Complex Systems
5.3 REVERSIBLE REACTIONS .1 Net Rate and Forms of Rate Law
5.3.4 Optimal T for Exothermic Reversible Reaction
An important characteristic of an exothermic reversible reaction is that the rate has an optimal value (a maximum) with respect to Tat a given composition (e.g., as measured by fA). This can be shown from equation 5.3-14 (with n = 1 and Keq = KC,eq). Since gf and g, are independent of T, and Y = r,,lvo (in equation 5.3-l),
(5.3-21)
(5.3-22)
since Keq = greqlgf,eq, and dKeqK.s
tive, and (gfkkgf,eqkreq) > 1 kf
= d InK,,. Since dk/dT is virtually always posi-
> gf,e4 and g, < gr,J, the first term on the right in equation 5.3-22 is positive. The second term, however, may be positive (endothermic reaction) or negative (exothermic reaction), from equation 3.1-5.
Thus, for an endothermic reversible reaction, the rate increases with increase in tem- perature at constant conversion; that is,
(drDldT), > 0 (endothermic) (5.3-23) For an exothermic reversible reaction, since AH” is negative, (drDldT), is positive or negative depending on the relative magnitudes of the two terms on the right in equation 5.3-22. This suggests the possibility of a maximum in r,, and, to explore this further, it is convenient to return to equation 5.3-3. That is, for a maximum in rb,
dr,idT = 0,and (5.3-24)
dk, dk,
gfz = ET,* (5.3-25)
Using equation 3.1-8, k = A exp(-E,lRT) for kf and k, in turn, we can solve for the temperature at which this occurs:
Topt = EAr iE”f ln(;;kf)l’ (5.3-26)
(a)
For the reversible exothermic first-order reaction A * D, obtain Topr in terms of fA, and, conversely, the “locus of maximum rates” expressing fA (at ro,,,,) as a function of T. Assume constant density and no D present initially.(b) Show that the rate (rn) decreases monotonically as fA increases at constant T, whether the reaction is exothermic or endothermic.
SOLUTION
(a) For this case, equation 5.3-3 (with r = rD) becomes
rD = kfCA - k,.c, (5.3-15)
That is,
and
gf = CA = cAo c1 - fA>
gr = cD = CAofA
Hence, from equation 5.3-26,
Topt = Ml 1qfyl
where fA = fA(%,max 7) and on solving equation (5.3-27) for fA, we have
fA(at rD,man) = [l + M2 exp(-MIW1-’
where
and
M2 = ArEArIAfEAf
(5.3-27)
(5.3-28)
(5.3-29)
(5.3-30)
(b) Whether the reaction is exothermic or endothermic, equation 5.3-15a can be written
rD = CA@f - (kf + kr)fAl (5.3-31)
from which
(d@fA)T = -c&f + k,) < o
(5.3-32) That is, ro decreases as fA increases at constant T.The optimal rate behavior with respect to T has important consequences for the design and operation of reactors for carrying out reversible, exothermic reactions. Ex- amples are the oxidation of SO, to SO, and the synthesis of NH,.
This behavior can be shown graphically by constructing the
rD-T-fA
relation from equation 5.3-16, in which kf, k,, and Ke4 depend on T. This is a surface in three- dimensional space, but Figure 5.2 shows the relation in two-dimensional contour form, both for an exothermic reaction and an endothermic reaction, withfA
as a function of T and (-rA) (as a parameter). The full line in each case represents equilibrium con- version. Two constant-rate ( -I~) contours are shown in each case (note the direction of increase in (- rA) in each case). As expected, each rate contour exhibits a maximum for the exothermic case, but not for the endothermic case.r
5 . 4 PARALhLkEACTIONS
A reaction network for a set of reactions occurring in parallel with respect to species A may be represented by
5.4 Parallel Reactions
101
1 . 0
0.8
0.6 fA
0 . 4
0.2
0.’ I I I
I 750 800 850 900 !
Temperature/K Temperature/K
(a) (b)
0.8
0.6 fA
0.4
/ (-r,j)l (-42
0.2
Figure 5.2 Typical (-rA)-T-fA behavior for reversible reactions: (a) exothermic reaction;
(b) endothermic reaction
IVAllA +
. ..%vpD+...I1.“4*lA + . ..%v.E+...
(5.4-1)
The product distribution is governed by the relative rates at which these steps occur.
For example, if the rate laws for the first two steps are given by
rDIvD = ( -c41)4v*1l = k*1(%41(C‘4~~
. .YlY4Il (5.4-2a)
and
IEl.V,E =
( -%2Y1?421 = kd%42k4~ * * MY421 (5.4-2b)
the relative rate at which D and E are formed is
rD _ vDVA2kAl(TkA1(CA~ * * *>
- -
~Ev*lkA2(~)g‘42(CA~~ * .>
(5.4-3)
rE
The product distribution depends on the factors (cA, . . . , T) that govern this ratio, and the design and operation of a reactor is influenced by the requirement for a favorable distribution.
From the point of view of kinetics, we illustrate here how values of the rate constants may be experimentally determined, and then used to calculate such quantities as frac- tional conversion and yields.
For the kinetics scheme
A+B+C; rB = kAICA
A+D+E; rD = lCAZCA
(5.4-4)
(a) Describe how experiments may be carried out in a constant-volume BR to mea- sure kAl and kA2, and hence confirm the rate laws indicated (the use of a CSTR is considered in problem 5-5);
(b) If kAl = 0.001 s-l and kA2 = 0.002 s-l, calculate (i) fA, (ii) the product distribution (c,, en, etc.), (iii) the yields of B and D, and (iv) the overall fractional yields of B and D, for reaction carried out for 10 min in a constant-volume BR, with only A present initially at a concentration CA, = 4 mol L-’ .
( c ) Using the data in (b), plot CA, cn and co versus t.
SOLUTION
(a) Since there are two independent reactions, we use two independent material balances to enable the two rate constants to be determined. We may choose A and B for this pur- pose.
A material balance for A results in
-dc,/dt = kAlCA + kA$A (5.4-5)
This integrates to
ln CA = ln CA0 - (kAl + kA& (5.4-6) In other words, if we follow reaction with respect to A, we can obtain the sum of the rate constants, but not their individual values.
If, in addition, we follow reaction with respect to B, then, from a material balance for B,
From equations 5.4-5 and -7,
dc,/dt = kAlCA (5.4-7)
-dCA/dC, = ('h + k&&l
which integrates to
cA = cAo - (1 + k,,&d(C, - '& (5.4-8) From the slopes of the linear relations in equations 5.4-6 and -8, kAl and kA2 can be de- termined, and the linearity would confirm the forms of the rate laws postulated.
(b) (i) From equation 5.4-6,
CA = 4eXp[-(0.001 + 0.002)10(60)] = 0.661 mol L-’
j-A = (4 - 0.661)/4 = 0.835
(ii) CA
is given in (i).From equation 5.4-8,
cB = cc = (4 - 0.661)/(1 + 0.002/0.001) = 1.113 mol L-’
From an overall material balance,
CD = CB = CA0 - CA - c, = 2.226 Ill01 L-’