome the attractive forces between molecules (review bonding Unit: intermolecular forces ppt) • ∆Hvap = energy needed to convert 1 mole of liquid to gas for a specific substance.

4. At higher temperatures, more molecules have the minimum Ek to escape the vapour phase and the rate of vapourisation will increase. • Causing an increase in pressure • A liquid boils when its surface pressure = its vapour pressure • Normal boiling point is the temp where VP is equal to standard atmospheric pressure (101.3 kPa) • As pressure is altered, so is the liquid’s BP

5. VP and BP

6. Homogeneous equilibrium: reactants and products are in the same phase • CO2(g) + H2(g) < - - > CO(g) + H2O(g) • Heterogeneous equilibirum: reactants and products are in two or more phases. • NH4HS(s) < - - > NH3(g) + H2S(g) • Kc = [NH3 ][H2S] ignore the solid, a complicated but interesting reason why is found here. Essentially (s) and (l) are considered a value of 1, so can be ignored.

7. Reaction Quotient, Qc • Identical expression to Kc, but its value is calculated using concentrations that aren’t necessarily found at equilibrium. • If Qc> Kc = shifts to reactants • If Qc< Kc = shifts to products • If Qc = Kc = equilibrium

8. Practice Problem: Calculate the Q for the following reaction using these equilibrium concentrations: [N2] = 0.10 M; [H2] = 0.30 M; and [NH3] = 0.20 M. Is the system at equilibrium if Kc of the reaction is 0.40? N2 + 3H2 < -- >2NH3 Answer: Not at equilibrium, Q>K, so shift to left (reactants)

9. Not on our topic 7 test E.12.1 Solve problems relating to the removal of heavy-metal ions, phosphates and nitrates from water by chemical precipitation

10. Solubility equilibrium is any chemical equilibrium between solid and dissolved states of a compound at saturation. • The equilibrium expresses the degree to which the solid is soluble in water • Called solubility product or Ksp

11. AgCl (s) <===> Ag+(aq) + Cl¯ (aq) • Writing the equilibrium expression for Ksp is similar to Kc. • You only include (aq), not (s). Anything (s) or (l) is denoted as worth 1. • Ksp=[Ag+ ][Cl¯] • The concentrations of the two ions are EQUAL because of 1:1 ratio, so if we know its Ksp value we can solve for the concentration of ions easily.

12. You treat the coefficients still as exponents Sn(OH)2 (s) <===> Sn2+ (aq) + 2 OH¯ (aq) Ksp = [Sn2+] [OH¯]2

13. Solubility Rules (know these) • All CO32- and PO43- compounds, except Li+, Na+, K+ and NH4+ are insoluble. • All S2- compounds, except Group 1,2 or NH4+ are insoluble. • All OH- and O2- compounds, except Sr2+, Ca2+ and Ba2+ are insoluble. • All Cl-, Br- and I- compounds, except Ag+, Hg22+, and Pb2+ are soluble. • All SO42- compounds, except Sr2+, Ca2+ , Pb2+ and Ba2+ are soluble. • Most nitrates (NO3-) are soluble • Most salts of Na+, K+ and NH4+ are soluble

14. Write the chemical equation showing how the substance dissociates and write the Ksp expression 1) Na3PO4 2) K2SO43) Na2S4) (NH4)3PO45) Sr(OH)2

15. Calculations The coefficient does two things: 1) it puts a power on the 'x' which represents that particular ion and2) it puts a coefficient in front of the 'x' Example Problem: Calculate the molar solubility (in mol/L) of a saturated solution of Sn(OH)2 knowing its Ksp is 5.45 x 10¯27

16. Solve • Sn(OH)2 <===> Sn2+ + 2 OH¯ • Ksp = [Sn2+] [OH¯]2 • One Sn2+ makes two OH¯. That means that if 'x' Sn2+ dissolves, then '2x' of the OH¯ had to have dissolved, so 5.45 x 10¯27 = (x) (2x)2 4x3 = 5.45 x 10¯27 x=1.11 x 10¯9 mol/L

17. Solve for Fe(OH)3 knowing its Ksp is 2.64 x 10¯39 • Iron(III) hydroxide • Fe(OH)3 <===> Fe3+ + 3OH¯ • Ksp = [Fe3+] [OH¯]3 • 2.64 x 10¯39 = (x) (3x)3 • 27 x4 = 2.64 x 10¯39 • x = 9.94 x 10¯11 M

18. Solubility and Ksp • Solubility of a substance is the quantity that dissolves to form a saturated solution • It can be measured in g/L or mol/L • Ksp is the constant which describes the equilibrium between the ionic solid and its ions in a saturated solution.

19. Flowchart Solubility of compound (g/L) Molar solubility of compound (mol/L) Molar conc. of ions (mol/L) Ksp

20. Precipitation and Separation of Ions • When two solutions are mixed to form a precipitate, we can use Ksp to see if the precipitate will actually form, or if it is slightly soluble enough in the particular scenario to dissolve. • If Q = Ksp, equilibrium exists • If Q > Ksp, precipitate will occur • If Q < Ksp, solid will dissolve

21. Practice: • Will a precipitate form if equal volumes of 3.0 x 10-3 M Pb(NO3)2 is added 5.0 x10-3 M Na2SO4 ? Figure out the double displacement reaction, then decide which is the insoluble/slightly soluble compound made. Determine the solubility product for that insoluble compound (Q) Compare it with the accepted value of Ksp

22. Answer to #1 Pb(NO3)2 + Na2SO4  2NaNO3 + PbSO4 The insoluble compound is PbSO4, so PbSO4 Pb+2 + SO4-2 Q= [Pb+2][SO4-2] Q= [3.0 x 10-3 ][5.0 x10-3 ] Q= 1.5 x 10-5 so, Q>Ksp, so precipitate Ksp = 1.8 x 10-8 occurs Look it up on chart!

23. Practice #2 • Will a precipitate form if 0.10 L of 3.0 x 10-3 M Pb(NO3)2 is added to 0.40L of 5.0 x10-3 M Na2SO4? PbSO4 Pb+2 + SO4-2 • Total volume of solution is 0.10 + 0.40 = 0.50 L • Find concentrations involved:

24. 0.10 L x 3.0 x 10-3 mol/L = 3.0 x 10-4 mol • 3.0 x 10-4 mol/ 0.50 L = 6.0 x 10-4 M • 0.40 L x 5.0 x10-3 mol/L = 2.0 x 10-3 mol • 2.0 x 10-3 mol/ 0.50 L = 4.0 x 10-3 M • Q= [6.0 x 10-4 ][4.0 x 10-3 ]=2.4 x 10-6 • Q > Ksp so precipitate will form