Calculating Equilibrium Composition Example Cl 2 (g) → 2Cl (g)

Calculating Equilibrium Calculating Equilibrium CompositionComposition

Example Example ClCl22 (g) (g) → 2Cl (g)→ 2Cl (g)


Example:Example: ClCl22 (g) (g) → 2Cl (g)→ 2Cl (g)

Initially, Initially, nn00 moles of Cl moles of Cl22 gas is placed in a closed gas is placed in a closed

reaction vessel.reaction vessel. The molecule partially dissociates into atoms.The molecule partially dissociates into atoms.


Example Example ClCl22 (g) (g) → 2Cl (g)→ 2Cl (g)

Both are gasses, so use partial pressures Both are gasses, so use partial pressures rather than concentrations.rather than concentrations.

We know that at equilibrium we have a We know that at equilibrium we have a definite mixture (definite mixture (i.ei.e., the composition of ., the composition of the mixture is not arbitrary), as the the mixture is not arbitrary), as the reactants and products are related.reactants and products are related.


What is the expression for the equilibrium What is the expression for the equilibrium quotient?quotient?


What is the expression for the equilibrium What is the expression for the equilibrium quotient?quotient?

We need expressions for the partial pressures We need expressions for the partial pressures at equilibrium.at equilibrium.

i ip x p


Initial No. of molesInitial No. of moles ClCl22 = ? = ?


Initial No. of molesInitial No. of moles ClCl22 = = nn00 2Cl = ?2Cl = ?


Initial No. of molesInitial No. of moles ClCl22 = = nn00 2Cl = 02Cl = 0



No. of moles at eq.No. of moles at eq. ClCl22 = ? = ?



No. of moles at eq.No. of moles at eq. ClCl22 = n = n0 0 – –



No. of moles at eq.No. of moles at eq. ClCl22 = n = n0 0 – – Cl = ?Cl = ?



No. of moles at eq.No. of moles at eq. ClCl22 = n = n0 0 – – Cl = 2Cl = 2



No. of moles at eq.No. of moles at eq. ClCl22 = n = n0 0 – – Cl = 2Cl = 2

Mole fractions at eq.Mole fractions at eq. ClCl22 = =



No. of moles at eq.No. of moles at eq. ClCl22 = = nn0 0 – – Cl = 2Cl = 2

Mole fractions at eq.Mole fractions at eq. ClCl22 = = nn0 0 – – nn00++Cl = Cl =

22//nn00++





22//nn00++

Partial pressures at eq.Partial pressures at eq. ClCl22 = ? = ?





22//nn00++

Partial pressures at eq.Partial pressures at eq. ClCl22 = ( = (nn0 0 – – nn00++))pp

2Cl = (22Cl = (2//nn00++))pp





22//nn00++


2Cl = (22Cl = (2//nn00++))pp

Now express Now express KKpp in terms of what we have. in terms of what we have.



2Cl = (22Cl = (2//nn00++))pp


In terms of partial pressures. Which, in terms of In terms of partial pressures. Which, in terms of and and nn00 is is

2

2.

.

eqCl

p eqCl

p

pK

p

p



2Cl = (22Cl = (2//nn00++))pp


Which in terms of a and the total pressure is Which in terms of a and the total pressure is

2

22.

0

.0

0

2eqCl

p eqCl

ppn pp

Kp n p

n pp



2Cl = (22Cl = (2//nn00++))pp


This can now be expresses in terms of This can now be expresses in terms of a a and and pp only only

2

22.

2 20

2 2.0 0 00

0

2

4 4

eqCl

p eqCl

ppn pp p p

Kn n p n pp n p

n pp


2 2

2 2 20

2

4 4

1

4

4

p

p p

p

p

p pK

n p p

pK K

p

or

Ka

pK

p

Thus, knowing Kp and the total pressure, we can calculate the equilibrium composition of the mixture.

Relative stability of Gases, Relative stability of Gases, Liquids, and SolidsLiquids, and Solids

Common experience: Common experience: Low Low TT favours solids favours solids High High TT favours gases favours gases Similarly for high and low pressure.Similarly for high and low pressure.

Hear we study the conditions under which two Hear we study the conditions under which two (or even three) phases co-exist in equilibrium, (or even three) phases co-exist in equilibrium, at a given at a given pp and and TT..

Phases: Solid, liquid, gases.Phases: Solid, liquid, gases. Gases exist in only one phase..Gases exist in only one phase..

Phases: Solid, liquid, gases.Phases: Solid, liquid, gases. Gases exist in only one phase..Gases exist in only one phase.. Liquids primarily also exist in only one Liquids primarily also exist in only one

phase..phase..


phase..phase.. Exception: supercritical liquidsException: supercritical liquids


phase..phase.. Exception: supercritical liquids.Exception: supercritical liquids.

Solids: can exist in several phases.Solids: can exist in several phases. E.g., crystal structures..E.g., crystal structures..

Water in a beaker, exists as a single Water in a beaker, exists as a single phase.phase.

Water and ice in a beaker = mixture of two Water and ice in a beaker = mixture of two distinct phases.distinct phases.

Conditions under which substances Conditions under which substances spontaneously form S, L, or Gspontaneously form S, L, or G

Common experienceCommon experience TT reduces from 300 to 250 K. reduces from 300 to 250 K. Water (liquid) turns to ice (solid)Water (liquid) turns to ice (solid)

T increases from 300 to 400 KT increases from 300 to 400 K Water turns to steam (gas)Water turns to steam (gas)


Solid COSolid CO22 at room temperature. at room temperature.

Sublimes: Turns from solid to gas, with out Sublimes: Turns from solid to gas, with out going through a liquid phase.going through a liquid phase.


What determines which phase is favoured What determines which phase is favoured (most thermodynamically stable) at a (most thermodynamically stable) at a given given pp and and TT??

What is the criterion for stability?What is the criterion for stability?


The minimising of the Gibb’s energy.The minimising of the Gibb’s energy.

For a pure substance,For a pure substance,

= chemical potential, = chemical potential, nn = mole fraction, = mole fraction,

, ,

?m

T p T p

nGG

n n


The minimising of the Gibb’s energy.The minimising of the Gibb’s energy.

For a pure substance,For a pure substance,

, ,

mm

T p T p

nGGG

n n


AsAs dd = =ddGGmm,,

m md S dT V dp


Thus, the variation of Thus, the variation of with with pp and and TT can be can be determined.determined.

m mp T

S and VT p


SSmm and and VVmm are always positive, thus; are always positive, thus;

decreases as decreases as TT increases, and increases, and Increases with increasing Increases with increasing pp..

m mp T

S and VT p


The entropy varies slowly with The entropy varies slowly with TT ( as ln ( as ln TT),), Thus, over a limited T range,Thus, over a limited T range, a plot of a plot of vv. . TT at const. at const. pp is a straight line of is a straight line of

negative slope.negative slope.


We know from experience that We know from experience that meltingmelting and and boilingboiling are endothermic.are endothermic.

Thus, Thus, SS = = HH//TT is positive for both of these is positive for both of these constant constant TT processes. processes.

We also know that Gasses, liquids and solids all We also know that Gasses, liquids and solids all have positive heat capacities.have positive heat capacities.


Therefore, Therefore,

.gas liq solidm m mS S S


The entropy of a phase is the magnitude of the slope The entropy of a phase is the magnitude of the slope of of

versusversus TT.. Recall Recall

mST


Thus, the functional relationship between Thus, the functional relationship between and and TT for for solids, liquids, and gasses (at a given solids, liquids, and gasses (at a given pp) can be ) can be expressed graphically.expressed graphically.

The stable state at any given The stable state at any given TT is the phase with the is the phase with the lowest lowest . .


Start in the solid phase and increase temperature.Start in the solid phase and increase temperature.

As As TT increases, increases, decreases with a certain slope..decreases with a certain slope..

Note the slopes for liquid and gas are greater.Note the slopes for liquid and gas are greater.

Therefore, they intersect.Therefore, they intersect.

The points of intersection of the solid/liquid and the The points of intersection of the solid/liquid and the liquid/gas are the melting and boiling temperatures, liquid/gas are the melting and boiling temperatures, respectively.respectively.


At the melting point (solid/liquid intersection) both At the melting point (solid/liquid intersection) both phases exist in equilibrium.phases exist in equilibrium.

However, a further, but small increase in However, a further, but small increase in TT results in results in complete melting.complete melting.

Why?Why?




Why?Why?

The Liquid phase has a lower The Liquid phase has a lower at at TTmm + d + dTT than the than the

solid phase.solid phase.




Why?Why?

The Liquid phase has a lower The Liquid phase has a lower at at TTmm + d + dTT than the than the

solid phase.solid phase.


Similarly, at Similarly, at TTbb both liquid and gas coexist at eq. both liquid and gas coexist at eq.

The system is a gas at The system is a gas at TT > > TTbb..


Note: the progression from solid to liquid to gas cam Note: the progression from solid to liquid to gas cam be fully explained only bybe fully explained only by

and and

mST

.gas liq solid

m m mS S S


What is we increase the temperature fast (too fast)?. What is we increase the temperature fast (too fast)?.



At a phase change, the system does not reach At a phase change, the system does not reach equilibrium, leading to super heating.equilibrium, leading to super heating.



At a phase change, the system does not reach At a phase change, the system does not reach equilibrium, leading to super heating (bumping).equilibrium, leading to super heating (bumping).

Similarly, rapid cooling leads to supercooling (e.g., Similarly, rapid cooling leads to supercooling (e.g., glass formation).glass formation).


What happens as a function of What happens as a function of p p at constant at constant TT??



m

T

Vp



Mostly, VMostly, VmmSolidSolid < V < Vmm

LiquidLiquid << V << VmmGasGas

m

T

Vp



Mostly, VMostly, VmmSolidSolid < V < Vmm

LiquidLiquid << V << VmmGasGas

Therefore, Therefore, versus versus TT changes more rapidly. changes more rapidly.

m

T

Vp


Note: Note: VVmmgasgas >> >> VVmm

liquidliquid >> 0. >> 0.

Therefore, increasing Therefore, increasing pp leads to an increase in the leads to an increase in the boiling point.boiling point.


Note: if Note: if VVmmliquidliquid > > VVmm

solidsolid

Increasing Increasing pp leads to melting point elevation leads to melting point elevation

if if VVmmliquidliquid < < VVmm

solidsolid

Increasing Increasing pp leads to melting point supression. leads to melting point supression.

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Calculating Equilibrium Composition Example Cl 2 (g) → 2Cl (g)