Heats of Reaction and Temperature.
Chapter 5: Standard free energy ∆GO for reaction at
temperature T
Temperature Effect
Heats of Reaction and Temperature.
Heats of Reaction and Temperature.
Equilibrium Conversion.
The equilibrium composition, as governed by the
The first problem is to evaluate the heat of reaction at temperature T2 equilibrium constant, changes with temperature, and
knowing the heat of reaction at temperature T1. This is found by the law of
conservation of energy as follows:
from thermodynamics the rate of change is given by:
When the heat of reaction DH, can be considered to be constant in the
In terms of specific heats temperature interval, integration yields
When the variation of DH, must be accounted for in the integration we have
Find the variation of the equilibrium
Temperature effects constant, hence, equilibrium
conversion, with temperature
• Irreversible reaction Equilibrium Conversion– Temperature: reversible reaction 1st
➢Endothermic: Conversion increases→ T decreases order
Decreasing of
reactant’s T
concentration decreases
when reaction
conversion rate
increases decreases
➢ Exothermic: Conversion increases→ T increases Figure 9.1 Effect of temperature on
equilibrium conversion as predicted
When X is low by thermodynamics (pressure fixed.)
The increasing of reaction rate due to the
increasing of T is higher than: ➢ The thermodynamic equilibrium constant is unaffected by the pressure of the system, by the
presence or absence of inerts, or by the kinetics of the reaction, but is affected by the
temperature of the system.
The decreasing of reaction rate due to the ➢ Though the thermodynamic equilibrium constant is unaffected by pressure or inerts, the
decreasing of reactant’s concentration equilibrium concentration of materials and equilibrium conversion of reactants can be
influenced by these variables.
Temperature effects Equilibrium Conversion– Temperature: reversible reaction 1st
Rate of irreversible reaction vs conversion in order
adiabatic reactor
Reaction rate
Reaction rate
Figure 9.1 Effect of temperature on
equilibrium conversion as predicted
by thermodynamics (pressure fixed.)
➢ K>> 1 indicates that practically complete conversion may be possible and that the reaction
Conversion Conversion can be considered to be irreversible. K <<1 indicates that reaction will not proceed to any
appreciable extent.
Endothermic Exothermic ➢ For an increase in temperature, equilibrium conversion rises for endothermic reactions and
drops for exothermic reactions.
➢ For an increase in pressure in gas reactions, conversion rises when the number of moles
decreases with reaction; conversion drops when the number of moles increases with reaction.
➢ A decrease in inerts for all reactions acts in the way that an increase in pressure acts for gas
reactions.
Equilibrium Conversion XAe XAe and Temperature
1 1
X Ae =
1 DH RX ( TR ) 1 1
endothermic K C ( T2 )
exp
R T − T + 1
A + heat B 2
XA,e
DH RX ( TR ) 1 1
Exothermic & DCP =0: DH RX ( TR ) 0, when T exp T − T & X Ae
exothermic R 2
A B + heat
0 Makes sense from Le Chatelier’s principle
T
Example) A⇌B CA0=1 CB0=0 Exothermic rxn produces heat→ A B + heat
increasing temp adds heat (product) & pushes rxn to left (lower conversion)
CBe CA0 ( 0 + X Ae ) X Ae Rearrange to
KC = = → KC =
CAe C A0 (1 − X Ae ) 1 − X Ae solve in terms of DH RX ( TR ) 1 1
Endothermic & DCp 0: DH RX ( TR ) 0, when T exp T − T & X Ae
XAe R 2
→ K C (1 − X Ae ) = X Ae → K C = X Ae + K C X Ae → K C = X Ae (1 + K C )
Makes sense from Le Chatelier’s principle A + heat B
KC This equation enables us to Heat is a reactant in an endothermic rxn→
→ =X
(1 + K C ) Ae express Xae as a function of T increasing temp adds reactant (heat) & pushes rxn to right (higher
conversion)
XAe and Temperature
KC DH RX ( TR ) 1 1
X Ae = K C ( T ) = K C ( T2 ) exp T − T KC
(1 + K C ) R 2 X Ae =
(1 + K C )
DH RX ( TR ) 1 1 Example 5.1:
K C ( T2 ) exp T − T
Substitute X =
Ae
R 2 Divide numerator EX 9.2 p.213
for KC: DH RX ( TR ) 1 1 & denominator
1 + K C ( T2 ) exp T − T by KC
R 2 1
1 = e− X
→ X Ae = eX
1 DH RX ( TR ) 1 1 Changed sign
K C ( T2 )
exp
R T − T + 1
2
DH RX ( TR ) 1 1
Exothermic: DH RX 0, when T exp T − T and X Ae
R 2
DH RX ( TR ) 1 1
Endothermic: DH RX 0, when T exp T − T and X Ae
R 2
General Graphical Design Procedure General Graphical Design Procedure
Temperature, composition, and reaction rate are uniquely related for any
single homogeneous reaction
The size of reactor required for a given duty and for a given temperature progression
is found as follows:
1. Draw the reaction path on the XA versus T plot. This is the operating line for the
operation.
2. Find the rates at various XA along this path.
3. Plot the 1/(-rA) versus XA curve for this path.
4. Find the area under this curve. This gives V/FAo.
For exothermic reactions we illustrate this procedure in Fig. 9.4 for three paths: path
AB for plug flow with an arbitrary temperature profile, path CD for nonisothermal plug
flow with 50% recycle, and point E for mixed flow. Note that for mixed flow the
operating line reduces to a single point.
➢ The first of these, the composition-temperature plot, is the most convenient so we will use
it throughout to represent data, to calculate reactor sizes, and to compare design
alternatives
General Graphical Design Procedure
General Graphical Design Procedure For exothermic reactions we illustrate this procedure in Fig. 9.4 for three paths: path
AB for plug flow with an arbitrary temperature profile, path CD for nonisothermal
➢ For a given feed (fixed CAo,CBo...) and using conversion of key component as a measure of the plug flow with 50% recycle, and point E for mixed flow. Note that for mixed flow the
operating line reduces to a single point.
composition and extent of reaction, the XA versus T plot has the general shape
Optimum Temperature Progression
❑ For irreversible reactions, the rate always increases with temperature at
any composition, so the highest rate occurs at the highest allowable
temperature. This temperature is set by the materials of construction or
by the possible increasing importance of side reactions.
Example 5.2: ❑ For endothermic reversible reactions a rise in temperature increases
both the equilibrium conversion and the rate of reaction. Thus, as with
EX 9.3 irreversible reactions, the highest allowable temperature should be used.
❑ For exothermic reversible reactions the situation is different, for here two
opposing factors are at work when the temperature is raised-the rate of
forward reaction speeds up but the maximum attainable conversion
decreases. Thus, in general, a reversible exothermic reaction starts at a
high temperature which decreases as conversion rises.
Optimum Temperature Progression
➢ We define the optimum temperature progression to be that progression which
minimizes V/FAo, for a given conversion of reactant.
➢ This optimum may be an isothermal or it may be a changing temperature: in time for
a batch reactor, along the length of a plug flow reactor, or from stage to stage for a
series of mixed flow reactors. It is important to know this progression because it is the
ideal which we try to approach with a real system. It also allows us to estimate Read in advanced:
how far any real system departs from this ideal.
➢ At any composition, it will always be at the temperature where the rate is Adiabatic/non adiabatic operations
a maximum.
Example 9.4 9.5 9.6
CSTR, PFR, Fig 9.14
Chapter: Nonideal Flow
Operating lines for minimum reactor size.
Heat efects Adiabatic operations
❑ When the heat absorbed or released by reaction can markedly change the
temperature of the reacting fluid, this factor must be accounted for in
design.
❑ Use both the material balance and energy balance expressions.
When C”p –C’p= 0:
❑ If the irreversible reaction is exothermic and if heat transfer is unable to
remove, all of the liberated heat, then the temperature of the reacting fluid
will rise as conversion rises.
which are straight lines in the Fig
❑ For irreversible endothermic reactions the fluid cools as conversion rises.
❑ Adiabatic operations
Graphical representation of energy balance equation for adiabatic operation. These are adiabatic
operating lines.
➢ This figure illustrates the shape of the energy balance curve for both
endothermic and exothermic reactions for both mixed flow and plug flow
reactors. This representation shows that whatever is the conversion at
any point in the reactor, the temperature is at its corresponding value on
the curve.
➢ For plug flow the fluid in the reactor moves progressively along the curve.
Adiabatic operations with large enough heat effect to cause a rise in ➢ For mixed flow the fluid immediately jumps to its final value on the curve.
temperature (exothermic) or drop in temperature (endothermic) in the ➢ A vertical line indicates that temperature is unchanged as reaction
reacting fluid. proceeds➔ the special case of isothermal reactions.
Adiabatic operations Adiabatic operations
❑Find the size of the reactor:
❑ Energy balance ➢ For plug flow tabulate the rate for various XA along this adiabatic
operating line, prepare the 1/(-rA) versus XA plot and integrate.
input = output + accumulation + disappearance by reaction
➢ For mixed flow simply use the rate at the conditions within the
reactor.
➢ (see figure 9.8)
Enthalpy of entering feed:
Enthalpy of leaving stream:
Energy absorbed by reaction:
At steady state, and rearranging:
the heat released by reaction just balances the heat necessary to raise the
reactants from T1 to T2
For CSTR For CSTR
V V X Af − X A0
• Solve these 3 eqs: = = (3.3)
FA0 . C A0 ( − rAf )
▪ Reacton rate V
= : Thoi gian the tich
▪ Material balance (Design equation)
V V 1 X − X A0 C (X − X A0 )
▪ Energy balance = = = Af → = A0 Af
FA0 . C A0 CA0 ( − rAf ) ( − rAf )
m t (T0 − Tf ) C p − (X Af − X A0 )DH r FA0 + KS(Tn − Tf ) = 0
hay m t (T0 − Tf ). C p − ( − rA ). V. DH r + K.S.(Tn − Tf ) = 0
For CSTR
For CSTR- 1st order reaction
• Given V, calculate X and T:
➔Determine T and composition of outlet feed: need to
solve 3 equations by trial method.
• Given X, Calculate T and V k 0 e − E/RT f
X Af = (5.2)
➔Solve independently energy balance eq →Determine 1 + k 0 e − E/RT f
T→ Determine reaction rate→ Find V
m t Cp
X Af = (Tf − T0 ) (5.3)
FA0 ( − DH r )
CSTR Stability R(T) > G(T) so For PFR
T gradually falls
G(T) > R(T) so T to T=SS3 Adiabatic operations
gradually rises to
G(T) R(T) T=SS3
R(T) > G(T) 3
G(T) & R(T)
so T gradually
falls to T=SS1
m t C P dT = FA0 (− ΔH r )dX A
G(T) > R(T) so 2
T gradually (5.13)
rises to T=SS1
1 − (DH r )FA0
Temperature T − T0 = (X A − X A0 ) (5.14)
T m t CP
3 steady states satisfy the TEB and BMB
• Suppose a disturbance causes the reactor T to drift to a T between SS1 & SS2
• Suppose a disturbance causes the reactor T to drift to a T between SS2 & SS3
• Suppose a disturbance causes the reactor T to drop below SS1
• Suppose a disturbance causes the reactor T to rise above SS3
SS1 and SS3 are locally stable (return to them after temp pulse)
SS2 is an unstable- do not return to SS2 if there is a temp pulse
❑ Best location for the adiabatic operating line.
For Batch reactor For plug flow, a trial and error search is needed to find this line; for
mixed flow, no search is needed
Adiabatic operations
➢ The best adiabatic operations of a single plug flow reactor are found by
Enery balance: shifting the operating line (changing the inlet temperature) to where the
dT dX A rates have the highest mean value. For endothermic operations this
m t CP = − (DH r )N A0 + KS(T0 − Ti ) (5.4) means starting at the highest allowable temperature. For exothermic
reactions this means straddling the locus of maximum rates as shown in
dt dt Fig.
➢ A few trials will locate the best inlet temperature, that which minimizes
V/FAo. For mixed flow the reactor should operate on the locus of
m t C P dT = − (DH r ) N A0 dX A (5.5) maximum rates.
− DH r N A0
T − T0 = (X A − X A0 ) (5.6)
m t CP
Nonadiabatic Operations How does one increase XA for adiabatic operation of
an exothermic reaction?
Reactor with Heating or Cooling with interstage cooling!
XA,EB4
XEB
final conversion
XA,EB3 Each reactor
operates
XA,EB2 adiabatically
XA,EB1
cooling process
T
Q = UA ΔT Sketch of the energy balance Cooling, C1 C2 C3
equation showing the shift in T0 Reactor 1 Reactor 2 Reactor 3 Reactor 4
adiabatic line caused by heat
exchange with surroundings.
Endothermic reactions are similar, but with heating instead of cooling
Inter-stage Cooler Inter-stage Cold Feed
Lowers Temp. Lowers Temp
Lowers Conversion
Exothermic Equilibria Exothermic Equilibria
Ways of approaching the ideal temperature profile with heat exchange
exothermic reaction
Ways of approaching the ideal temperature profile with heat
exchange: (c) endothermic reaction.
