Page 92 - Introduction to chemical reaction engineering and kinetics

74 Chapter 4: Development of the Rate Law for a Simple System

SOLUTION

From the rate law and the material balance equation 2.2-10, the equation to be integrated
is

- dCA = k,&
CiCB

The result is rather tedious to obtain, but the method can be the same as that in Example
3-5: use of the stoichiometric relationship and the introduction of 5, followed by integration
by partial fractions and reversion to CA and cn to give

(4.3-3) /

where M = VBCA~ - VACB~. The left side is a linear function oft; kA can be determined
from the slope of this function.

Suppose the following data were obtained for the homogeneous gas-phase reaction 2A +
2B --) C + 2D carried out in a rigid 2-L vessel at 8OO’C.

PO1 (dPldt),l
Wa XAO (lcFa)min-’

46 0.261 -0.8
70 0.514 -7.2
80 0.150 -1.6

Assuming that at time zero no C or D is present, obtain the rate law for this reaction,
stating the value and units of the rate constant in terms of L, mol, s.

SOLUTION
From equation 4.2-6, in terms of A and initial rates and conditions, and an assumed form
of the rate law, we write

h-Ap)o = -(dPddt), = ‘&P&P,, P (1)

Values of (dpA/dt), can be calculated from the measured values of (dP/dt),, as shown in
Example 4-1. Values of PA0 and Pa0 can be calculated from the given values of P, and
XA,, (from equation 4.2-1). The results for the three experiments are as follows:

&Ad PBoI (dPddt),l
kPa kFa kFa rnin-’
12 34 -1.6
36 34 - 14.4
12 68 -3.2