GASEOUS EQUILIBRIUM CH. 12 EQUILIBRIUM

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1 EQUILIBRIUM I. EQUILIBRIUM IS REACHED WHEN BOTH THE FORWARD AND REVERSE REACTIONS ARE OCCURRING AT THE SAME RATE. A. DYNAMIC EQUILIBRIUM: BOTH REACTIONS ARE STILL OCCURRING BUT THE CONCENTRATION OF REACTANTS ARE CONSTANT. EX. REACTION: 2NO2(g) N2O4(g) B. WHEN THE REACTION STARTS, THE FOLLOWING HAPPEN SIMULTANEOUSLY: 1. REACTANT CONC. 2. FORWARD REACTION RATE 3. PRODUCT CONC. 4. REVERSE REACTION RATE C. AS EQUILIBRIUM IS REACHED, THE DEGREE AT WHICH THE VARIABLES CHANGE (RATE OF THE RATE) DECREASES. D. ONCE EQUILIBRIUM IS REACHED 1. ALL VARIABLES ARE CONSTANT 2. FORWARD AND REVERSE REACTIONS ARE STILL OCCURRING AT A CONSTANT RATE. 3. ALL CONCENTRATIONS ARE CONSTANT. E. EQUILIBRIUM IS SHOWN USING A DOUBLE ARROW: OR. THIS SHOWS THAT BOTH REACTIONS ARE BEING CARRIED OUT. II. MORE ABOUT EQUILIBRIUM A. EQUILIBRIUM CAN BE REACHED FROM ANY DIRECTION 1. IF YOU START WITH ALL PRODUCTS, ALL REACTANTS, OR A PROPORTION OF BOTH, THE RATIO OF PRODUCTS AND REACTANTS WILL REMAIN CONSTANT.

2 EQUILIBRIUM CONSTANT (Keq) I. FIRST, A BIT ABOUT RATE LAWS (EX. RXN: 2NO2(g) N2O4(g)): A. RATE LAW OF FORWARD RXN: RATEf = kf(no2) 2 B. RATE LAW OF REVERSE RXN: RATEr = kr(n2o4) C. AT EQUILIBRIUM, RATEf = RATEr, THEREFORE 1. kf(no2) 2 = kr(n2o4) 2. AFTER REARRANGING, YOU GET: 3. THE RATIO OF RATE CONSTANTS AT A CERTAIN TEMPERATURE IS THE EQUILIBRIUM CONSTANT AT THAT TEMPERATURE THEREFORE, II. IN GENERAL, FOR THE RXN: aa + bb cc + dd A. THE EQUILIBRIUM CONSTANT EXPRESSION IS: OR B. EQUILIBRIUM EXPRESSION NOTES: 1. PURE SOLIDS AND LIQUIDS ARE NOT INCLUDED IN THE EXPRESSION 2. YOU CAN USE PRESSURES OR CONCENTRATIONS TO WRITE THE EXPRESSION a. Keq FOR PRESSURES (ALL GASES- USE PARENTHESES) IS SHOWN AS Kp b. Keq FOR CONC. (SOLUTIONS- USE BRACKETS) IS SHOWN AS Kc c. Keq FOR ACIDS/BASES IS SHOWN AS Ka AND Kb RESPECTIVELY d. Keq FOR PRECIPITATE REACTIONS IS SHOWN AS Ksp 3. IN GENERAL, IT S WRITTEN AS PRODUCTS OVER REACTANTS BUT WHAT THE PRODUCTS AND REACTANTS ARE IS RELATIVE. 4. K CHANGES WITH TEMP.!!! C. EXAMPLES: 1. WRITE THE EQUILIBRIUM EXPRESSIONS FOR THE FOLLOWING: III. CALCULATING USING THE EXPRESSION A. FOR EVERY REACTION, WRITE THE EQUILIBRIUM EXPRESSION. B. FIND OUT WHAT YOU ARE LOOKING FOR AND WHAT S GIVEN. C. PLUG EVERYTHING IN TO FIND YOUR UNKNOWN. D. BE CAREFUL WITH THE EXPONENTS.

3 E. IMPORTANT NOTES: 1. K OF THE FORWARD REACTION IS EQUAL TO THE RECIPROCAL OF THE REVERSE REACTION. N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) 2. IF THE COEFFICIENTS ARE CHANGED, THE EXPONENTS CHANGE! N2O4(g) 2NO2(g) 2N2O4(g) 4NO2(g) 3. THE Keq FOR A MULTISTEP REACTION IS THE PRODUCT OF THE CONSTANTS OF THE INDIVIDUAL STEPS!!! a. Given the following information, determine the value of K c for the reaction IV. RELATING KC TO KP A. REMEMBER: CONCENTRATION IS B. USE PV= nrt TO RELATE THEM C. IF TOTAL MOLES OF REACTANTS AND PRODUCTS (IN THE EQUATION) AND VOLUME DOES NOT CHANGE, KC WILL BE EQUAL TO KP D. PLUGGING IN PV= nrt INTO KC AND KP, YOU GET: K p = K c (RT) Δn 1. V. REACTION QUOTIENT, Q A. OBTAINED BY PLUGGING IN INITIAL CONC. OR PRESS. INTO THE EXPRESSION. B. USED TO PREDICT THE DIRECTION OF THE REACTION. 1. IF Q=K, SYSTEM IS AT EQUILIBRIUM 2. IF Q>K, THE REACTION WILL MOVE FROM RIGHT TO LEFT BECAUSE THE AMOUNT OF PRODUCTS IS TOO LARGE AND THE AMOUNT OF REACTANTS IS TOO SMALL. 3. IF Q<K, THE REACTION WILL MOVE FROM LEFT TO RIGHT BECAUSE THE AMOUNT OF REACTANTS IS TOO LARGE AND THE AMOUNT OF PRODUCTS IS TOO SMALL.

4 C. EXAMPLE: At 448 C the equilibrium constant K c for the reaction is Predict in which direction the reaction will proceed to reach equilibrium at 448 C if we start with mol of HI, mol of H 2, and mol of I 2 in a L container. D. Another one: At 1000 K the value of K p for the rxn is Calculate the value for Q p, and predict the direction in which the rxn will proceed toward equilibrium if the initial partial pressures are VI. SIGNIFICANCE OF Keq A. IF Keq IS >1, EQUIL. POSITION IS TOWARDS THE PRODUCTS. B. IF Keq IS <1, EQUIL. POSITION IS TOWARDS THE REACTANTS. C. IF Keq IS >>>1, RXN IS CLOSE TO COMPLETION. VII. FINDING EQUILIBRIUM CONCENTRATIONS: A. MANY TIMES, WHAT YOU RE GIVEN ISN T WHAT YOU USE UP B. INTRODUCING: THE ICE CHART 1. I = INITIAL; C=CHANGE; E=EQUILIBRIUM 2. USE THIS CHART TO FIND HOW MUCH REACTANT IS USED OR HOW MUCH PRODUCT IS FORMED. 3. THE CHANGE ROW MAINTAINS STOICHIOMETRIC RATIOS 4. SOMETIMES, THE QUADRATIC EQUATION MAY BE NEEDED! a. SOLVER ON YOUR GRAPHING CALC. MAY PROVE USEFUL! b. YOU WILL GET TWO ANSWERS BUT ONLY ONE WILL MAKE SENSE. C. EXAMPLE: 1. A closed system initially containing M H 2 and x10 3 M I 2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is M. Calculate K c at 448 C for the reaction taking place, which is 2. Sulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel is charged at 1000 K with SO 3 (g) at a partial pressure of atm. At equilibrium the SO 3 partial pressure is atm. Calculate the value of K p at 1000 K. (ans )

5 3. USING INITIAL CONC.: A L klask is killed with mol of H 2 and mol of I 2 at 448 C. The value of the equilibrium constant K c for the reaction at 448 C is What are the equilibrium concentrations of H 2, I 2, and HI in moles per liter? LE CHÂTELIER S PRINCIPLE I. THIS STATES: IF A SYSTEM IS DISTURBED BY A CHANGE IN TEMPERATURE, PRESSURE, OR CONCENTRATION, THE SYSTEM WILL SHIFT ITS EQUILIBRIUM POSITION TO COUNTERACT THE DISTURBANCE. II. REMEMBER THE SEE- SAW MODEL A. DEPENDING ON WHAT IS DONE, THE SEE- SAW WILL BE UNBALANCED AND THE SYSTEM WILL SHIFT TO A CERTAIN DIRECTION TO RE- ESTABLISH EQUILIBRIUM. III. CHANGE IN CONC. EX: 2 NO 2 (g) N 2 O 4 (g) H = kj: Change to System Heavy Side of See-Saw Equil. Shift By... Add NO2 Add N2O4 Remove NO2 A. BOTTOM LINE: EQUILIBRIUM SHIFTS AWAY FROM WHAT IS ADDED AND TOWARDS WHAT IS REMOVED. IV. CHANGE IN TEMP.: EX: 2 NO 2 (g) N 2 O 4 (g) H = kj: Change to System Heavy Side of See-Saw Equil. Shift By... Decrease temp. Increase temp. A. REMEMBER THAT CHANGING TEMPERATURE CHANGES K! THIS IS BECAUSE CONCENTRATION AND PRESSURE ARE HELD CONSTANT. B. TREAT ENERGY AS A PRODUCT OR A REACTANT TO DETERMINE EQUILIBRIUM SHIFT

6 V. CHANGE IN PRESSURE: EX: 2 NO 2 (g) N 2 O 4 (g) H = kj: A. ONLY APPLIES TO GASES!!! B. INCREASING PRESSURE FAVORS THE SIDE WITH LEAST AMOUNT OF MOLES, THUS THE EQUILIBRIUM POSITION SHIFTS THAT DIRECTION (RIGHT, PRODUCING N2O4 AND HEAT). C. DECREASING PRESSURE FAVORS SIDE WITH MORE MOLES (LEFT, PRODUCING NO2 AND ABSORBING HEAT). D. VOLUME HAS THE OPPOSITE EFFECT SINCE IT IS INVERSELY PROPORTIONAL TO PRESSURE. VI. CATALYSTS: THEY LOWER THE ACTIVATION ENERGY OF A REACTION. A. THIS AFFECTS THE RATE ONLY B. THEREFORE, ADDING A CATALYST DOES NOT AFFECT EQUILIBRIUM POSITION, THUS NO SHIFT WILL OCCUR 2. BUT, THE TIME IT TAKES TO REACH EQUILIBRIUM IS LESS! VII. INERT GASES: A. INERT MEANS THAT IT IS NOT REACTIVE 1. THUS, INERT GASES ARE NOT INVOLVED IN THE REACTION. B. THEY WILL NOT AFFECT REACTION SO NOTHING WILL HAPPEN TO THE EQUILIBRIUM POSITION OR REACTION RATE. VIII. ADDITION OF ACIDS/BASES: A. STRONG ACIDS AND BASES FULLY DISSOCIATE RESULTING IN H + (OR H3O + )AND OH - IONS RESPECTIVELY. B. THESE IONS WILL REACT WITH THE OPPOSITE ION IF IT IS PRESENT IN THE EQUATION C, EXAMPLE: TO THE FOLLOWING: NH3 + H2O NH4 + + OH - 6M NITRIC ACID IS ADDED. 1. ACID REACTS WITH OH -, CAUSING THE REMOVAL OF IT. 2. REACTION SHIFTS TO THE RIGHT, CREATING MORE NH4 + AND OH -