chemical equilibrium

Chemical EquilibriumChemical Equilibrium

AP ChemistryJ.M.Soltmann

AP ChemistryJ.M.Soltmann

A Brief Review Question


• Write out the theoretical rate law for the reaction: • 3 H2 (g) + N2 (g) <--> 2 NH3 (g)





• Rate = k[H2]3[N2]


• Rate = k[H2]3[N2]

What about this?What about this?

• Write out the theoretical rate law for the reaction: • 2 NH3 (g) <--> 3 H2 (g) + N2 (g)


What about this?What about this?


• Rate = k[NH3]2


• Rate = k[NH3]2

Let’s consider that…Let’s consider that…

• 3 H3 H22 (g) + N (g) + N22 (g) (g) <--> 2 NH<--> 2 NH33 (g) (g)

• Rate = k[HRate = k[H22]]33[N[N22]]

• If we mix If we mix hydrogen and hydrogen and nitrogen, we can nitrogen, we can produce ammonia.produce ammonia.

• 3 H3 H22 (g) + N (g) + N22 (g) (g) <--> 2 NH<--> 2 NH33 (g) (g)

• Rate = k[HRate = k[H22]]33[N[N22]]

• If we mix If we mix hydrogen and hydrogen and nitrogen, we can nitrogen, we can produce ammonia.produce ammonia.

• 2 NH2 NH33 (g) <--> (g) <--> 3 H3 H22 (g) + N (g) + N22 (g)(g)

• Rate = k[NHRate = k[NH33]]22

• If we heat If we heat ammonia, we can ammonia, we can decompose it to decompose it to hydrogen and hydrogen and nitrogen.nitrogen.

• 2 NH2 NH33 (g) <--> (g) <--> 3 H3 H22 (g) + N (g) + N22 (g)(g)

• Rate = k[NHRate = k[NH33]]22

• If we heat If we heat ammonia, we can ammonia, we can decompose it to decompose it to hydrogen and hydrogen and nitrogen.nitrogen.

Reversible reactionsReversible reactions

• The two reactions we looked at our opposites, and we call them reversible reactions. Not all reactions are reversible.

• If we put nitrogen and hydrogen into a closed container, they will make ammonia. However, over time the ammonia will decompose into hydrogen and nitrogen.

• At some point the rate of production of ammonia will be equal to the rate of consumption of ammonia and the reaction appears to stop.

• The two reactions we looked at our opposites, and we call them reversible reactions. Not all reactions are reversible.

• If we put nitrogen and hydrogen into a closed container, they will make ammonia. However, over time the ammonia will decompose into hydrogen and nitrogen.

• At some point the rate of production of ammonia will be equal to the rate of consumption of ammonia and the reaction appears to stop.

Why did you say “appears?”

Why did you say “appears?”

• The reaction never stops on a molecular level, even at equilibrium.

• However if one molecule is made at the same time one molecule is broken the overall effect means that there is no change in concentration.

• The reaction never stops on a molecular level, even at equilibrium.

• However if one molecule is made at the same time one molecule is broken the overall effect means that there is no change in concentration.

QuickTime™ and a decompressor

are needed to see this picture.

Chemical EquilibriumChemical Equilibrium

• In a closed system, when the rate of the forward reaction is equal to the rate of the reverse reaction the reaction appears to stop. It has reached equilibrium.

• In a closed system, when the rate of the forward reaction is equal to the rate of the reverse reaction the reaction appears to stop. It has reached equilibrium.



Equilibrium ConstantEquilibrium Constant

• Going back to the previous example, at equilibrium: ratef=rater.

• Ratef = kf[H2]3[N2]

• Rater = kr[NH3]2

• kf[H2]3[N2] = kr[NH3]2

• kf/kr = [NH3]2/[H2]3[N2]

• Keq = [NH3]2/[H2]3[N2]

• Going back to the previous example, at equilibrium: ratef=rater.

• Ratef = kf[H2]3[N2]

• Rater = kr[NH3]2

• kf[H2]3[N2] = kr[NH3]2

• kf/kr = [NH3]2/[H2]3[N2]

• Keq = [NH3]2/[H2]3[N2]

More on KeqMore on Keq

• Keq = [products]/[reactants]

• Keq is not given units!

• Keq is a constant for a given reaction, at a given temperature.

• Since solids and liquids have constant concentrations (called densities), they are not included in the products or reactants; solids and liquids are part of the Keq value.

• Keq = [products]/[reactants]

• Keq is not given units!

• Keq is a constant for a given reaction, at a given temperature.

• Since solids and liquids have constant concentrations (called densities), they are not included in the products or reactants; solids and liquids are part of the Keq value.

Write out the expression for Keq in

the following:


the following:

• H2 (g) + I2 (s) <--> 2 HI (s)

• 2 H2 (g) + O2 (g) <--> 2 H2O (l)

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)

• H2 (g) + I2 (s) <--> 2 HI (s)

• 2 H2 (g) + O2 (g) <--> 2 H2O (l)

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)


the following:


the following:• H2 (g) + I2 (s) <--> 2 HI (s)

• Keq = 1/[H2]

• 2 H2 (g) + O2 (g) <--> 2 H2O (l)• Keq = 1/[H2]2[O2]

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)• Keq = [CO2]4[H2O]6/[C2H6]2[O2]7

• H2 (g) + I2 (s) <--> 2 HI (s)• Keq = 1/[H2]

• 2 H2 (g) + O2 (g) <--> 2 H2O (l)• Keq = 1/[H2]2[O2]

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)• Keq = [CO2]4[H2O]6/[C2H6]2[O2]7

Calculating KeqCalculating Keq

• When we calculated the rate of reaction, we used the initial concentrations because we were really only interested in the initial rate.

• Now we are looking at the end, at equilibrium. When we calculate Keq, we need to know the equilibrium concentrations.

• When we calculated the rate of reaction, we used the initial concentrations because we were really only interested in the initial rate.

• Now we are looking at the end, at equilibrium. When we calculate Keq, we need to know the equilibrium concentrations.

For example,For example,

• Ethane was combusted in a closed container with oxygen. At equilibrium the following information was found:

• [C2H6] = .300 atm [O2] = .450 atm

• [CO2] = 4.4 atm [H2O] = 6.6 atm

• What is the value of Keq?

• Ethane was combusted in a closed container with oxygen. At equilibrium the following information was found:

• [C2H6] = .300 atm [O2] = .450 atm

• [CO2] = 4.4 atm [H2O] = 6.6 atm


We already looked at the RXN

We already looked at the RXN

• 2C2H6 (g) + 7O2 (g)< --> 4CO2 (g) + 6H2O (g)

• Keq = [CO2]4[H2O]6/[C2H6]2[O2]7

•Keq = [4.4]4[6.6]6/[.30]2[.45]7

•Keq = 9.21 x 1010

• The fact that Keq is much greater than 1 tells us that products are heavily favored in this reaction.

• 2C2H6 (g) + 7O2 (g)< --> 4CO2 (g) + 6H2O (g)

• Keq = [CO2]4[H2O]6/[C2H6]2[O2]7

•Keq = [4.4]4[6.6]6/[.30]2[.45]7

•Keq = 9.21 x 1010

• The fact that Keq is much greater than 1 tells us that products are heavily favored in this reaction.

Try this oneTry this one

• Sodium bicarbonate is mixed with sulfuric acid. At equilibrium, the following concentrations were measured:

• [HCO3-] = .200 M [H+]

= .250 M

• [CO2] = .6 atm [H2O] = 55.6 M


• Sodium bicarbonate is mixed with sulfuric acid. At equilibrium, the following concentrations were measured:

• [HCO3-] = .200 M [H+]

= .250 M

• [CO2] = .6 atm [H2O] = 55.6 M


Write out the net ionic equation

Write out the net ionic equation

• 2NaHCO3 (aq) + H2SO4 (aq) <--> 2 H2O (l) + 2 CO2 (g) + Na2SO4 (aq)

• HCO3- (aq) + H+ (aq) <--> H2O (l) + CO2

(g)

• Keq = [CO2]/[HCO3-][H+]

• Keq = [.6]/[.2][.25]

• Keq = 12

• Are products or reactants favored?

• 2NaHCO3 (aq) + H2SO4 (aq) <--> 2 H2O (l) + 2 CO2 (g) + Na2SO4 (aq)

• HCO3- (aq) + H+ (aq) <--> H2O (l) + CO2

(g)

• Keq = [CO2]/[HCO3-][H+]

• Keq = [.6]/[.2][.25]

• Keq = 12

• Are products or reactants favored?

What if we don’t know the equilibrium concentrations?

What if we don’t know the equilibrium concentrations?

• If we know the initial values, we can’t determine Keq without some extra experimental information.

• Let’s look at an example of this type of scenario.

• If we know the initial values, we can’t determine Keq without some extra experimental information.

• Let’s look at an example of this type of scenario.

Finding Keq from initial concentrations

Finding Keq from initial concentrations

• 2.00 atm of propane are mixed in a closed vessel with 8.00 atm of oxygen. At equilibrium, the concentration of oxygen was .500 atm.


• 2.00 atm of propane are mixed in a closed vessel with 8.00 atm of oxygen. At equilibrium, the concentration of oxygen was .500 atm.


ICE timeICE time

• Here’s what was given:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 00

• C: • E: .500

• Here’s what was given:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 00

• C: • E: .500

ICE timeICE time

• Now we work on the change of O2:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -7.50• E: .500

• Now we work on the change of O2:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -7.50• E: .500

ICE timeICE time

• Now do the Stoichiometry:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -1.50 -7.50 +4.50 +6.00

• E: .500

• Now do the Stoichiometry:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -1.50 -7.50 +4.50 +6.00

• E: .500

ICE timeICE time

• Next, bring it on down:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -1.50 -7.50 +4.50 + 6.00

• E: .50 .500 4.506.00

• Next, bring it on down:

• C3H8 (g) + 5 O2 (g) <--> 3 CO2 (g) + 4 H2O (g)

• I: 2.00 8.00 0 0• C: -1.50 -7.50 +4.50 + 6.00

• E: .50 .500 4.506.00

Now we can calculate Keq

Now we can calculate Keq

• Keq = [CO2]3[H2O]4/[C3H8][O2]5

• Keq = [4.5]3[6.0]4/[.5][.5]5

• Keq = [CO2]3[H2O]4/[C3H8][O2]5

• Keq = 7558272 or 7.56 x 106

• Keq = [CO2]3[H2O]4/[C3H8][O2]5

• Keq = [4.5]3[6.0]4/[.5][.5]5

• Keq = [CO2]3[H2O]4/[C3H8][O2]5

• Keq = 7558272 or 7.56 x 106

Going the other wayGoing the other way

• Generally more useful to Chemists, we also can use a known value of Keq to help us determine the actual yield or the actual concentrations of a substance at equilibrium.

• Generally more useful to Chemists, we also can use a known value of Keq to help us determine the actual yield or the actual concentrations of a substance at equilibrium.

For Example:For Example:

• 5.00 M nitric acid is heated until it decomposes:• 2 HNO3 (aq) <--> H2O (l) + N2O5 (g)

• If Keq for this reaction is .100, then how what concentration of dinitrogen pentoxide will be formed?

• 5.00 M nitric acid is heated until it decomposes:• 2 HNO3 (aq) <--> H2O (l) + N2O5 (g)

• If Keq for this reaction is .100, then how what concentration of dinitrogen pentoxide will be formed?

The ProcessThe Process

• Let’s start our ICE table:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0

• Let’s start our ICE table:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0


• We don’t know changes, so we will

• Assume the change in N2O5 is x:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0• C: -2x +x +x

• We don’t know changes, so we will

• Assume the change in N2O5 is x:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0• C: -2x +x +x


• Now we bring it down:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0• C: -2x +x +x• E: 5 - 2x x x

• Now we bring it down:

• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M 0 0• C: -2x +x +x• E: 5 - 2x x x


• But wait, water is a liquid, so we don’t

• need to include it.• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M - 0• C: -2x - +x• E: 5 - 2x - x

• But wait, water is a liquid, so we don’t

• need to include it.• 2 HNO3 (aq) --> H2O (l) + N2O5 (g)

• I: 5.00 M - 0• C: -2x - +x• E: 5 - 2x - x


• Write out the equation for Keq

• and substitute in, then solve.• 2 HNO3 (aq) <--> H2O (l) + N2O5 (g)

• Keq = [N2O5]/[HNO3]2

• .1 = (x)/(5-x)2 = x/(25-10x+x2)• 2.5-x+.1x2 = x• .1x2 - 2x + 2.5 = 0

• Write out the equation for Keq

• and substitute in, then solve.• 2 HNO3 (aq) <--> H2O (l) + N2O5 (g)

• Keq = [N2O5]/[HNO3]2

• .1 = (x)/(5-x)2 = x/(25-10x+x2)• 2.5-x+.1x2 = x• .1x2 - 2x + 2.5 = 0


• .1x2 - 2x + 2.5 = 0 is a quadratic. • To solve this, we will have to either graph it or use the quadratic equation.

• The choice is yours.• You will get two answers:

• X = 1.34 M or X = 18.7 M

• Since 18.7 M is not chemically possible, we know that the equilibrium concentration of dinitrogen pentoxide is 1.34 M.

• .1x2 - 2x + 2.5 = 0 is a quadratic. • To solve this, we will have to either graph it or use the quadratic equation.

• The choice is yours.• You will get two answers:

• X = 1.34 M or X = 18.7 M

• Since 18.7 M is not chemically possible, we know that the equilibrium concentration of dinitrogen pentoxide is 1.34 M.

Manipulating Keq of Reversible ReactionsManipulating Keq of Reversible Reactions

• Think about this. Let’s assume that the rxn A + 2B--> C has Keq =5.

• What is the Keq of the rxn C-->A + 2B?

• Think about this. Let’s assume that the rxn A + 2B--> C has Keq =5.

• What is the Keq of the rxn C-->A + 2B?

Manipulating Keq of Reversible ReactionsManipulating Keq of Reversible Reactions

• A + 2B--> C has Keq =5.

• So Keq1 = 5 = [C]/[A][B]2

• Now look at C-->A + 2B?• Keq2 = [A][B]2/[C]

• The two equations are reciprocals:• [C]/[A][B]2 = ([A][B]2/[C])-1

• So, Keq2 = (Keq1)-1 = 5-1 = .2

• A + 2B--> C has Keq =5.

• So Keq1 = 5 = [C]/[A][B]2

• Now look at C-->A + 2B?• Keq2 = [A][B]2/[C]

• The two equations are reciprocals:• [C]/[A][B]2 = ([A][B]2/[C])-1

• So, Keq2 = (Keq1)-1 = 5-1 = .2

Manipulating Keq of Mulitplied ReactionsManipulating Keq of Mulitplied Reactions

• Now think about this. We will still say that the rxn A + 2B--> C has Keq =5.

• What is Keq for the rxn 2A + 4B--> 2 C ?

• Now think about this. We will still say that the rxn A + 2B--> C has Keq =5.

• What is Keq for the rxn 2A + 4B--> 2 C ?

Manipulating Keq of Mulitplied ReactionsManipulating Keq of Mulitplied Reactions

• A + 2B--> C has Keq =5.

• So Keq1 = 5 = [C]/[A][B]2

• For the rxn 2A + 4B --> 2C?• Keq2 = [A]2[B]4/[C]2

• [A]2[B]4/[C]2 = ([C]/[A][B]2)2

• Thus Keq2 = (Keq1)2 = 5 2 = 25

• A + 2B--> C has Keq =5.

• So Keq1 = 5 = [C]/[A][B]2

• For the rxn 2A + 4B --> 2C?• Keq2 = [A]2[B]4/[C]2

• [A]2[B]4/[C]2 = ([C]/[A][B]2)2

• Thus Keq2 = (Keq1)2 = 5 2 = 25

Manipulating Keq of Additive ReactionsManipulating Keq of Additive Reactions

• This time let’s say we have two rxns. X + Q --> J has Keq = 12.

• J + Q --> D has Keq = 2

• What is Keq for the rxn X + 2Q--> D?

• This time let’s say we have two rxns. X + Q --> J has Keq = 12.

• J + Q --> D has Keq = 2

• What is Keq for the rxn X + 2Q--> D?

Manipulating Keq of Additive ReactionsManipulating Keq of Additive Reactions

• X + Q --> J • Keq1 = 12 = [J]/[X][Q]

• J + Q --> D • Keq2 = 2 = [D]/[J][Q]

• X + 2Q--> D• Keq3 = [D]/[X][Q]2

• [D]/[X][Q]2 =([J]/[X][Q])*([D]/[J][Q])

• So, Keq3 = Keq1 * Keq2 = 12*2 = 24

• X + Q --> J • Keq1 = 12 = [J]/[X][Q]

• J + Q --> D • Keq2 = 2 = [D]/[J][Q]

• X + 2Q--> D• Keq3 = [D]/[X][Q]2

• [D]/[X][Q]2 =([J]/[X][Q])*([D]/[J][Q])

• So, Keq3 = Keq1 * Keq2 = 12*2 = 24

Can you put it all together?

Can you put it all together?

• Given:

• N2 (g) + O2 (g) 2 NO (g) Keq = 6.00

• 2 NO (g) + 2 O2 (g) 2 NO2 (g) Keq = .500

• 2 N2O (g) O2 (g) + 2 N2 (g) Keq = 15.0

• Calculate Keq for the reaction:

• N2O (g) + NO2 (g) 3 NO (g)

• Given:

• N2 (g) + O2 (g) 2 NO (g) Keq = 6.00

• 2 NO (g) + 2 O2 (g) 2 NO2 (g) Keq = .500

• 2 N2O (g) O2 (g) + 2 N2 (g) Keq = 15.0

• Calculate Keq for the reaction:

• N2O (g) + NO2 (g) 3 NO (g)

It’s like Hess’s Law all over again!!

It’s like Hess’s Law all over again!!

• N2 (g) + O2 (g) 2 NO (g) Keq = 6.00• This rxn and Keq stays the same.

• 2 NO (g) + 2 O2 (g) 2 NO2 (g) Keq = .500• This reaction is reversed and halved, so Keq = (.5)-.5

• 2 N2O (g) O2 (g) + 2 N2 (g) Keq = 15.0• This reaction is halved, so Keq = (15).5

• Now multiply all 3 Keq values.

• N2O (g) + NO2 (g) 3 NO (g) Keq = 32.9

• N2 (g) + O2 (g) 2 NO (g) Keq = 6.00• This rxn and Keq stays the same.

• 2 NO (g) + 2 O2 (g) 2 NO2 (g) Keq = .500• This reaction is reversed and halved, so Keq = (.5)-.5

• 2 N2O (g) O2 (g) + 2 N2 (g) Keq = 15.0• This reaction is halved, so Keq = (15).5

• Now multiply all 3 Keq values.

• N2O (g) + NO2 (g) 3 NO (g) Keq = 32.9

LeChatelier’s Principle

LeChatelier’s Principle

• When a change is made to a system at equilibrium, the system shifts the equilibrium position to counter the change (partially).

• Basically this means that the system wants to undo the stress applied to it.

• For example:

• When a change is made to a system at equilibrium, the system shifts the equilibrium position to counter the change (partially).

• Basically this means that the system wants to undo the stress applied to it.

• For example:

If we increase the pressure:

(or decrease volume)

If we increase the pressure:

(or decrease volume)• The system responds by shifting towards the side with less pressure (this is generally the side with less gas moles).

• In the RXn 3 H2 (g) + N2 (g) <--> 2NH3 (g), which side is favored?

• The system responds by shifting towards the side with less pressure (this is generally the side with less gas moles).

• In the RXn 3 H2 (g) + N2 (g) <--> 2NH3 (g), which side is favored?



If we add energy:If we add energy:

• If energy is added, the system wants to use the energy, so it shifts equilibrium to the side of highest enthalpy. In other words, the endothermic reaction is favored.

• 3 H2 (g) + N2 (g) <--> 2NH3 (g) + heat

• If energy is added to this reaction, would equilibrium favor reactants or products?

• If energy is added, the system wants to use the energy, so it shifts equilibrium to the side of highest enthalpy. In other words, the endothermic reaction is favored.

• 3 H2 (g) + N2 (g) <--> 2NH3 (g) + heat

• If energy is added to this reaction, would equilibrium favor reactants or products?

If we increase the concentration of a

substance:


substance:• When more of a chemical is added to a reaction at equilibrium, the system tries to use up the extra substance, shifting equilibrium to the other side.

• Let’s look at that a bit more closely.

• When more of a chemical is added to a reaction at equilibrium, the system tries to use up the extra substance, shifting equilibrium to the other side.

• Let’s look at that a bit more closely.


substance


substance• 3 H2 (g) + N2 (g) <--> 2NH3 (g) + heat

• If I were to add extra hydrogen gas, there would be an increase in collisions between hydrogen and nitrogen. I would use up more hydrogen and more nitrogen, thus producing more ammonia. Overall the hydrogen changes little, but the [N2] is noticeably less and the [NH3] is noticeably greater.

• 3 H2 (g) + N2 (g) <--> 2NH3 (g) + heat

• If I were to add extra hydrogen gas, there would be an increase in collisions between hydrogen and nitrogen. I would use up more hydrogen and more nitrogen, thus producing more ammonia. Overall the hydrogen changes little, but the [N2] is noticeably less and the [NH3] is noticeably greater.

If we add a catalystIf we add a catalyst

• nothing happens to equilibrium.• Why? Well, the point of a catalyst is to speed up a reaction, by lowering the activation energy. Since the forward and reverse reaction both happen faster, there is no net change.

• nothing happens to equilibrium.• Why? Well, the point of a catalyst is to speed up a reaction, by lowering the activation energy. Since the forward and reverse reaction both happen faster, there is no net change.

Applying LeChatelierApplying LeChatelier

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)

• Explain what happens to the equilibrium concentrations of each chemical in the rxn if:

• The pressure is increased• The volume is increased• The temperature is increased• More oxygen is added• A palladium catalyst is used

• 2C2H6 (g) + 7O2 (g) <--> 4CO2 (g) + 6H2O (g)

• Explain what happens to the equilibrium concentrations of each chemical in the rxn if:

• The pressure is increased• The volume is increased• The temperature is increased• More oxygen is added• A palladium catalyst is used

chemical equilibrium

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