Page 325 - Advanced thermodynamics for engineers

314 CHAPTER 14 CHEMICAL KINETICS

But
k f1 ½NŠ ½NOŠ ¼ k b1 ½N 2 Š ½OŠ ¼ R 1 ;
e
e
e
e
so the net rate for Eqn (14.19) becomes abR 1 þ R 1 .
Using a similar procedure for Eqns (14.20), (14.21) and (14.24) involving NO gives the following
expression, which allows for the change in volume over a time step,
1 d
ð½NOŠVÞ ¼ aðbR 1 þ R 2 þ R 3 þ 2aR 6 Þ þ R 1 þ bðR 2 þ R 3 Þ þ 2gR 6 : (14.26)
V dt
where V is the volume of the products zone.

14.4.1.2 Expression for Atomic Nitrogen, N
Equations (14.19)–(14.21), which all involve N, give
1 d
ð½NŠVÞ¼ bðaR 1 þ R 2 þ R 3 Þþ R 1 þ aðR 2 þ R 3 Þ: (14.27)
V dt

14.4.1.3 Expression for Nitrous Oxide N 2 O
Equations (14.22)–(14.25) all involve N 2 O and can be combined to give
1 d 2
ð½N 2 OŠVÞ¼ gðR 4 þ R 5 þ R 6 þ R 7 Þþ R 4 þ R 5 þ a R 6 þ R 7 : (14.28)
V dt
A finite time is required for the reactions to reach their equilibrium values; this is called the
relaxation time. It has been found (Lavoie et al. (1970)) that the relaxation times of reactions (14.27)
and (14.28) are several orders of magnitude shorter than those of reaction (14.26), and hence it can be
assumed that the [N] and [N 2 O] values are at steady state, which means that the right-hand sides of
Eqns (14.27) and (14.28) are zero. Then, from Eqn (14.27)

R 1 þ aðR 2 þ R 3 Þ
b ¼ ; (14.29)
ðaR 1 þ R 2 þ R 3 Þ
and, from Eqn (14.28)

2
R 4 þ R 5 þ a R 6 þ R 7
g ¼ : (14.30)
ðR 4 þ R 5 þ R 6 þ R 7 Þ
These values can be substituted into Eqn (14.26) to give
1 d 2 R 1 R 6
ð½NOŠVÞ¼ 2 1 a þ (14.31)
V dt 1 þ a½R 1 =ðR 2 þ R 3 ފ 1 þ½R 6 =ðR 4 þ R 5 þ R 7 ފ
This is the rate equation for NO that is solved in computer programs to evaluate the level of NO in
the products of combustion. It should be noted that the variation in the molar concentration is a first-
order differential equation in time and relates the rate of change of [NO] to the instantaneous ratio of
the actual concentration of NO to the equilibrium value (i.e. a). When the actual level of [NO] is at the
equilibrium level then a ¼ 1 and the rate of change of [NO] ¼ 0.