Reaction+Heat+Effects+and+Chemical+Equilibrium

Chapter 3:

Reaction Heat Effects and Chemical Equilibrium

Reading Assignment: Chapter 3 of Textbook

Structure of the Course

Batch CSTR Plug Flow Fixed-Bed Catalytic ReactorTypes of Reactors

Reaction Rate

Interaction of physical and chemical processes

Integration of the reaction equations;

determination of kinetic constants

Stoichiometry Thermostatics Chemical Kinetics

Chapter Objectives• Heat effects:

• Energy changes occurring during reactions Very important for energy balances

• Chemical equilibrium:

• Even though reactors are almost never at equilibrium, we need to study chemical reaction equilibrium in order to find the limits of reactor performance.

• Or, to find which changes in design or operating conditions will remove these restrictions and lead to improved reactor performance.

Heat of Formation• The difference between the enthalpy of one mole of a pure chemical

compound and the total enthalpy of the chemical elements of which is composed is called the heat of formation ΔHf of the compound.

• It represents the energy changes that come about when atoms are combined to form a molecule.

• Example: When 2 g of hydrogen and 16 g of oxygen (both gaseous) react to form 18 g of liquid water at 1 atm and 25°C, then 285.8 kJ will be released. H2 g( )+ 1

2O2 g( ) = H2O l( ) ΔH f = −285.8 kJ/mol

Problem

• The standard heats of formation of gaseous and liquid water from gaseous hydrogen and oxygen are respectively −241.82 kJ/mol and −285.8 kJ/mol. What does this difference represent?

Heat of Reaction

• The standard heat of reaction (at constant pressure) ΔH is defined as the difference between the enthalpy of the products and that of the reactants.

• Since heat is released when the enthalpy of the products is smaller than that of the reactants, reactions with negative ΔH are called exothermic.

• Those with positive ΔH absorb heat and are called endothermic.

PbO s( )+ S s( )+ 32

O2 g( ) = PbSO4 s( ) ΔH o = −700.5 kJ/mol

Heat of Reaction

General reaction: ν jAj = 0j=1

n

∑

If hj is the partial molal enthalpy of Aj : ΔH = ν jhjj=1

n

∑

Total enthalpy of a mixture: H = N jhjj=1

n

∑When the reaction is taking place: N j = N j0 +ν j XThen:

∂H∂X

=∂N j

∂Xhj + N j

∂hj∂X

⎧⎨⎩

⎫⎬⎭j=1

n

∑ = ν jhjj=1

n

∑ = ΔH †( )

This term vanishes

Heat of ReactionTo prove that the term in †( )vanishes, we will first show that

N j

∂hj∂Nkj=1

n

∑ = 0

Note that hj =∂H∂N j

and H = N jhj∑ . The partial molal enthalpy

hj will depend on N j as well as on the pressure and temperature, but

hk =∂

∂Nk

N jhj P,T ,N1,...,Nn( )j=1

n

∑ = hk + N j

∂hj∂Nkj=1

n

∑

∴ N j

∂hj∂Nkj=1

n

∑ = 0

Heat of ReactionFinally

N j

∂hj∂Xj

∑ = N jj∑ ∂hj

∂Nk

∂Nk

∂Xk∑ = ν k N j

∂hj∂Nkj

∑k∑ = 0

We conclude that

∂H∂X

=∂N j

∂Xhj + N j

∂hj∂X

⎧⎨⎩

⎫⎬⎭j=1

n

∑ = ν jhjj=1

n

∑ + N j

∂hj∂Xj=1

n

∑ ⇒

∂H∂X

= ν jhjj=1

n

∑ = ΔH

Heat of Reaction

Since ΔH f( ) j is the molal enthalpy of A j at standard conditions,

we have for ideal mixtures:

ΔH= ν j ΔH f( ) jj=1

n

∑

Thus, the heat of reaction can now be calculated from tables ofheats of formation.

Heat of ReactionIf U is the internal energy, V is the volume and P the pressure, the enthalpy is given by

H =U + PVAt constant pressure, dH = dU + PdVand from the first Law of Thermodynamics dU = dQ − dW( )

dH = −dQwhere Q is the heat released. Thus, ΔH is the heat that would be adsorbed by the reaction at constant pressure. In other words, ifthe reaction goes to an infinitesimal extent dX at constant pressurethe heat evolved would be

−ΔH( )dX

Heat Work

Heat of Reaction

To calculate how heat is evolved in a constant volume reaction, we maysuppose that

(a) the reaction is allowed to proceed to an infinitesimal extentdX at constant pressure and

(b) the volume is then restored isothermally.If υ j is the volume of one mole of Aj , the volume change will be

d N jυ jj∑⎛

⎝⎜⎞

⎠⎟= N jυ j

j∑⎛

⎝⎜⎞

⎠⎟dX = ΔV( )dX

Heat of ReactionThe internal energy change associated with an isothermal volumechange of dV is ∂U ∂V( )T dV and the second operation requires a heat removal of

dQ = dU + PdV = ∂U∂V

⎛⎝

⎞⎠ T

+ P⎧⎨⎩

⎫⎬⎭dV = ∂U

∂V⎛⎝

⎞⎠ T

+ P⎧⎨⎩

⎫⎬⎭ΔV( )dX

Thus, the net heat generated when the extent of reaction changes by dX at constant volume is

−ΔUdX = − ΔH − ∂U∂V

⎛⎝

⎞⎠ T

+ P⎡

⎣⎢

⎤

⎦⎥ ΔV( )⎧

⎨⎩

⎫⎬⎭dX

Heat of Reaction

The net heat generated when the extent of a reaction changes by dX is

At constant pressure: dQ = −ΔH( )dX

At constant volume: dQ = − ΔH − ∂U∂V

⎛⎝

⎞⎠ T

+ P⎡

⎣⎢

⎤

⎦⎥ ΔV( )⎧

⎨⎩

⎫⎬⎭dX

The standard entropy change ΔS is related to the change in Gibbs free energy ΔG and ΔH by ΔG = ΔH −TΔS

Key variable for chemical equilibrium

Variation of heat of reactionHeat capacities for mixture and Aj :

∂H∂T

⎛⎝

⎞⎠ P

= CP and ∂hj∂T

⎛⎝⎜

⎞⎠⎟ P

= cPj

∂H∂P

⎛⎝

⎞⎠ T

= T ∂S∂P

⎛⎝

⎞⎠ T

+V =V −T ∂V∂T

⎛⎝

⎞⎠ P

=V 1− a( )

For an equation of state: PV = ZNRT

ΔH − ΔH 0 = ∂ΔH∂P

⎛⎝

⎞⎠ TdP

1

P

∫ + ∂ΔH∂T

⎛⎝

⎞⎠ PdT

298

T

∫

Compressibility factor

Coefficient of thermal expansion

Standard value at 1 atm and 25°C

Rate of heat generation

At constant pressure:dQ = −ΔH( )dX

dQdt

= −ΔH( ) dXdt

= −ΔH( )R*

q = 1VdQdt

= −ΔH( ) R*

V= −ΔH( )r

Rate of heat generationFor simultaneous reactions: ν ijAj

j=1

n

∑ i = 1,2,...,m

we have

q = 1VdQdt

= −ΔH( )ii=1

m

∑ ri

If the pressure varies while the volume is held constant:

q = 1VdQdt

= −ΔH( )υi=1

m

∑ ri

where −ΔH( )υ = − ΔH + ∂U∂V

⎛⎝

⎞⎠ T

+ P⎧⎨⎩

⎫⎬⎭

ν j∑ υ j

⎡

⎣⎢

⎤

⎦⎥

Chemical Equilibrium

• State variables: Temperature, pressure and number of moles of each component.

• The difference in Gibbs energy between two states with slightly different temperatures, pressures and number of moles of each component is:where µj is the chemical potential of component Aj.

dG = −SdT +VdP + µ jdN j∑

Chemical Equilibrium

At constant P and T and with composition changes restrained bydN j = ν jdX

the condition for equilibrium becomes: ν jµ jj=1

n

∑ = 0

For ideal mixtures: µ j = µ j0 + RT lnPj

At equilibrium: RT ν j lnPjj=1

n

∑ = − ν jµ j0

j=1

n

∑

Chemical Equilibrium

ν j lnPjj=1

n

∑ = − 1RT

ν jµ j0

j=1

n

∑

Taking the exponentials of both sides:

P1ν1 ⋅P2

ν2 ⋅...⋅Pnνn = exp −

ν jµ j0

j=1

n

∑RT

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Pjν j

j=1

n

∏ = KP

Chemical EquilibriumIn the non-ideal case, partial pressures are replaced by partial fugacities:

µ j = µ j0 + RT ln f j

At equilibrium: f jν j

j=1

n

∏ = K T( ) = exp −ν jµ j

0

j=1

n

∑RT

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Fugacity and pressure are related by: f = Pexp υRT

− 1P

⎛⎝

⎞⎠0

P

∫ dP = γ P

If the fugacity coefficients are known: Pjν j

j=1

n

∏ = K T( ) γ j−ν j

j=1

n

∏

Chemical Equilibrium

Using activities:

µ j = Gj0 + RT ln aj( ) = Gj

0 + RT lnf jf j

0

⎛

⎝⎜⎞

⎠⎟

At equilibrium: ν jµ jj∑ = ν jGj

0 + RTj∑ ν j ln aj( )

j∑ = 0 ⇒

− ΔG0

RT= ν j ln aj( )

j∑ ⇒

exp − ΔG0

RT⎛⎝⎜

⎞⎠⎟= aj

ν j

j∏ ⇒ K = aj

ν j

j∏

ExampleFor the reaction isobutane + 1-butene ⎯→⎯←⎯⎯ 2,2,3-trimethylpentane

determine the equilibrium composition at a pressure of 2.5 atmand temperature 400°K, assuming that the gases are ideal. Thestandard Gibbs energy for this reaction at 400°K is-3.72 kcal/mol.

For ideal gases: f j = Pj and aj =Pjf j

0 =Pj

1 atm

Then: K = aPaIaB

⇒ K = PPPIPB

= yPyI yBP

Also: yI + yB + yP = 1Three unknowns

But only two equations

Example

• We need additional information to calculate the equilibrium composition.

• Use mole balances and the reaction extent to derive an algebraic equation for the extent of reaction starting from the equilibrium equation:

• Note that K is a function of the system temperature, but not a function of the system pressure or composition.

K = exp − ΔG0

RT⎛⎝⎜

⎞⎠⎟= aj

ν j

j∏

Gibbs energy• The equilibrium state is a minimum in the Gibbs energy and this

minimum is unique.

• The Gibbs energy of a single phase system can be computed from the chemical potentials and the composition:

G = µ jN jj∑

Since µ j = Gj0 + RT ln yj + lnP⎡⎣ ⎤⎦

G = N jGj0 + RT N j ln yj + lnP⎡⎣ ⎤⎦∑

j∑

Modified Gibbs energy

Since N jGj0 =∑ N j0Gj

0 + X ν j∑∑ Gj0

N jGj0 =

j∑ N j0Gj

0 + XΔG0

j∑

Modified Gibbs energy: !G T ,P, ′ξ( ) =G − N j0Gj

0

j∑N0RT

!G = − ′ξ lnK + 1+ν ′ξ( )lnP + yj0 +ν j ′ξ( )j∑ ln

yj0 +ν j ′ξ1+ν ′ξ

⎛⎝⎜

⎞⎠⎟

Gibs energy changeStandard Gibbs energy of formation: Gj

0f = Gj

0 − ajlGEl0

l∑

Gibbs energy change for the reaction at 25°C: ΔGi0 = ν jGj

0f

j=1

n

∑

Temperature dependence: ∂G∂T

⎛⎝

⎞⎠ P,N j

= −S and ∂Gj

0

∂T⎛

⎝⎜⎞

⎠⎟ P,N j

= −Sj0

For a single reaction:∂ ν jGj

0( )∂T

= − ν jS j0

j∑

j∑ ⇒ ∂ΔG

0

∂T= −ΔS0

Multiple reactions

For multiple reactions: dN j = ν ijdXii=1

m

∑Substitute in the equilibrium condition dG = 0 to get

ν ijµ jdXi = 0j=1

n

∑i=1

m

∑If the reactions are independent, this expression will vanish

for arbitrary dXi if ν ijµ j = 0j=1

n

∑ i = 1,2,...,m

Thus, for each reaction: Ki = ajνij

j∏ with Ki = exp − ΔGi

0

RT⎛⎝⎜

⎞⎠⎟