Question 4.3: Consider air (N2, N, O2, O, NO) and the particular nitrogen ...

N_{2}+N_{2} \rightarrow 2N + N_{2}        (4.50)

\nu ^{\prime}_{N_{2}}=2,\nu ^{\prime}_{O_{2}}=\nu ^{\prime}_{NO}=\nu ^{\prime}_{N}=\nu ^{\prime}_{O}=0        (4.51)

\nu ^{\prime\prime}_{N}=2,\nu ^{\prime\prime}_{N_{2}}=1 ,\nu ^{\prime\prime}_{O_{2}}=\nu ^{\prime \prime}_{NO}=\nu ^{\prime\prime}_{O}=0       (4.52)

Note that, unlike in equilibrium chemistry analysis, we must now consider the catalyst particle M in dissociation–recombination reactions:

N_{2} + M ⇔ N + N + M                   (4.53)

because, at the molecular level, two particles must collide for dissociation to
occur.

To describe the chemical composition, we will use species concentrations:

[X_{s}]  {mol}/{m^{3}}            (4.54)

The rate of formation of a product is given by

\frac{d[X_{s}]}{dt}= \nu ^{\prime \prime}_{s}k_{f}(T)\prod\limits_{s}{[X_{s}]^{Z^{\prime}_{s}}}         (4.55)

where k_{f} (T ) is the forward rate coefficient and is usually expressed in modified Arrhenius form:

k_{f}(T)= C_{f}T^{\eta f} exp \left(-\frac{\theta _{d}}{T} \right)         (4.56)

The rate of depletion of a reactant is similarly given by

\frac{d[X_{s}]}{dt}= -\nu ^{\prime}_{s}k_{f}(T)\prod\limits_{s}{[X_{s}]^{Z^{\prime}_{s}}}        (4.57)

Note that the order of reaction, Z^{\prime}_{1} + Z^{\prime}_{2} +···= Z^{\prime}  that  means  that  k_{f}  has  units  \left(\frac{mol}{m^{3}} \right)^{1-Z^{\prime}} \frac{1}{sec}.

As a specific example, consider dissociation–recombination of nitrogen.
(a) Dissociation:

Here, z^{\prime} = 2, so we have a second-order reaction. The time rates of change of the atom and molecule concentrations can be written

\frac{d[N]}{dt} = 2k_{f}(T)[N_{2}][M]       (4.58)

\frac{d[N_{2}]}{dt} = -k_{f}(T)[N_{2}][M]       (4.59)

(b) Recombination:

Considering both dissociation and recombination, the overall rate of change in concentration of atomic nitrogen is

\frac{d[N]}{dt}=\left(\frac{d[N]}{dt}\right)_{f}+\left(\frac{d[N]}{dt}\right)_{b}=2k_{f}(T)[N_{2}][M]+\left(\frac{d[N]}{dt}\right)_{b}    (4.60)

Under conditions of chemical equilibrium, the net rate of change of all variables including species concentrations is zero:

\Rightarrow \left(\frac{d[N]}{dt}\right)^{\divideontimes }_{b} = -2k_{f}(T)[{N_2}]^{\divideontimes }[M]^{\divideontimes }    (4.61)

where we use ∗ to indicate equilibrium. Recall the equilibrium constant, which we here express in terms of concentrations instead of pressures,

k_{c}(T)=\frac{([N]^\divideontimes )^{2}}{[N_{2}]^\divideontimes }       (4.62)

Thus,

\left(\frac{d[N]}{dt}\right)^{\divideontimes }_{b} = -2\frac{k_{f}(T)}{K_{c}(T)} [N]^{\divideontimes 2}[M]^\divideontimes     (4.63)

We now make the assumption that chemical processes in any one direction proceed at rates determined by the temperature alone and are independent of whether or not equilibrium exists. This same approach is employed in experiments when changes in concentrations are measured and used to deduce rate coefficients. Thus

\left(\frac{d[N]}{dt}\right)^{\divideontimes }_{b} = -2\frac{k_{f}(T)}{K_{c}(T)} [N]^{2}[M]=-2k_{b}(T)[N]^2[M]        (4.64)

where the backward rate coefficient is

K_{b}(T)=\frac{k_{f}(T)}{K_{c}(T)}       (4.65)

In general, we can write

K_{c}(T)=C_{c}T^{\eta c}exp\left(-\frac{\theta _{d}}{T} \right)     (4.66)

so that

K_{b}(T)=\frac{C_{f}}{C_{c}} T^{\eta_{f}-\eta_{c}} =C_{b}T^{\eta_{b}}       (4.67)

Note that the activation energy for recombination is zero, and that recombination has order z^{\prime} = 3, so the units of kb are \frac{m^{6}}{mol^{2}.sec} .

(c) Combined system
Let us now consider the effects of all reactions leading to concentration changes in a relatively simple system consisting of molecular (N_{2}) and atomic (N) nitrogen.

\frac{d[N]}{dt} = 2k_{f}(T)[M]\left\{[N_{2}]-\frac{1}{K_{c}(T)}[N]^{2}\right\}      (4.68)

By inspection:

If [N]>[N]^{\divideontimes } \Rightarrow \frac{[N]^{2}}{K_{c}}> [N]_{2} \Rightarrow \frac{d[N]}{dt} < 0  because  K_{c}(T)= \frac{([N]^\divideontimes )^2 }{[N_{2}]^\divideontimes }

If [N]<[N]^{\divideontimes } \Rightarrow \frac{d[N]}{dt} > 0

Therefore, the system always moves toward equilibrium.
In the overall analysis, we must account for all possible catalysts: M=N_{2}, N; therefore

\frac{d[N]}{dt}=2\left\{k_{f1}(T)[N_{2}]+k_{f2}(T)[N]\right\}\left\{[N_{2}]-\frac{[N]^{2}}{K_{c}(T)} \right\}             (4.69)

where we use

The equilibrium constant is evaluated from

K_{c}(T)= \frac{K_{p}(T)}{KT\hat{N}} \approx C_{c}T^{\eta _{c}}exp\left(-\frac{\theta _{d}}{T} \right)           (4.70)

As shown in Fig. 4.8, the following approximate expression for nitrogen agrees well with more exact calculations that are based on detailed analysis of the partition functions (Park 1990)

K_{C}(T)=18  exp\left(-\frac{113,000}{T} \right) {mol}/{cm^{3}}       (4.71)