Chemical Systems & Equilibrium Chapter 7. Factors affecting an equilibrium Temperature, pressure and alterations to concentration adjust equilibria. These

Chemical Systems & Equilibrium

Chapter 7

Factors affecting an equilibrium

Temperature, pressure and alterations to concentration adjust equilibria.

These changes result in a favour to either the forward or reverse reaction and cause the equilibrium concentrations to “shift” to establish a new stable state.

This principle was studied in 1884 by Henry Louis Le Châtelier.

Le Châtelier’s Principle

When a chemical equilibrium is disturbed by a change in property, (temperature, pressure, concentration), the system adjusts in a manner that relieves the change.

An equilibrium shift is an adjustment that results in changes to the concentrations of the equilibrium reactants and products.

Original equilibrium established

“Equilibrium Shift” to new equilibrium concentrations

Change in concentration of reactants

Time (s)

Con

cent

ratio

n (m

ol/L

)2 CO2(g) + energy D 2 CO(g) + O2(g)

[O2(g)]

[CO2(g)]

[CO(g)]

[CO2(g)] added

Shift a

Change in concentration of reactants

An increase in the concentration of reactants results in an increase in the concentration of the products.

Notice that the [CO2] is reduced and the [CO] and [O2] is increased during the adjustment.

The reaction shifts to “counter” the increase in CO2.

Change in concentration & Kinetic Theory

The addition of a substance increases the likelihood of a collision and thereby increases the rate of reaction in that direction.

The opposing reaction lags the initial response but eventually “catches up” and establishes an new equilibrium.

Change in temperature

Adding heat favours the endothermic reaction – it is a “reactant”.

Reactants + energy products Consider energy as a “product” or a

“reactant” in predicting the shift. The equilibrium shifts in the direction that

ABSORBS the heat.



Change in temperature

Time (s)

Con

cent

ratio

n (m

ol/L


[O2(g)]

[CO2(g)]

[CO(g)]

heat added

Shift a

Change in temperature & Kinetic Theory

Adding heat increases the rate of reaction due to the more frequent and energetic collisions.

The reaction that uses heat as a reactant (endothermic) will be affected quickest and thereby shift the equilibrium in that direction.

The increased rate produces more products, which are reactants for the reverse reaction.

The increased heat and reactants result in the reverse reaction “catching up” and establishing new equilibrium concentrations.

Change in pressure/volume According to Boyle’s law, the concentration of a

gas is directly proportional to the pressure of the gas.

An increase in pressure reduces the volume and increases the concentration of the gas.

Therefore, changing the pressure or volume of an equilibrium involving gases may affect the equilibrium.



Change in Pressure or Volume

Time (s)

Con

cent

ratio

n (m

ol/L


[O2(g)]

[CO2(g)]

[CO(g)]

Increased pressure or reduced volume

Shift \

Change in pressure/volume An increase in pressure, or a decrease in

volume favours the reaction that produces fewer moles of gas.

They occupy less volume and thereby relieve the stress of the added pressure.

An increase in pressure adds energy to the system and therefore, reacts similar to the addition of heat.

Change in pressure/volume & Kinetic Theory

Reducing the volume (increasing pressure) increases the likelihood of collision, thereby increasing the rate of reaction.

The reaction that has more particles (increased number of moles) will have more collisions and increase at a quicker rate.

As the shift occurs the products of the quicker reaction will increase and improve the rate of the reverse reaction.

Quantitative analysis of equilibria

Using Le Châtelier’s Principle & the Equilibrium Law one can quantitatively assess;The equilibrium constant (Keq)The position or progress of the equilibrium

system with the reaction quotient (Q)The equilibrium concentrations

Determining Keq

To determine the value for Keq, one must;Use the balanced chemical equation and the

Equilibrium Law to produce and equilibrium expression.

Substitute given values for equilibrium concentrations into the equilibrium expression to solve for Keq.

The equilibrium concentrations for the chemicals are; [N2] = 0.40 mol/L, [H2] = 0.15 mol/L & [NH3]=0.20 mol/L. Determine the value for the equilibrium constant (Keq) under these conditions. Balanced equation for the reaction

N2 (g) + 3 H2 (g) 2 NH3 (g)

Generate the equilibrium expression

Substitute equilibrium concentrations into the expression to solve for Keq.

Determining the position or status of an equilibrium system. Calculate the reaction quotient using the

equilibrium concentrations.

Compare the reaction quotient value to the equilibrium constant value.

Determine whether the reaction is reactant or product rich and adjust the equilibrium to establish the equilibrium.

At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Balanced equation for the reaction

N2 (g) + 3 H2 (g) 2 NH3 (g)

Generate the equilibrium expression and reaction quotient expression

At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Substitute the supplied concentrations into the

expression to solve for Q.

Compare the Q value to the known Keq value.

Q < Keq 14.8 < 25.0

At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? Use the Q and Keq values to determine the position of the

equilibrium system.

Since Q < Keq, the system has fewer products than the equilibrium state. Therefore the reaction needs to progress toward the products to attain equilibrium. (i.e.- Shift to the right.)

Calculating equilibrium concentrations

To determine the equilibrium concentrations, one must use:A balanced equation for the equilibrium

reactionAn equilibrium expression and constant, Keq

An Initial concentration, Change in concentration, and Equilibrium concentration (ICE) table.


In cases where the Keq value is very small the addition and subtraction of the change value (x) may be insignificant and thereby omitted. Use the ratio of the smallest initial concentration and the

Keq to assess the affect of the change.

clueionconcentrat initialsmallest

eqK


If clue > 500, the addition or subtraction of “x” is insignificant and may be ignored.

If 100<clue<500, the addition or subtraction of “x” is probably insignificant, but should be checked.

If clue<100, the addition or subtraction of “x” is significant and must be included in the calculations.

clueionconcentrat Initial smallest

eqK


In cases where the Keq expression results in the resolution of a quadratic equation, the use of the quadratic formula my prove helpful.

When the quadratic formula produces two answers select the one that is viable. Remembering that you can not have a negative concentration!

a

acbbx

2

42

Example – At high temperatures, as with lightning, nitrogen and oxygen will react to produce nitrogen monoxide. A chemist puts 0.085 moles of N2(g) and 0.038 mol of O2(g) in a 1.0 L flask at high temperature, where the Keq= 4.2 x 10-8. What is the concentration of the NO(g) in the mixture at equilibrium? Strategy

1.Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.

2.Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.

3.Set up an ICE table letting “x” represent the change in concentrations.

4.Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

5.Calculate the required value(s).

1. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.

58

100.9102.4

038.0 x

x


eqK

The clue is far greater than 500 so we can ignore the changes in N2(g) and O2(g).

2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.

N2(g) + O2(g) NO(g)N2(g) + O2(g) 2 NO(g)

8-

22

22 10 x 4.2 HN

NO eqK

3. Set up an ICE table letting “x” represent the change in concentrations.

Concentration (mol/L)

N2(g) + O2(g) 2 NO(g)

Initial concentration 0.085 0.038 0

Change in concentration -x -x 2x

Equilibrium concentration

0.085 – x ≈0.085

0.038 – x ≈0.038 2x

4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

6

2

28-

22

22

10 x 82.5

4

00232.0

00232.0

4

)038.0)(085.0 (

)2( 10 x 4.2

HN

NO

x

x

x

x

Keq

The negative value is impossible as one can

not have a negative concentration.

5. Calculate the required value(s).

5

6

102.1

)1082.5(2

2

x

x

[NO] xWhat is the NO(g) concentration at equilibrium?

The NO(g) concentration at equilibrium is 1.2 x 10-5 mol/L.

Example – In a 1.00 L flask, 2.00 mol of H2(g) is combined with 3.00 mol of I2(g) which produces HI(g). The Keq for the reaction is 25 at 1100 K. What is the What is the concentration of each gas in the mixture at equilibrium?

Strategy1.Divide the smallest initial concentration by the Keq to

determine if the change in concentration can be ignored.2.Write a balanced chemical reaction for the equilibrium

reaction and determine the equilibrium expression. 3.Set up an ICE table letting “x” represent the change in

concentrations.4.Substitute the equilibrium concentrations (ICE Table) into

the expression and solve for “x”.5.Calculate the required value(s).

1. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.

08.025

00.2


eqK

The clue is much less than 500 so we can not ignore the changes in H2(g), I2(g) and HI(g).

2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.

H2(g) + I2(g) HI(g)H2(g) + I2(g) 2 HI(g)

52

IH

HI

22

2

eqK

3. Set up an ICE table letting “x” represent the change in concentrations.

Concentration (mol/L)

H2(g) + I2(g) 2 HI(g)

Initial concentration 2.00 3.00 0

Change in concentration -x -x 2x

Equilibrium concentration 2.00 – x 3.00 – x 2x


600.5840.00

56

4

003002

225

IH

HI

2

2

2

2

22

2

xx

xx

x

x).-x)(.(

x)(

Keq


7.13.4

)6(2

)6)(84.0(4)5()5(

2

4

600.5840.00

2

2

2

or

x

a

acbbx

xx


4.3 and 1.7 are determined through the quadratic formula calculation, but only one is correct!

3.00 – 4.3 = negative value so 1.7 mol/L is the correct value. (A negative concentration is impossible!)

Therefore the final equilibrium concentrations are: [H2]eq = 2.00 – 1.7 = 0.3 mol/L[I2]eq = 3.00 – 1.7 = 1.3 mol/L[HI]eq = 2( 1.7) = 3.4 mol/L

The Solubility Equilibrium

Remember from SPH3U: Solubility is the amount of solute that dissolves

in a given amount of solvent (usually water) at a particular temperature. Measured in g/100 mL or mol/L (molar concentration) Very soluble - > 10 g/100 mL

Soluble – 1 – 10 g/100 mLSlightly soluble – 0.1 – 1 g/100 mLInsoluble - < 0.1 g/100 mL


Remember from SPH3U: Produced solubility curves

Most solids increased in solubility as the temperature increased. While, most gases decreased in solubility as the temperature increased.

Terminology – Saturated, Unsaturated, Supersaturated

The Solubility ProcessS

olu

bil

ity

(gm

/100

ml)

Temperature (oC)

Saturated solution

Unsaturated solution

Supersaturated solution


Remember from SPH3U: Dissociation of ionic compounds into ions

Ionic equations, net ionic equations, spectator ions

Electrolytes, non-electrolytes Solubility charts

Predict precipitates in ionic solution combinations (Pg 801 & Pg 487 Table 4)

The Solubility EquilibriumA solution equilibrium can be treated as a chemical equilibrium.

The solubility equilibrium must be saturated (has excess salt (solute)). Therefore, they occur with low solubility compounds.

Remember, solids have a constant concentration and are not used in the equilibrium law/expression. The equilibrium constant is simplified to the product of the ion equilibrium concentrations raised to their proper exponents.

The equilibrium constant (Keq) can be considered a solubility product constant (Ksp) in a solubility equilibrium.

The Solubility Product ConstantA solution equilibrium can be treated as a chemical equilibrium.

The solubility equilibrium must be saturated and have excess solute (salt), therefore they occur with low solubility compounds.

The balanced chemical reaction (dissociation equation) can be used to generate an equilibrium expression.

Remember, solids have a constant concentration and are not used in the equilibrium law expression. The equilibrium constant is simplified to the product of the ion concentration raised to their proper exponents.

The equilibrium constant (Kc) can be considered a solubility product constant (Ksp) in an solubility equilibrium.

The Solubility Product ConstantA solution equilibrium can be treated as a chemical equilibrium.

PbI2 (s) Pb 2+(aq) + 2 I 1-

(aq) (saturated sol’n)

PbI2 is a solid so its concentration does not alter and is therefore removed from the equilibrium expression.

PbI

IPb 1

2

2-112

eqK

2-12speq ][I ]Pb[K K

The Solubility Product Constant

A general application of the solubility equilibria results in the following Ksp expression.

BC(s) D b B+(aq) + c C-

(aq)

Where BC(s) is a slightly soluble salt, and B+(aq)

and C-(aq) are dissociated aqueous ions.

Ksp values are reported in reference tables such as those in the appendix. (Pg 802)

cb

sp CBK

The Solubility Product Constant usage

1. Calculating solubility constants from solubility data. (i.e.- solubility of 0.0003 g/100 mL @ 25oC)

2. Predicting whether a precipitate will form in a solubility equilibrium.

1. Trial ion product, Q– the reaction quotient applied to the ion concentrations.

2. Q > Ksp (supersaturated) precipitate will formQ = Ksp (saturated) precipitate will not formQ < Ksp (unsaturated) precipitate will not form

The Solubility Product Constant usage

3. The common ion effect The reduction in the solubility of a salt

caused by the addition of a second salt with a common ion.

Le Chatelier’s principle is applied and results in a shift of the solubility equilibrium.

Use equilibrium (ICE) table to solve for new concentrations.

Calculating solubility constants from solubility dataDetermine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X 10-10.

Strategy

1. Balanced chemical equation for the system.

2. Equilibrium Law expression.

3. Initial, Change in & Equilibrium Concentration table.

4. Sub-n-solve.


Strategy


AgCl(s) D Ag +1(aq) + Cl-1(aq)


2. Equilibrium Law expression.

Ksp = [Ag+1] [Cl-1] = 1.8 x 10-10


3. Initial, Change in & Equilibrium Concentration table.

AgCl (s) D Ag+1(aq) + Cl+1(aq)

[Initial] [constant] 0 0

[change in] [constant] +x +x

[Equilibrium] [constant] 0+x = x 0+x = x


4. Sub-n-solve.

Lmol 10 x 34.1

10 x 8.1

][10 x 8.1

]][[10 x 8.1

]Cl][Ag[

5-

10-

210-

10-

1(aq)

1(aq)

x

x

x

xx

K sp


4. Check assumptions

Lmol 10 x 34.1

][

]][[

]Cl][Ag[

5-

2

1(aq)

1(aq)

x

x

x

xx

K sp

Determining if a precipitate will formStrategy


2. Identify the concentrations to be considered.

3. Use the equilibrium law expression to generate the “Trial Ion Product (Qsp)” expression.

4. Use given concentration and the expression to solve for Qsp.

Determining if a precipitate will formStrategy

5. Compare the Qsp value to the known Ksp value to determine if the situation is at equilibrium.

Qsp = Ksp the solution is saturated & at equilibrium so no precipitate will form.

Qsp < Ksp the solution is unsaturated so no precipitate will form.

Qsp > Ksp the solution is “supersaturated” so a

precipitate will form.

Determining if a precipitate will formIf 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4.

NiCl2 (aq) + Na2CO3 (aq) → NiCO3 (s) + 2 NaCl (aq)

(Use the solubility rules to identify which of the products is most likely to form a solid.)

NiCO3 (s) ⇌ Ni 2+(aq) + 2 Cl-(aq)


NiCl2 (aq) → Ni 2+(aq) + 2 Cl- (aq)

[Ni2+]= 0.200 mol/L(50 mL)/(100 mL) = 0.100 mol/L

Na2CO3 (aq) → 2 Na+(aq) + CO3

2- (aq)

[CO32-]= 0.0500 mol/L(50 mL)/(100 mL) = 0.0250 mol/L


NiCO3 (s) ⇌ Ni 2+(aq) + CO3

2- (aq)

0025.0

)mol0250.0)(Lmol 100.0(

]][CONi[ -23

2

L

Qsp


Qsp > Ksp (0.0025 > 8.2 x 10-4 )

In this case the solution is either supersaturated or a precipitate will be formed.

Documents

Chemical Systems & Equilibrium Chapter 7. Factors affecting an equilibrium Temperature, pressure and alterations to concentration adjust equilibria. These