Modern Engineering Thermodynamics

632 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.15 (Continued)
a. To determine the required results for part a of this problem, we set p m = p° =0:1 MPa, and then K e can be found in
Table C.17 for various temperatures. Our task is now to pick several reaction temperatures, look up their corresponding
K e values, and solve the previous equation for y.
Squaring both sides of the preceding reversible equilibrium dissociation reaction equation and rearranging gives a
cubic equation in y:
where
y 3 +
3α
1 − α y − 2α
1 − α = 0
α = Ke
2 p°
p m
A cubic equation in y has a simple algebraic solution. If we let
a =
3α
1 − α
and b = −
2α
1 − α
then the cubic equation becomes y 3 + ay + b = 0, and the three roots of this equation are
where
y 1
y 2
y 3
= A + B
= − A + B + A − B pffiffiffiffiffiffiffiffi
− 3
2 2
= − A + B − A − B pffiffiffiffiffiffiffiffi
− 3
2 2
r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1/3 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1/3
A = − b 2 + b 2
4 + a3 and B = − b 27
2 − b 2
4 + a3
27
Note that, in evaluating these expressions, the cube root of a negative number is computed as the negative of the cube
root of the positive value of the number; that is, (−8) 1/3 = −(8 1/3 ) = −2.
If α is less than 1, then b 2 /4 + a 3 /27 > 0 and there is just one real root, given by y 1 .Butifα is greater than 1, then
b 2 /4 + a 3 /27 < 0 and there are three real roots. The following procedure is a simple way of calculating these roots. Let
J = [(− (a/3) 3 ] 1/2 and K = J 1/3 . Then calculate L = arccos[−b/(2J)], M = cos(L/3), and N = (3 1/2 )sin(L/3). The three roots are
then given by
y 1 = 2K cosðL/3Þ
y 2 = − KðM + NÞ
y 3 = − KðM − NÞ
Note that the only valid root here is the one between 0 and 1. Table 15.8 lists several typical results using these solution
equations with p°=0.1 MPa.
These results can then be plotted either by hand or with the use of computer software. The final plot showing
the variation in the degree of dissociation (y) versus reaction equilibrium temperature is shown in Figure 15.10.
b. With the solution from part a, it is a simple matter to generate reaction equations for various total pressures at T = 3000. K.
The curve in Figure 15.11 shows the results of this effort and provides a graphical representation of the effect of total
pressure on the degree of dissociation (y) at 3000. K.
Table 15.8 Typical Values for the Degree of Dissociation of Carbon Dioxide at Various
Temperatures and Pressures with p°=0.100 MPa
Reaction
Temperature (K)
Reaction Pressure
p m (MPa) log 10 (K e ) K e y
2000. K 0.100 MPa 2.863 0.0014 0.0154
3000. K 0.100 MPa 0.469 0.3396 0.4436
5000. K 0.100 MPa +1.387 24.378 0.9770
3000. K 1.00 MPa 0.469 0.3396 0.2453
3000. K 10.0 MPa 0.469 0.3396 0.1235