Chapter 9, Problem 9.3P

(Flame retardants) Hydrogen radicals are important to sustaining combustion reactions. Consequently, if chemical compounds that can scavenge the hydrogen radicals are introduced, the flames can be extinguished. While many reactions occur during the combustion process, we shall choose CO flames as a model system to illustrate the process (S. Senkan et al., Combustion and Flame, 69, 113). In the absence of inhibitors

${\text{O}}_2\to \text{O}\cdot +\text{O}\cdot$            (P9-3.1)

${\text{H}}_2\text{O}+\text{O}\cdot \to 2\text{OH}\cdot$        (P9-3.2)

$\text{CO}+\text{OH}\cdot \to {\text{CO}}_2+\text{H}\cdot$    (P9-3.3)

$\text{H}\cdot +{\text{O}}_2\to \text{OH}\cdot +\text{O}\cdot$        (P9-3.4)

The last two reactions are rapid compared to the first two. When HCl is introduced to the flame, the following additional reactions occur:

$\begin{array}{l}\text{H}\cdot +\text{HCl}\to {\text{H}}_2+\text{Cl}\cdot \\ \text{H}\cdot +\text{Cl}\cdot \to \text{HCl}\end{array}$

Assume that all reactions are elementary and that the PSSH holds for the O·, OH·, and Cl· radicals.

1. (a) Derive a rate law for the consumption of CO when no retardant is present.
2. (b) Derive an equation for the concentration of H· as a function of time, assuming constant concentration of O₂, CO, and H₂O for both uninhibited combustion and combustion with HCl present. Sketch H· versus time for both cases.

Interpretation Introduction

Interpretation:

Rate law for the consumption of $CO$ in the absence of retardant has to be given.

Concept Introduction:

Rate law or rate equation: Rate law:

It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

  $\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

Explanation of Solution

Uninhibited combustion can be given as:

  $\begin{array}{l}{O}_2\stackrel{k_1}{\to }2O·\\ \\ O·+H_{2}O\stackrel{k_2}{\to }2OH·\\ \\ CO·+OH·\stackrel{k_3}{\to }C{O}_2+H·\\ \\ H·+O_2\stackrel{k_4}{\to }OH·+O·\end{array}$

Reaction when $HCl$ can be given as:

  $\begin{array}{l}HCl+H·\stackrel{k_5}{\to }{H}_2+Cl·\\ \\ H·+Cl·\stackrel{k_6}{\to }HCl\end{array}$

Rate law for the consumption of $CO$ in the absence of retardant follows as:

$-r_{CO}=r_{C{O}_2}=k_3\left(CO\right)\left(OH·\right)$

Interpretation Introduction

Interpretation:

Equation for the concentration of $H·$ as a function of time by assuming constant concentration of $O_2,CO,and{H}_{2}O$ for both uninhibited combustion and combustion with $HCl$ present has to be given. Also $H·$ versus time for both cases has to be sketched.

Concept Introduction:

Rate law or rate equation: Rate law:

It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

  $\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

Explanation of Solution

Uninhibited combustion can be given as:

  $\begin{array}{l}{O}_2\stackrel{k_1}{\to }2O·\\ \\ O·+H_{2}O\stackrel{k_2}{\to }2OH·\\ \\ CO·+OH·\stackrel{k_3}{\to }C{O}_2+H·\\ \\ H·+O_2\stackrel{k_4}{\to }OH·+O·\end{array}$

Reaction when $HCl$ can be given as:

  $\begin{array}{l}HCl+H·\stackrel{k_5}{\to }{H}_2+Cl·\\ \\ H·+Cl·\stackrel{k_6}{\to }HCl\end{array}$

Rate law for the consumption of $CO$ in the absence of retardant follows as:

$-r_{CO}=r_{C{O}_2}=k_3\left(CO\right)\left(OH·\right)$

Rate of formation of active intermediates is zero. Here, $\text{O}·,\text{OH}·\text{,}\text{H}·and\text{Cl}·$ are the active intermediates.

Hence,

  $\begin{array}{l}{r}_{O·}=0=2k_1\left(O_2\right)-k_2\left(O·\right)\left(H_{2}O\right)+k_4\left(H·\right)\left(O_2\right)\\ \\ \left(O·\right)=\frac{2k_1\left(O_2\right)+k_4\left(H·\right)\left(O_2\right)}{k_2\left(H_{2}O\right)}\end{array}$

Also,

  $\begin{array}{l}{r}_{OH·}=0=2k_2\left(O·\right)\left(H_{2}O\right)-k_3\left(CO\right)\left(OH·\right)+k_4\left(H·\right)\left(O_2\right)\\ \\ \left(OH·\right)=\frac{2k_2\left(O·\right)\left(H_{2}O\right)+k_4\left(H·\right)\left(O_2\right)}{k_3\left(CO\right)}\end{array}$

Substitute $\left(O·\right)$ as follows:

  $\begin{array}{c}\left(OH·\right)=\frac{2k_2\left(\frac{2k_1\left(O_2\right)+k_4\left(H·\right)\left(O_2\right)}{k_2\left(H_{2}O\right)}\right)\left(H_{2}O\right)+k_4\left(H·\right)\left(O_2\right)}{k_3\left(CO\right)}\\ \\ =\frac{4k_1\left(O_2\right)+3k_4\left(H·\right)\left(O_2\right)}{k_3\left(CO\right)}\end{array}$

  $\begin{array}{l}-r_{CO}=r_{C{O}_2}=k_3\left(CO\right)\left(OH·\right)\\ \\ -r_{CO}=4k_1\left(O_2\right)+3k_4\left(H·\right)\left(O_2\right)\end{array}$

$r_{H·}and{r}_{Cl·}$ is given as:

  $\begin{array}{c}{r}_{H·}=k_3\left(CO\right)\left(OH·\right)-k_4\left(H·\right)\left(O_2\right)-k_6\left(Cl·\right)\left(H·\right)\\ \\ r_{Cl·}=0=k_5\left(HCl\right)\left(H·\right)-k_6\left(H\right)\left(Cl·\right)\end{array}$

Therefore,

  $Cl·=\frac{k_5\left(HCl\right)}{k_6}$

Substitute equation for $Cl·and-r_{CO}$ and this in the equation of $r_{H·}$ as follows:

$\begin{array}{l}{r}_{H·}=4k_1\left(O_2\right)+3k_4\left(H·\right)\left(O_2\right)-k_4\left(H·\right)\left(O_2\right)-2k_5\left(HCl\right)\left(H·\right)\\ \\ r_{H·}=4k_1\left(O_2\right)+2\left[k_4\left(O_2\right)-2k_5\left(HCl\right)\right]\left(H·\right)\\ \\ r_{H·}=a+b\left(H·\right)\end{array}$

When volume is constant, concentration of $H·$ with respect to time can be given as:

$e^{-bt}\frac{d{C}_{H·}}{dt}-C_{H·}\times b\times e^{-bt}=a{e}^{-bt}$

By integration factor

  $\begin{array}{l}\frac{d\left(e^{-bt}{C}_{H·}\right)}{dt}=a{e}^{-bt}\\ \\ C_{H·}=-\frac{a}{b}+K_l{e}^{bt}\end{array}$

When $t=0$, $C_{H·}=0$ and hence $K_l=\frac{a}{b}$

Therefore,

  $C_{H·}=\frac{a}{b}\left(e^{bt}-1\right)$

When $k_4\left(O_2\right)>k_5\left(HCl\right)$, then “b” becomes positive.

When $k_4\left(O_2\right)<k_5\left(HCl\right)$, then “b” becomes negative.

Concentration of $CO$ with respect to time can be written as:

  $\begin{array}{c}\frac{d{C}_{CO}}{dt}=r_{CO}=-\left[4k_1\left(O_2\right)+3k_4\left(H·\right)\left(O_2\right)\right]\\ \\ =-a-p\left(C_{H·}\right)\\ \\ =-a-p\left(\frac{a}{b}{e}^{bt}-1\right)\end{array}$

Where,

  $\begin{array}{c}p\left(C_{H·}\right)=3k_4\left(O_2\right)C_{H·}\\ \\ p=3k_4\left(O_2\right)\\ \\ p=4k_1\left(O_2\right)\end{array}$

When $t=0$, $C_{CO}=C_{C{O}_0}$

Therefore,

  $C_{CO}=C_{C{O}_0}-at-p\frac{a}{b^2}{e}^{bt}+pt$