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# Equilibrium - PowerPoint PPT Presentation

Equilibrium. --another concept that the mentally unbalanced struggle with. When you set up a reaction:. A + B  C + D The reactants in the flask start to make products. [A] and [B] fall The rate of the reaction begins to slow as reactants are used. Almost any reaction is reversible.

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### Equilibrium

--another concept that the

mentally unbalanced struggle with.

A + B  C + D

• The reactants in the flask start to make products.

• [A] and [B] fall

• The rate of the reaction begins to slow as reactants are used.

C + D A + B

• The rate of the reverse reaction starts at 0 M/s

• This rate begins to increase as [C] and [D] increase

• As C and D appear, they re-make the original reactants

A + B C + D

• Will come to a dynamic equilibrium.

• When the rate of the forward reaction falls to the same rate that the reverse reaction rises to, both rates stabilize.

• The forward rate quits falling as the reverse reaction re-makes products fast enough

• The reverse rate quits rising as the forward reaction quits increasing product concentrations

• At equilibrium, concentrations become constant, unless the system is disturbed

• The rates of the forward and reverse reactions are equal.

• Products and reactants are both being used and formed at the same rate

• Concentrations are stable

Equilibrium does not mean:

• The forward and reverse reactions stop.

• Product and reactant molecules are stable

• Product and reactant concentrations are equal

Good foils for multiple choice!

This rate depends on…

This rate depends on…

• Temperature

• Concentrations (considered together)

• Ea of forward and reverse reactions

Products

Ea

DG

Reactants

• The reaction that starts off fastest is the one whose reactants were present

• The reaction whose products end up more common (*) is the easier one to carry out (lower Ea)

• An exothermic reaction has an endothermic reverse. The endothermic direction has a higher Ea

• For a general reaction,

aA + bB cC + dD

…all substances being (aq) or (g),

K= [C]c[D]d

[A]a[B]b

…is a constant at constant T

1) the combustion of ammonia to form nitrogen monoxide and water vapor

2) the conversion of carbon monoxide and hydrogen to methanol

3) the association of ammonia and water to form ammonium and hydroxide ions

• --putting reactants on top

• --using coefficients inside the brackets

• --adding instead of multiplying concentrations

• --multiplying by coefficients, instead of raising to the power

• --including liquids and solids.

Avoid these errors!

1) For the combustion (in .30 M O2) of ammonia (.05 M) to form nitrogen monoxide (at .20 M) and water vapor (at .40 M)?

2) For the conversion of carbon monoxide (at.20 M) and hydrogen (at .20 M) to methanol (at .03 M) ?

3) For the association of ammonia (at .10 M) and water (at 55.5 M) to form ammonium (at .0013 M) and hydroxide ions (at .0013 M)?

1) For the combustion (in .40 M O2) of ammonia (.06 M) to form nitrogen monoxide (at .15 M), what is the concentration of water vapor?

2) For the conversion of carbon monoxide (at what concentration?) and hydrogen (at .25 M) to methanol (at .07 M)?

3) For the association of ammonia (at ? M) and water (at 55.5 M) to form ammonium (at .0019 M) and hydroxide ions (at .0003 M)?

• Given original concentrations, and K, what are the final concentrations?

• Ex. For the synthesis of ammonia from its elements, K=.034 at some temperature. If you start with [N2]=[H2]=[NH3]=.50M, what are the final concentrations?

• Step 1: Write and balance

• Step 2: Write eq. const. expression.

• Step 3: Plug in initial values to find Q, the reaction quotient

--if Q>K, the reaction shifts left

--if Q<K, the reaction shifts right

--if Q=K, (yeah, right. Like that’ll ever happen)

• Step 4: Make an ICE table

[A] [B] [C]

• Initial [Ao] [Bo] [Co]

• Change -ax -bx +cx

• Equilibrium [Ao] -ax [Bo] -bx [Co] +cx

(reverse signs on changes if Q>K)

• Step 5: Plug the eq. values, solve for x.

• Step 1: N2(g) + 3H2(g) 2NH3(g)

• Step 2: K= [NH3] 2

[N2] [H2]3

• Step 3:Q= .5M2=4 (greater than K)

.5M x .5M3

…so the reaction will shift left

More ammonia will decompose than will be made—but how much more?

• Step 4 Make an ICE table

[N2] [H2] [NH3]

• Initial .5M .5M .5M

• Change + +x +3x -2x

• Equilibrium .5+x M .5+3x M .5-2x M

• Step 5: Plug the equilibrium values into the expression, solve for x.

• K=.034=[NH3] 2 = (.5-2x)2

[N2] [H2]3 (.5+x)(5+3x)3

• These are equal where x=.172

[N2] [H2] [NH3]

• Equilibrium .5+xM .5+3xM .5-2xM

=.672M =1.016M =.156M

• Check:

• K= .156M2/.672M x 1.016M3=.035

(aka: “Close enough for government work.”)

• The reaction proceeds to the left .172 times through the reaction as written.

• It proceeds .172 moles:

--farther to the left than to the right

--for a substance with a coefficient of one

--at which point, the rates equalize.

1) [O2 ]=.40 M,[NH3 ]=.02 M,[NO] =.18 M, [H2O]=.20M

2) [CO ]=.03M,[H2]= .15 M,[CH3OH]=.02M

3) [NH3]=.12M,[NH4+]=.0019 M,[OH-]=.0005 M

• Well, we could, but we usually measure pressure

• Use PV=nRT to convert.

• If N2(g) + 3H2(g) 2NH3(g)

K= [NH3] 2

[N2] [H2]3 & M=n/V=P/RT

Then:

• K= (PNH3/RT)2= PNH32 (RT) 2

(PN2/RT)(PH2/RT)3 PN2PH23

• Kc -- calculated by concentrations (M)

• Kp -- calculated by pressures (atm.)

Kp=Kc(RT)Dn

Dn is the change in number of moles of gas

Kp=Kc(RT)Dn

• ‘Bout says it all, doesn’t it?

• No?

• A pressure constant is related to a concentration constant only differing by the RT factor represented by the unequal moles of gas that don’t cancel. This reaction lost 2 moles of gas from reactants to products.

• Oxygen gas can dissociate into individual atoms,

O2 2 O

• where K=[O]2/[O2], is a very small value at normal temperatures, larger at higher T

• Oxygen gas can dissociate into individual atoms,

O2 2 O

• where K=[O]2/[O2], is a very small value at normal temperatures, larger at higher T

OK?

1) For the combustion (in .30 M O2) of ammonia (.05 M) to form nitrogen monoxide (at .20 M) and water vapor (at .40 M)?

2) For the conversion of carbon monoxide (at.20M) and hydrogen (at .20 M) to methanol (at .03 M) ?

3) For the association of ammonia with water— Not a gas phase reaction. No Kp

• Dn = kDM (const. vol.)

—used w/ Kc calculations

• Dalton’s Law of partial pressures implies Dn = kDP

—used w/ Kc calculations

1) PO2 =1.3 atm,PNH3=2.1 atm, PNO =1.8 atm, PH2O=2.0 atm

2) PCO =4.3 atm, PH2= 2.5 atm, PCH3OH=.80atm

3) PNH3=9.5 atm, PNH4+=1.1 atm, POH-=2.3 atm

(Just joking on that last one—that really is in an aqueous environment—don’t try to use pressures)

?

• If A + B C + D has an equilibrium constant, K1….

• --then C + DA + B has an equilibrium constant, K2=1/K1

(products and reactants are reversed)

• --and 2A + 2B 2C + 2D has an equilibrium constant, K3=K12

(all of the concentrations are squared)

• If a solid or a liquid are part of a reversible reaction, they do not affect the equilibrium except by their absence.

• If they are present—the system can be in equilibrium. If they are not present, the system is not in equilibrium

Henri LeChatelier attended the Ecole Polytechique, in the Latin Quarter of Paris, where his principal, Monsieur Bouvier, always made a point of….

If a system in equilibrium is subjected to a stress, the system will shift in the direction that will relieve that stress

• “Shift right”

--forward reaction is faster,

--more of all products are formed

--all reactants are used

• “Shift left”

--reverse reaction is faster,

--more of all reactants are formed

--all products are used

• An aqueous or gas substance in the reaction added—shift away to use it up

• Increasing pressure—shift toward side with fewer moles of gas to relieve pressure

• Increasing temperature—shift in the endothermic direction to absorb heat

• An aqueous or gas substance in the reaction removed—shift towards to replace it

• Decreasing pressure—shift toward side with more moles of gas to fill the space

• Decreasing temperature—shift in the exothermic direction to produce heat

Add N2(g)

Add H2(g)

Add NH3(g)

Increase P (compress)

Increase T

Add a catalyst

Remove N2(g)

7. Remove H2(g)

8. Remove NH3(g)

9. Decrease P (allow to expand)

10. Decrease T

11. Increase pressure by adding He (g)

N2(g) + 3H2(g) 2NH3(g) + D Which way would the equilibrium shift if you:

Add CO2(g)

Add CaO(s)

Add CaCO3(s)

Increase P (compress)

Increase T

Add a catalyst

Remove CO2(g)

7. Remove CaO(s)

8. Remove CaCO3(s)

9. Decrease P (allow to expand)

10. Decrease T

11. Increase pressure by adding He (g)

CO2(g) +CaO(s) CaCO3(s) + D Which way would the equilibrium shift if you:

• An equilibrium will not shift if:

--a solid or liquid is added or removed

--a catalyst is used

--gasses are compressed/allowed to expand and Dn=0

--forward/reverse rates are affected equally

--a noble gas is added (Pt increases, partial pressures do not change)