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CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub

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CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART II Chapter Description 14 Chemical Equilibrium 15 Acids and Bases 16 Acid-Base Equilibrium 17 Solubility and Complex-Ion Equilibrium 19 Electrochemistry 20 Nuclear Chemistry CH. 14 CHEMICAL EQUILIBRIUM • DYNAMIC EQUILIBRIUM: WHEN FORWARD AND REVERSE REACTION RATES ARE THE SAME. *PURE SOLIDS AND LIQUIDS DO NOT APPEAR IN THE EQUILIBRIUM CONSTANT EXPRESSION (ANY K OR Q) • THE EQUILIBRIUM CONSTANT KC: THE VALUE OBTAINED FOR THE EQUILIBRIUM-CONSTANT EXPRESSION WHEN EQUILIBRIUM CONCENTRATIONS ARE SUBSTITUTED. • THE EQUILIBRIUM CONSTANT KP: THE EQUILIBRIUM CONSTANT EXPRESSION FOR A GASEOUS REACTION IN TERMS OF PARTIAL PRESSURES. ΔN FOR GASES: KP = KC (RT) , WHERE ΔN = DIFFERENCE OF COEFFICIENT. Ch. 14 Chemical Equilibrium CALCULATING VALUES OF K • IF 2 OR MORE REACTIONS ARE ADDED TO ACHIEVE A GIVEN REACTION, THE EQUILIBRIUM CONSTANT FOR THE GIVEN EQUATION EQUALS THE PRODUCT OF THE EQUILIBRIUM CONSTANTS (K) OF THE ADDED EQUATIONS. K1xK2 • IF REACTION IS REVERSED YOU TAKE THE INVERSE OF ORIGINAL K OF REACTION. 1/K • IF MULTIPLIED OR DIVIDED, RAISE K TO THAT POWER. EX: MULTIPLY BY 2 K2 EX: DIVIDE BY 3 K1/3 Ch. 14 Chemical Equilibrium USING QC TO DETERMINE THE DIRECTION OF EQUILIBRIUM • REACTION QUOTIENT (QC): THE INITIAL REACTION RATE OF A REACTION. ITS NOT A CONSTANT BUT DYNAMIC. • THE RELATIONSHIP BETWEEN QC AND KC GIVES THE DIRECTION OF THE EQUILIBRIUM. • IF KC > QC EQUILIBRIUM IS IN THE FORWARD DIRECTION. • IF KC = QC REACTION IS IN EQUILIBRIUM. • IF KC < QC EQUILIBRIUM IS IN THE REVERSE DIRECTION. • IF QC = 0 ONLY REACTANTS ARE PRESENT. • IF QC = ∞ ONLY PRODUCTS ARE PRESENT. Ch. 14 Chemical Equilibrium 1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide: N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K a) What is the value of Kc for this reaction? b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g) c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)? Ch. 14 Chemical Equilibrium 1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide: N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K a) What is the value of Kc for this reaction? b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g) c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)? Δn 2 a) Kp = Kc (RT) b) Reaction reversed then take the c) (PNO2) Kp inverse of Kp Kp = _________________ K = _______ c 1/ K = 1/ 0.660 (PN2O4 ) (RT)Δn p (0.223) 2 0.660 0.660 = ___________ Kc = ________________ 2-1 (0.0821 x 319 ) (PN2O4 ) Ch. 14 Chemical Equilibrium 2. If a 2.50 L vessel at 1000˚C containes 0.525 mol CO2, 1.25 mol CF4, and 0.75 mol COF2, in what direction will a net reaction occur to reach the equilibrium? CO2 (g) + CF4 (g) ↔ 2COF2 (g) Kc = 0.50 @ 1000 K Solution: How do you determine the direction of a reaction in order to reach equilibrium? 2 [COF2] Qc = ___________ [CO2] [CF4] [CO2] = 0.525 mol / 2.5 L = 0.21 M [CF4] = 1.25 mol / 2.5 L = 0.5 M [COF2] = 0.75 mol / 2.5 L = 0.3 M [0.3]2 Qc = ___________ [0.21] [0.5] Kc = 0.50 Ch. 14 Chemical Equilibrium 3. Starting with 0.100 mol each of CO and H2O in a 5.00 L flask, equilibrium is established in the following reaction at 600K: CO(g) + H2O(g) ↔ CO2 (g) + H2 (g) Kc = 23.2 at 600K What is the concentration of hydrogen at equilibrium? Ch. 14 Chemical Equilibrium Solution 2 [X] [CO] = 0.100 mol/ 5 L = 0.02 M 23.2 = _________________ [H2O] = 0.100 mol / 5 L = 0.02 M 2 [0.02-X] Balanced CO(g) + H2O(g) ↔ CO2 (g) H2 (g) Equation *Take the square root of both sides: Initial 0.02 0.02 0 0 [X] 4.82 = _________________ Change - x -x +x +x [0.02-X] Equilibrium 0.02-x 0.02-x x x 0.096 -4.82X = X [CO2] [H2] 0.096 = X + 4.82X 5.82X = 0.096 c _________________ K = X = 0.096/ 5.82 [CO] [H2O] X = 0.0165 SO [H2]= X at equilibrium [X] [X] 23.2 = ______________ [0.02-X] [0.02-X] Ch. 14 Chemical Equilibrium -7 4. The Keq of a reaction is 4x10 . At equilibrium_____. a) The products are favored b) The reactants are favored c) The reactants and products are present in equal amounts d) The rate of the forward and reverse reaction are the same Ch. 14 Chemical Equilibrium -7 4. The Keq of a reaction is 4x10 . At equilibrium_____. a) The products are favored c) The reactants and products are present in equal amounts d) The rate of the forward and reverse reaction are the same Explanation: Here Keq= 0.0000004 Keq<< 1 Therefore Reactants are favored!! What if Keq >> 1????? Ch. 14 Chemical Equilibrium -7 4. The Keq of a reaction is 4x10 . At equilibrium_____. a) The products are favored c) The reactants and products are present in equal amounts d) The rate of the forward and reverse reaction are the same Explanation: Here Keq= 0.0000004 Keq<< 1 Therefore Reactants are favored!! What if Keq >> 1????? Products would be favored Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Remove Cl2 Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar No effect Decrease V (or increase P) Increase T Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar No effect Decrease V (or increase P) Left High to Low #of moles of Gases 1 mol ← 2 mol Increase T Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar No effect Decrease V (or increase P) Left High to Low #of moles of Gases 1 mol ← 2 mol Increase T Right Add Catalyst Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar No effect Decrease V (or increase P) Left High to Low #of moles of Gases 1 mol ← 2 mol Increase T Right Add Catalyst No effect on equilibrium *Catalysts Lowers Activation Energy only Ch. 14 Chemical Equilibrium 5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill in the following table: Changes Shifts RX'N which way? Add PCl5 Right Remove Cl2 Right Add Ar No effect Decrease V (or increase P) Left High to Low #of moles of Gases 1 mol ← 2 mol Increase T Right Add Catalyst No effect on equilibrium *Catalysts Lowers Activation Energy only Which of the top changes would cause a change in K? • ONLY a change in Temperature can cause a change in K Ch. 14 Chemical Equilibrium CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH-] Example: Identify following as Arrhenius Acid/Base HCl NaOH Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH-] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: Identify following as Bronsted-Lowry Acid/Base :NH3 HCl Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH-] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: HCl (Acid) :NH3 (Base) Lewis Definitions: 1) Acid: is an electron pair acceptor. 2) Base: is an electron pair donor. Example: Identify following as Lewis Acid/Base + Ag :NH3 Ch. 15 Acids and Bases CH. 15 ACIDS AND BASES Arrhenius Definitions: 1) Acid: substance that when dissolved in water increases [H+] 2) Base: substance that when dissolved in water increases [OH-] Example: HCl (Acid) NaOH (Base) Brønsted-Lowry Definitions: 1) Acid: substance that donates a proton, H+, in a reaction 2) Base: substance that accepts a proton, H+, in a reaction Example: HCl (Acid) :NH3 (Base) Lewis Definitions: 1) Acid: is an electron pair acceptor.
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