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 Equilibrium for a Multi- step Mechanism A + 2B + D E A + 2B C

k1F k1R

C + D E

k2F k2R At equilibrium forward and backward rates for the first and second steps k1F[A][B]2=k1R[C] k2F[C][D] = k2R[E] and Multiply equation 1 by equation 2 = K1K2 = Keq (for the overall reaction) [C] [A][B]2 [E] [C][D] x [E] [A][B]2[D]= Keq = [C] [A][B]2 k1F k1R (1) = K1 = [E] [C][D] k2F k2R (2) = K2 The equilibrium constant (Keq) for a multistep process can 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 amount reactants Q =

• is equal to Keq at equilibrium

Reaction quotient for reactions in solution Pb2+(aq) +2Cl-(aq) PbCl2 (s) 1 Q = ____________ [Pb2+][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

2C8H18(g) +25O2(g) 16CO2(g) + 18H2O (g) (PCO2 )16(PH2O)18 Q = ____________________ (PC8H18)2(PO2 )25

• for gases the partial pressures appear in Q
• each is raised to power of n

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Uses of Q

• calculating Q is like taking a snapshot of

the reaction

• Tells how far it is from equilibrium
• if Q less than Keq reaction will be moving

forward (towards products)

• if Q = Keq then reaction is at equilibrium
• if Q greater than Keq the reaction will be

moving backwards (towards reactants)

Keq

Equilibrium constant (Keq)

[C][D] [A][B] Kc =

c stands for concentration in [Molarity]

PCPD PAPB Kp =

p stands for partial pressure equilibrium constant ALWAYS CAPITAL K

if it applies to solutions if it applies to gases c

Keq

p

Meaning of the size of Keq

• if Keq is large the reaction tends to have

more products than reactants at equilibrium

• if Keq is small the reaction tends to have

more reactants at equilibrium

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Manipulating Keq

• Keq for the reverse of a reaction

Keq = Keq

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Keq (reverse) = 1 Keq (forward) double reaction Keq = Keq

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triple reaction etc.

(old ) (new) (new) (old )

• when reactions are added Keq for overall

reaction is the product of each of the steps Keq values

Keq for an overall reaction

A + B  C K1 = 10 C  D + E K2 = 5 A + B  D + E Koverall = K1  K2 = 50

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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

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disturbance = Cl increased response = system decreases [Cl] by moving towards products

Le Chateliers’ Principle

What effect will adding Cl ions have? Ag+ (aq) + Cl (aq)  AgCl (s)

(changes in reactants/products)

Le Chateliers’ Principle - If a system at equilibrium is disturbed it will respond by moving in a direction to counteract the disturbance

Le Chateliers’ Principle

Hb (aq) + O2 (aq)  HbO2 (aq)

hemoglobin = Hb

• xygenated hemoglobin = HbO2

disturbance = O2 decreased response = system responds to raise [O2 ] by moving towards reactants

this occurs when the hemoglobin reaches a cell that has a lower O2 concentration due to using it in

• metabolism. The shift in equilibrium

is the “release” of oxygen to the cell

(changes in reactants/products)

Le Chateliers’ Principle - If a system at equilibrium is disturbed it will respond by moving in a direction to counteract the disturbance

What will be the effect of reducing the partial pressure of O2 ?

Hemoglobin, O2 and equilibrium

Hb (aq) + (aq)  (aq) lungs [O2] high

cell

Hb (aq) + O2 (aq) cell [O2] low

lungs

[O2]

high [O2] shifts reaction towards HbO2

hemoglobin = Hb

• xygenated hemoglobin = HbO2

low [O2] shifts reaction towards Hb + O2

O2 HbO2

HbO2 (aq) 

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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

H2CO3 (aq)  CO2 (g) + H2O (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 CO2 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 NH4Cl (s) + heat  NH3 (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

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LCP and partial pressures

N2 (g) + 3H2 (g)  2NH3 (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

Kp = pNH3

2

PN2PH2

3 Say applied P was doubled V will be halved Each partial pressure will be doubled

Q = 22 2  23

= 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

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How far will it go?

• Reactant and products concentrations

stop changing when Q = Keq 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

A + 2B  D + 3E initial 0.055 0.28 0 0 (M) change (M)

Do the initial values have to add up to or multiply out to any particular quantity? Do the changes in concentration have add up to, or multiply out to

• etc. any particular quantity?

No, they can be whatever we decide to make them. Yes, they must be in the same ratio as the balanced equation 0.025 moles A x 2 moles B 1 mole A = 0.05 moles B react x moles A x 2 moles B 1 mole A = 2x moles B react

• x -2x

say for instance 0.025 moles A react then if x moles A react 1 2 1 2 1 2 1 3

+x +3x

1 2

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Initial-Change-Equilibrium table

A + 2B  D + 3E initial 0.055 0.28 0 0 (M) change (M) equilibrium 0.055-x 0.28-2x x 3x (M)

• x -2x

+x +3x the initial values plus changes are the equilibrium values + + + +

the equilibrium values if plugged into the reaction quotient will be equal to Keq

plug into Q Keq Q = [D][E]3 [A][B]2

Initial-Change-Equilibrium table

A + 2B  D + 3E initial 0.055 0.28 0 0 (M) change (M) equilibrium 0.055-x 0.28-2x x 3x (M)

• x -2x

+x +3x + + + +

If the initial concentrations and Keq are known the equilibrium concentrations can be calculated The problem is getting x by itself

Keq = [D][E]3 [A][B]2

=

x*3x (.055-x)(.28-2x)

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CO + 2H2  CH3OH

initial (M) change (M) equilibrium (M)

Fe3+ + SCN-  FeSCN2+

initial (M) change (M) equilibrium (M)

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H2 (g) + I2 (g)  2HI (g)

initial (M) change (M) equilibrium (M)

2NO + Br2  2NOBr

initial (atm) change (atm) equilibrium (atm)