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Chapter 18: Chemical Equilibria

Au 2 O 3 (s) + 2 Fe (s) 2 Au (s) + Fe 2 O 3 (s). Ex:. Chapter 18: Chemical Equilibria. Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up. “Only” the forward reaction occurs.

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Chapter 18: Chemical Equilibria

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  1. Au2O3 (s) + 2 Fe (s) 2 Au (s) + Fe2O3 (s) Ex: Chapter 18: Chemical Equilibria Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up. “Only” the forward reaction occurs. Reversible reactions: Both the forward and the reverse reaction occur to a significant degree. Forward: 3 H2(g) + N2(g) 2 NH3(g) 3 H2(g) + N2(g) 2 NH3(g) Reverse: 3 H2(g) + N2(g)⇋2 NH3(g) anim Equilibrium: A reaction is at equilibrium when Rate forward rxn = Rate reverse rxn

  2. General form of equilibrium constant, Keq: aA + bB ⇋cC + dD If K > 1, then more products than reactants present at equilibrium If K < 1, then more reactants than products present at equilibrium

  3. [NH3] Keq = [N2] [H2] Concentration Time ⇋ N2(g) + 3 H2(g) 2 NH3(g) 2 H2 3 N2 NH3

  4. Heterogeneous equilibria • Occurs when the reactants and products are in more than one state • Because the molar concentration of solids and liquids do not change, SOLIDS AND LIQUIDS NEVER appear in equilibrium constant calculations (they have a concentration value of 1). Ex: BaCl2 (s)⇋ Ba2+(aq) + 2 Cl-(aq) Ksp • Since this particular type of equilibrium involves the solubility of the product, it is given a special designation: • Ksp = solubility product constant

  5. [CO][H2] Keq = [H2O] [CO2] Keq = [CO] Heterogeneous equilibria (cont.) Ex: H2O (l)⇋ H2O (g) • What are the equilibrium constants for the following: • C10H8(s)⇋ C10H8(g) • CaCO3(s)⇋ CaO (s) + CO2(g) • C (s) + H2O (g)⇋ CO (g) + H2(g) • FeO (s) + CO (g)⇋ Fe (s) + CO2(g) Keq = [C10H8] Keq = [CO2]

  6. The equilibrium constant for the reaction below at 700K is 0.44. What is the concentration of the carbon dioxide if [H2O] = 0.16 M, [CO] = 0.15 M and [H2] = 0.14 M? H2O(g) + CO(g) ⇋ H2(g) + CO2(g) 1. Set up the equilibrium expression 2. Substitute in the known values and solve for the unknown. [CO2] = 0.075 M

  7. Molar solubility: the number of moles of the solute per liter in a saturated solution Ex 1: What is the molar solubility of lead(II) iodide if its Ksp is 8.7 x 10–9 ? PbI2(s)  Pb2+ + 2 I– x = the molar solubility x 2x Ksp = [Pb2+][I–]2 8.7 x 10–9 = [x][2x]2 8.7 x 10–9 = 4x3 x = 1.3 x 10–3 M Ex 2: What is the molar solubility of PbI2 if 0.2 M KI is added? [I–] = 0.2 M 8.7 x 10–9 = [x][0.2]2 x = 2.2 x 10–7 M

  8. Ex 3: What is the Ksp of Pb3(PO4)2 if it has a molar solubility of 1.5 x 10-9 M? 1) Write out reaction equation: Pb3(PO4)2 3 Pb2+ + 2 PO43- 2x 3x 2) Write out Ksp expression: Ksp = [Pb2+]3 [PO43-]2 Ksp = (3x)3 (2x)2 Ksp = (27x3)(4x2) Ksp = 108x5 x = 1.5 x 10-9 Ksp = (108)(1.5 x 10-9 )5 Ksp = 8.2 x 10-43

  9. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe+3(aq) + SCN-1(aq) ⇋ FeSCN+2(aq) Colorless ⇋ Dark red Reactants Products Left shift = more reactants (color is lighter) Right shift = more products (color is darker) Initial color

  10. Other ways to cause a Le Châtelier Shift: 3 H2(g) + N2(g) + heat ⇋ 2 NH3(g) What kind of shift would you see if: *Pressure increased? Right shift  [NH3] increased?  Left shift Right shift  Heating temperature increased? *A gas–phase equilibrium will shift to the side of the reaction that takes up less space (smaller coefficient sum) when pressure is increased.

  11. CH4(g) + 2 Cl2(g)⇋ CCl4(g) + 2 H2(g) + heat What kind of shift would you see if: Pressure increased? No Change Heating temperature increased?  Left shift

  12. 2 drops 0.05 M NaSCN 2 drops 0.01 M Fe(NO3)3 3 drops H2O 1 2 3 4 + 2 drops H2O + 1 drop 1 M NaNO3 H2O 0.01 M Fe(NO3)3 0.1 M Fe(NO3)3 0.05 M NaSCN 1 M NaNO3 + 2 drops 0.1 M Fe(NO3)3 + 2 drops 0.05 M NaSCN

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