Chemical Reactions 2: Equilibrium & Oxidation-Reduction

1. Chemical Reactions 2: Equilibrium & Oxidation-Reduction

2. Equilibrium Constant The equilibrium constant (Kc) is defined as the ratio between the concentrations of the products and the reactants, each raised to the power corresponding to its coefficient in the balanced equation. aA(g) + bB(aq) cC(g) + dD(aq) Products Reactants

3. Equilibrium Constant The equilibrium constant (Kc): _ONLY changes value if we change Temperature _Does NOT have units _ONLY includes gaseous or aqueous species at equilibrium _Solid and Liquids are not included in Kc’s expression as their concentrations do not change _If Kc > 1 (Product favoured equilibrium) _If Kc < 1 (Reactants favoured equilibrium)

4. Equilibrium Constant Write the equilibrium constant (Kc) for the following reactions: • CO(g) + H2O(g) CO2(g) + H2(g) • 2NO2(g) 2NO(g) + O2(g) • 4HCl(g)+ O2(l) 2H2O(g) + 2Cl2(g) • CaCO3(s) CaO(s) + CO2(g)

5. Equilibrium Constant Calculate the equilibrium constant of the reaction below, if we know that at a given temperature the concentration of each substance is: [NO] = 0.062 M, [H2] = 0.012 M, [N2] = 0.019 M and [H2O] = 0.138 M 2NO(g) + H2(g) N2(g) + 2H2O(g) 653.7

6. Equilibrium Constant Calculate the concentration of C2H4 at the equilibrium described by the reaction below, if you know that the equilibrium constant is 300, and the concentration of the other substances are: [C2H5OH] = 0.1500 M and [H2O] = 0.0225 M C2H4(g) + H2O(g) C2H5OH(g) = 0.0222M

7. Equilibrium Constant We inject 1 mole of N2O4 into a 2-L container, and raise the temperature to 120°C. When equilibrium is reached, the gaseous mixture contains 0.4 moles of N2O4. Calculate the value of Kc. (N2O4(g) 2NO2(g)) Initially 1 mole 0 mole Conversion 1-x 2x (1:2 ratio) Equilibrium 0.4 mole 1.2 mole (x = 0.6 mole) Concentration 0.4 mole/2L = 0.2M 1.2 mole/2L = 0.6M Kc = [NO2]2 / [N2O4] Kc = (0.6M)2/ (0.2M) Kc = 1.8

8. Equilibrium Constant 276g of NO2 injected in a 4-L container, and temperature elevated to 120°C. When equilibrium is reached, the gaseous mixture contains 1.2 moles of NO2. Calculate the value of Kc. (2NO2(g) N2O4 (g)) Initially 6 mole (NO2= 46 g/mol) 0 mole Conversion 6-x x/2 (2:1 ratio) Equilibrium 1.2 mole 2.4 mole (x = 4.8 mole) Concentration1.2 mole/4L = 0.3M 2.4 mole/4L = 0.6M Kc = [N2O4] / [NO2]2 Kc = (0.6M) / (0.3M)2Kc = 6.67

9. Equilibrium Constant A technician introduces 2.5 moles of PCl5into a 2-L container, and temperature elevated to 250°C. At this point Kc is 0.042. Calculate the concentration of each substance at equilibrium. (PCl5(g) PCl3(g)+ Cl2(g)) Initially 1.25M (2.5 moles/2L) 0M 0M Conversion -x+ x + x Equilibrium 1.25 - xx x Kc = [PCl3][Cl2] / [PCl5]0.042 = x2/ (1.25 – x) [PCl5] = 1.04M x2 + 0.042x – 0.0525 = 0 [PCl3] = 0.209M Using x = (- b ± √b2-4ac) / 2a [Cl2] = 0.209M x = - 0.251 (impossible) or x = 0.209

10. Equilibrium Constant The solubility product constant (Ksp) is defined as the product of the concentrations of the dissolved ions, each raised to the power corresponding to its coefficient in the balanced equation. aA(s) cC+(aq) + dD-(aq)

11. Equilibrium Constant The solubility product constant (Ksp): _ONLY changes value if we change Temperature _Does NOT have units _ONLY includes aqueous species (ions) at equilibrium _Dissolution equation must be balanced (charges*) _If Ksp > 1 (very soluble) _If Ksp< 1 (rather insoluble)

12. Equilibrium Constant

13. Equilibrium Constant Write the solubility product constant (Ksp) for the following reactions:

14. Equilibrium Constant _The solubility of a compound is the maximum concentration of solute that the solvent can dissolve (saturated solution) _Solubility ONLY changes with temperature _The solubility of a given solid compound determines the value of the solubility product constant at a given temperature

15. Equilibrium Constant

16. Equilibrium Constant Rank the following salts in increasing order of solubility: ZnS, CuS, FeS, CdS, MgCO3, CaCO3 CuS, CdS, ZnS, FeS, CaCO3,MgCO3

17. Equilibrium Constant Determine the value of Ksp of silver chloride if, at 25°C its solubility is 1.3 x 10-5 M. AgCl(s) Ag+(aq) + Cl-(aq) Ksp= [Ag+][Cl-] Ksp= (1.3 x 10-5)(1.3 x 10-5) Ksp= 1.69 x 10-10

18. Equilibrium Constant Determine the value of Ksp of CaF2 if, at 25°C its solubility is 3.5 x 10-4 M. CaF2(s) Ca2+(aq) + 2F-(aq) Ksp= [Ca2+][F-]2 Ksp= (3.5 x 10-4)(2*3.5 x 10-4)2 Ksp= 1.71 x 10-10

19. Equilibrium Constant Determine the value of Ksp of SrF2 if, at 25°C its solubility is 0.0073g in 100 ml. SrF2(s) Sr2+(aq) + 2F-(aq) MM(SrF2) = (1*88) + (2 * 19) = 126 g/mol 126g _____1 mol 0.0073g ______X X = 5.79 x 10-5mol C = n/V = 5.79 x 10-5mol / 0.1 L = 5.79 x 10-4mol/L Ksp= [Sr2+][F-]2 Ksp= (5.79 x 10-4)(2*5.79 x 10-4)2 Ksp= 7.76 x 10-10

20. Equilibrium Constant Precipitation: Separation technique based on the solubility of a given compound. Uses “common ion effect”. aA(s) cC+(aq) + dD-(aq) _Increasing the concentration of either C+ or D-, will shift the equilibrium toward the precipitation of A(s). _Soluble compounds containing the common ion (C+ or D- ) will be used to achieve the separation of the solid. _Precipitation is a technique used to determine ions present in solution

21. Equilibrium Constant You have two colourless solutions of PbCl2 and AgCl. Which of the following solutions would you rather use to identify the PbCl2 using the common ion effect? Explain your reasons. _NaNO3 _HCl _PbI2 Common ion Pb2+ would increase its concentration in solution, displacing the equilibrium toward precipitation of PbCl2. HCl is not useful because it could also precipitate AgCl, and we are only interested in PbCl2.