Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

CHAPTER 15 Chemical Equilibrium Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K) Reaction Quotient (Q) Kc vs Kp ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle van’t Hoff Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Le Chatelier’s Principle Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium positions Combination of concentrations that allow Q = K Le Chatelier Definition Equilibrium positions Combination of concentrations that allow Q = K Infinite number of possible equilibrium positions Le Châtelier’s principle System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress System said to “shift to right” when forward reaction is dominant (Q < K) System said to “shift to left” when reverse direction is dominant (Q > K) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Q = K reaction at equilibrium Q < K reactants go to products Le Chatelier Q & K Relationships Q = K reaction at equilibrium Q < K reactants go to products Too many reactants Must convert some reactant to product to move reaction toward equilibrium Q > K products go to reactants Too many products Must convert some product to reactant to move reaction toward equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Equilibrium mixture is blue-green Add excess Cl– (conc. HCl) Equilibrium shifts to products Makes more yellow CuCl42– Solution becomes green Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Add Ag+ Removes Cl–: Ag+(aq) + Cl–(aq)  AgCl(s) Equilibrium shifts to reactants Makes more blue Cu(H2O)42+ Solution becomes increasingly more blue Add H2O? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration: Example Le Chatelier Change in Concentration: Example For the reaction 2SO2(g) + O2(g) 2SO3(g) Kc = 2.4 × 10–3 at 700 °C Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture? Towards the products Towards the reactants No change will occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Concentration Le Chatelier Change in Concentration When changing concentrations of reactants or products Equilibrium shifts to remove reactants or products that have been added Equilibrium shifts to replace reactants or products that have been removed Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Pressure or Volume Le Chatelier Change in Pressure or Volume Consider gaseous system at constant T and n 3H2(g) + N2(g) 2NH3(g) If volume is reduced Expect pressure to increase To reduce pressure, look at each side of reaction Which has less moles of gas Reactants = 3 mol + 1 mol = 4 mol gas Products = 2 mol gas Reaction favors products (shifts to right) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Pressure or Volume Le Chatelier Change in Pressure or Volume Consider gaseous system at constant T and n H2(g) + I2(g) 2HI(g) If pressure is increased, what is the effect on equilibrium? nreactant = 1 + 1 = 2 nproduct = 2 Predict no change or shift in equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Pressure or Volume Le Chatelier Change in Pressure or Volume 2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g) If you decrease volume of reaction, what is the effect on equilibrium? Reactants: All solids, no moles gas Products: 2 moles gas Decrease in V, causes an increase in P Reaction shifts to left (reactants), as this has fewer moles of gas Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Change in Pressure or Volume Le Chatelier Change in Pressure or Volume Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can Increasing pressure Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids Substances are already almost incompressible Changes in V, P and [X ] effect position of equilibrium (Q), but not K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Le Chatelier Change in Temperature Ice water Boiling water Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Reaction endothermic Adding heat shifts equilibrium toward products Cooling shifts equilibrium toward reactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Add heat energy, shift reaction right Le Chatelier Change in Temperature Hf°=+6 kJ (at 0 °C) Energy + H2O(s) H2O(l ) Energy is reactant Add heat energy, shift reaction right 3H2(g) + N2(g) 2NH3(g) Hrxn= –47.19 kJ 3 H2(g) + N2(g) 2 NH3(g) + energy Energy is product Add heat, shift reaction left H2O(s) H2O(l) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier Change in Temperature Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Increase in temperature of exothermic reaction makes K smaller Le Chatelier Change in Temperature Changes in T change value of mass action expression at equilibrium, so K changed K depends on T Increase in temperature of exothermic reaction makes K smaller More heat (product) forces equilibrium to reactants Increase in temperature of endothermic reaction makes K larger More heat (reactant) forces equilibrium to products Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Catalyst lowers Ea for both forward and reverse reaction Le Chatelier Change with Catalyst Catalyst lowers Ea for both forward and reverse reaction Change in Ea affects rates k r and k f equally Catalysts have no effect on equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Addition of Inert Gas at Constant Volume Le Chatelier Addition of Inert Gas at Constant Volume Inert gas One that does not react with components of reaction e.g. argon, helium, neon, usually N2 Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products Since it doesn’t react with anything No change in concentrations of reactants or products No net effect on reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

How To Use Le Chatelier’s Principle Write mass action expression for reaction Examine relationship between affected concentration and Q (direct or indirect) Compare Q to K If change makes Q > K, shifts left If change makes Q < K, shifts right If change has no effect on Q, no shift expected Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq) Group Problem Consider: H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq) What will happen if PO43– is removed? Q is proportional to [PO43–] Decrease [PO43–], decrease in Q Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic. Group Problem The reaction H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic. What will happen if system is cooled? heat Since reaction is exothermic, heat is product Heat is directly proportional to Q Decrease in T, decrease in Q Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pink blue Group Problem The equilibrium between aqueous cobalt ion and the chlorine ion is shown: [Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pink blue It is noted that heating a pink sample causes it to turn violet. The reaction is: endothermic exothermic cannot tell from the given information Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products? Ka = 2.2 × 10–3 Ka = 1.8 × 10–5 Ka = 4.0 × 10–10 Ka = 6.3 × 10–3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E