Lecture 4 Professor Hicks Inorganic Chemistry (CHE152) Add the - PDF document

Lecture 4 Professor Hicks Inorganic Chemistry (CHE152) Add the following homework problems Chapter 14: 61, 63, 69, 71 At equilibrium forward and backward Equilibrium for a Multi- rates for the first and second steps step Mechanism k 1F [A][B] 2 =k 1R [C] k 1F [C] C k 1F A + 2B = = K 1 (1) [A][B] 2 k 1R k 1R and k 2F C + D E k 2F [C][D] = k 2R [E] k 2R A + 2B + D E [E] k 2F = = K 2 (2) [C][D] k 2R Multiply equation 1 by equation 2 [C] [E] x = K 1 K 2 = K eq (for the overall reaction) [A][B] 2 [C][D] [E] The equilibrium constant (K eq ) for a multistep process can [A][B] 2 [D]= K eq be obtained from the overall reaction It is equal to the ratio of products to reactants raised to their stoichiometric numbers

Reaction quotient (Q) • Q monitors progress of a reaction • Q = 0 at moment reactants mixed • increases as products form amount products Q = amount reactants • is equal to K eq at equilibrium Reaction quotient for reactions in solution Pb 2+ (aq) +2Cl - (aq)  PbCl 2 (s) 1 Q = ____________ [Pb 2+ ][Cl - ] 2 For dissolved substances: 1) molarity appears in Q 2) pure substances (solids and liquids that are not dissolved) never appear in Q 3) molarities raised to power of n in the overall reaction Reaction quotient (Q) for gases 2 C 8 H 18 (g) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O (g) (P CO2 ) 16 (P H2O ) 18 Q = ____________________ (P C8H18 ) 2 (P O2 ) 25 • for gases the partial pressures appear in Q • each is raised to power of n

Uses of Q • calculating Q is like taking a snapshot of the reaction • Tells how far it is from equilibrium • if Q less than K eq reaction will be moving forward (towards products) • if Q = K eq then reaction is at equilibrium • if Q greater than K eq the reaction will be moving backwards (towards reactants) Equilibrium constant (K eq ) equilibrium constant ALWAYS CAPITAL K c K eq if it applies to solutions [C][D] c stands for concentration in [Molarity] K c = [A][B] K eq p if it applies to gases P C P D p stands for partial pressure K p = P A P B Meaning of the size of K eq • if K eq is large the reaction tends to have more products than reactants at equilibrium • if K eq is small the reaction tends to have more reactants at equilibrium 3

Manipulating K eq • K eq for the reverse of a reaction 1 K eq (reverse) = K eq (forward) triple reaction double reaction 2 K eq = K eq 3 K eq = K eq (new) (old ) (new) (old ) etc. K eq for an overall reaction • when reactions are added K eq for overall reaction is the product of each of the steps K eq values A + B  C K 1 = 10 C  D + E K 2 = 5 K overall = K 1  K 2 = 50 A + B  D + E 4

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Le Chateliers’ Principle • If a system at equilibrium is disturbed it will move in a direction to counteract the disturbance • LCP is used to predict the direction a reaction will move in response to changes in temperature, pressure, or amounts of reactants/products 7

Le Chateliers’ Principle (changes in reactants/products) What effect will adding Cl  ions have? Ag + (aq) + Cl  (aq)  AgCl (s) disturbance = Cl  increased Le Chateliers’ Principle - If a system at equilibrium is disturbed it will respond response = system decreases by moving in a direction to counteract the [Cl  ] by moving towards products disturbance Le Chateliers’ Principle (changes in reactants/products) What will be the effect of reducing the partial pressure of O 2 ? Hb (aq) + O 2 (aq)  HbO 2 (aq) hemoglobin = Hb oxygenated hemoglobin = HbO 2 this occurs when the hemoglobin reaches a cell that has a lower O 2 concentration due to using it in metabolism. The shift in equilibrium disturbance = O 2 decreased is the “release” of oxygen to the cell Le Chateliers’ Principle - If a system response = system responds at equilibrium is disturbed it will respond to raise [O 2 ] by moving towards by moving in a direction to counteract the reactants disturbance Hemoglobin, O 2 and equilibrium lungs [O 2 ] high hemoglobin = Hb oxygenated hemoglobin = HbO 2 Hb (aq) + (aq)  HbO 2 [O 2 ] O 2 (aq) high [O 2 ] shifts reaction towards HbO 2 cell [O 2 ] low lungs cell HbO 2 (aq)  Hb (aq) + O 2 (aq) low [O 2 ] shifts reaction towards Hb + O 2 8

Le Chateliers’ Principle (changes in applied pressure or volume) • If pressure is increased the system will shift towards the side that has smaller volume to reduce the pressure • C(s, graphite)  C(s, diamond) larger volume smaller volume putting graphite under large pressures reduces the volume and causes it to turn into diamond Le Chateliers’ Principle (changes in applied pressure or volume) • If pressure is decreased (or V is increased) the system will respond by shifting towards the side that has larger volume (more moles of gas) attempting to increase the pressure H 2 CO 3 (aq)  CO 2 (g) + H 2 O (l) less moles gas more moles gas pressure is decreased when champagne is uncorked the system responds by trying to increase the pressure. This means shifting the equilibrium towards the side with more moles gas  releasing CO 2 gas Le Chateliers’ Principle (changes in temperature) • raising temperature can be thought of as adding heat in order to “remove the heat” the reaction will move in the direction that consumes heat NH 4 Cl (s) + heat  NH 3 (g) + H + (aq) + Cl- (aq) heating will drive reaction towards products reactions that absorb heat (  H positive) are said to be endothermic reactions that release heat (  H negative) are said to be exothermic 9

LCP and partial pressures N 2 (g) + 3H 2 (g)  2NH 3 (g) If the applied pressure is increased on this system it would tend to decrease the volume – the LCP response is to move to right The reaction quotient is disturbed by this - that is why it responds to find a new equilibrium position 2 2 p NH3 2 Q = Say applied P was doubled K p = V will be halved 2  2 3 Each partial pressure will be doubled 3 P N2 P H2 = ¼ of previous value If the total pressure is changed by adding a gas not involved in the reaction at constant volume none of the partial pressures will be changed so there will be no need to re-equilibrate 10

How far will it go? • Reactant and products concentrations stop changing when Q = K eq Q: How do we figure out [products] and [reactants] at equilibrium? A: The algebra to calculate [products] and [reactants] is usually organized in a table • Initial-Change-Equilibrium (ICE) are the three concentrations involved Initial-Change-Equilibrium (ICE) table 2 1 3 1 A + 2B  D + 3E initial 0.055 0.28 0 0 1 2 (M) change -x -2x +x +3x (M) Do the initial values have to add up to or No, they can be whatever we multiply out to any particular quantity? decide to make them. Do the changes in concentration Yes, they must be in the same have add up to, or multiply out to ratio as the balanced equation etc. any particular quantity? say for instance 0.025 moles A react then if x moles A react x moles A x 2 moles B 0.025 moles A x 2 moles B = 2x moles B 1 mole A = 0.05 moles B 1 mole A react react 1 2 1 2 11

Initial-Change-Equilibrium table A + 2B  D + 3E initial 0.055 0.28 0 0 (M) + + + + change -x -2x +x +3x (M) equilibrium 0.055-x 0.28-2x x 3x (M) plug into Q Q = [D][E] 3 [A][B] 2 K eq the equilibrium values if plugged into the reaction quotient will be equal to K eq the initial values plus changes are the equilibrium values Initial-Change-Equilibrium table A + 2B  D + 3E initial 0.055 0.28 0 0 (M) + + + + change -x -2x +x +3x (M) equilibrium 0.055-x 0.28-2x x 3x (M) K eq = [D][E] 3 x*3x = [A][B] 2 (.055-x)(.28-2x) If the initial concentrations and K eq are known the equilibrium concentrations can be calculated The problem is getting x by itself 12

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 CO + 2H 2 CH 3 OH initial (M) change (M) equilibrium (M) Fe 3+ + SCN -  FeSCN 2+ initial (M) change (M) equilibrium (M) 16

 H 2 (g) + I 2 (g) 2HI (g) initial (M) change (M) equilibrium (M)  2NO + Br 2 2NOBr initial (atm) change (atm) equilibrium (atm) 17