Complicated reaction can be broken down into elementary steps.
Consider the combustion of methane shown below.
Reversible Steps
When we consider an elementary reaction, we often ignore the reverse reaction.
$$ \begin{align}
\mathrm{A} & \xrightarrow{k_1} \mathrm{B} \\
\mathrm{B} & \xrightarrow{k_{-1}} \mathrm{A}
\end{align} $$
If the reverse reaction is extremely slow, it could be ignored
Usually, we must include the reverse steps in out analysis.
We're going to first analyze the simplest case, a reversible unimolecular reaction
$$ \mathrm{A} \xrightleftharpoons{k_1}{k_{-1}} \mathrm{B} $$
Let write the rate law for this reaction
$$ \begin{align}
\frac{d\conc{A}}{dt} & = -k_1 \conc{A} + k_{-1}\conc{B} \\
\frac{d\conc{B}}{dt} & = k_1 \conc{A} - k_{-1}\conc{B}
\end{align} $$
Give initial concentrations of \( \conc{A}_0 \) and \( \conc{B}_0 \) we can define the extent of the reaction, \( x \)
$$ x = \conc{B} - \conc{B}_0 = \conc{A}_0 - \conc{A} $$
note that the initial concentrations are constant.
Developing the Integrated Rate Law
We need to use our rate law to develope an integrated rate law
At equilibrium the rate is zero so we would have
$$ -\frac{d\conc{A}}{dt} = \frac{d\conc{B}}{dt} = k_1 \conc{A}_{eq} - k_{-1} \conc{B}_{eq} = 0 $$
Doing some quick math we find that
$$ K_{eq} =\frac{\conc{B}_{eq}}{\conc{A}_{eq}} = \frac{k_1}{k_{-1}} $$
Parallel Reactions
Competing (parallel) reactions draw from the same reactant(s) to form different products
$$ \chem{A \xrightarrow{k_1} B},\; \chem{A \xrightarrow{k_2} C} $$
The overall reaction velocity will be
$$ v = -\frac{d\conc{A}}{dt} = k_1 \conc{A} + k_2 \conc{A} = \left( k_1 + k_2 \right) \conc{A} $$
Using the same method we have used several times now, we can generate the corresponding integrated rate law
$$ \conc{A}=\conc{A}_0e^{-\left( k_1 + k_2 \right)t} $$
The compound \(\beta\)-pinene isomerizes at 650 K to either 4-isopropenyl-1-methylcyclohexene (call it IMC), with a rate constant of \( k_1 = 0.22\,\mathrm{s}^{-1} \), or to myrcene, with a rate constant of \( k_2 = 0.13\,\mathrm{s}^{-1} \). If the reaction is carried out under kinetic control, what is the ratio of IMC to myrcene, and what is the half-life of the \(\beta\)-pinene?
Sequential Reactions
Sequential reactions are the basis of multi-step reactions
Products made in one step can be used in these next
There are chemical intermediates
The simplest set of sequential reactions is
$$ \chem{A} \xrightarrow{k_1} \chem{B} \xrightarrow{k_2} \chem{C} $$
Let's start by assuming that \( \conc{A}_0\ne 0 \) and \( \conc{B}_0 = \conc{C}_0 = 0 \)
So, for \(\chem{A}\) it is easy
$$ -\frac{d\conc{A}}{dt}=k_1\conc{A}\;\;\;\mathrm{and}\;\;\;\conc{A}=\conc{A}_0 e^{-k_1 t} $$
It is not the purpose of this class to drag you through the relatively difficult task of solving the differential equation. It is in your book on page 484 if you are interested.
The result is that
$$ \conc{B} = \frac{\conc{A}_0k_1}{k_2-k_1}\left( e^{-k_1t}-e^{-k_2t} \right) $$
Using these together we can find that
$$ -\frac{d\conc{A}'}{dt} = \left( 1-\frac{k_{-1}}{k_{-1}+k_2} \right) k_1 \conc{A}'\left( \conc{B}_0 - \conc{A}_0 + \conc{A}' \right) $$
Integrating this we arrive at
$$ \begin{align}
\frac{1}{\conc{B}_0-\conc{A}_0} & \left[ \ln \left( \frac{\conc{B}_0-\conc{A}_0+\conc{A}'}{\conc{A}'}-\ln \left( \frac{\conc{B}_0}{\conc{A}_0} \right) \right) \right] \\
& = k_1 \left( 1-\frac{k_{-1}}{k_{-1}+k_2} \right) t
\end{align} $$
Doing a lot of algebra we find that
$$ \begin{multline}
\conc{A}' = \conc{A}_0 \left( 1-\frac{\conc{A}_0}{\conc{B}_0} \right) \\ \left\{ \exp \left[ \left( \conc{B}_0 - \conc{A}_0 \right) \left( 1-\frac{k_{-1}}{k_{-1}+k_2} \right)k_1 t \right] - \frac{\conc{A}_0}{\conc{B}_0} \right\}^{-1}
\end{multline} $$
Example 14.2
Show that the last equation correctly predicts the rate law of an irreversible bimolecular reaction \( \chem{A+B} \xrightarrow{k_1} \chem{D} \) in the limit that \( k_{-1} \ll k_1 \)
If \( k_{-1} \gg k_2 \) then the fast-equilibrium approximation may be applicable.
Due to the relative sizes of the rate constants, A, B, and C quickly reach their equilibrium values.
$$ \frac{\conc{C}}{\conc{A}{\conc{B}}} \approx K_1 = \frac{k_1}{k_{-1}} $$
Solving this we find that
$$ \conc{C}_{fe} = \frac{k_1}{k_{-1}}\conc{A}\conc{B} $$
The Initial Rate Approximation
We can estimate the value of a derivative by numerical differentiation
$$ \frac{d\conc{A}}{dt} \approx \frac{\conc{A}_{t+\delta t}-\conc{A}_t}{\delta t} $$
Therefore, we can measure the concentrations very soon after starting the reaction and solve for the rate law without having to do calculus
If a reaction has two reactants A and B then all we know is that
$$ -\frac{d\conc{A}}{dt} = k\conc{A}^n\conc{B}^m $$
\(n \) and \(m \) can be non-integers for complex reactions
Experiment
To find the initial rate, we run three experiments
The initial concentrations are each \( 1.0\,\mathrm{mol}\,\mathrm{L}^{-1} \). The concentration of A drops to \( 0.90\,\mathrm{mol}\,\mathrm{L}^{-1} \) at 22.4 seconds after starting the reaction.
We double the initial concentration of B to \( 2.0\,\mathrm{mol}\,\mathrm{L}^{-1} \) and now find that A reaches \( 0.90\,\mathrm{mol}\,\mathrm{L}^{-1} \) at 11.1 seconds.
We double the initial concentration of A to \( 2.0\,\mathrm{mol}\,\mathrm{L}^{-1} \), (with B back at \( 1.0\,\mathrm{mol}\,\mathrm{L}^{-1} \)) and find that A reaches \( 1.90\,\mathrm{mol}\,\mathrm{L}^{-1} \) at 5.3 seconds.
We are going to replace \( d\conc{A} \) in the rate law by the in concentration of A.
Solving the Experiment
$$ \begin{align}
-\frac{d\conc{A}}{dt} = & k\conc{A}^n\conc{B}^m \\
-\frac{\Delta \conc{A}}{\Delta t} \approx & k\conc{A}_0^n\conc{B}_0^m \\
\frac{0.10}{22.4} \approx & k\left( 1.0 \right)^n \left( 1.0 \right)^m \\
\frac{0.10}{11.1} \approx & k\left( 1.0 \right)^n \left( 2.0 \right)^m \\
\frac{0.10}{5.3} \approx & k\left( 2.0 \right)^n \left( 1.0 \right)^m \\
\end{align} $$
Dividing the first equation into the second and third we find that
$$ 2\approx \left(2.0\right)^m \;\;\; 4 \approx \left(2.0\right)^n $$
Thus \(m=1\) and \(n=2\).
Solving the Experiment Continued
Our rate law has become
$$ -\frac{d\conc{A}}{dt}=k\conc{A}^2 \conc{B} $$
We can now estimate the rate constant:
$$ k=\frac{0.10\,\mathrm{mol}\,\mathrm{L}^{-1}}{22.4\,\mathrm{s}\left( 1.0\,\mathrm{mol}\,\mathrm{L}^{-1} \right)^3}=0.0045\,\mathrm{L}^2\,\mathrm{mol}^{-2}\,\mathrm{s}^{-1} $$
We can apply the steady-state approximation to all the compounds since they are all intermediates
$$ \begin{align}
\frac{d\conc{O}}{dt} &= 2j_1\conc{O_2}-k_2\conc{O}\conc{O_2}\conc{M}+j_3\conc{O_3}-k_4\conc{O}\conc{O_3} \\
&= 0 \\
\frac{d\conc{O_2}}{dt} &= -j_1\conc{O_2}-k_2\conc{O}\conc{O_2}\conc{M}+j_3\conc{O_3}+2k_4\conc{O}\conc{O_3} \\
&= 0 \\
\frac{d\conc{O_3}}{dt} &= k_2\conc{O}\conc{O_2}\conc{M}-j_3\conc{O_3}-k_4\conc{O}\conc{O_3} \\
&= 0
\end{align} $$
The abundance of \(\chem{N_2}\) is not affected by this series of reaction so we can assume that \(\conc{M}\) is constant, \(\conc{M}\approx 6.7\times 10^{17}\,\mathrm{cm}^{-3} = 1.1\times 10^{-6}\,\mathrm{mol}\,\mathrm{cm}^{-3}\)
First Analysis Continued
Adding the second and third of the rate equations together we find
$$ -j_1\conc{O_2}+k_4\conc{O}\conc{O_3}=0 $$
We can solve this for \(\conc{O}\)
$$ \conc{O}=\frac{j_1\conc{O_2}}{k_4\conc{O_3}} $$
Using this with the first rate equation
$$ \begin{align}
\conc{O_3} &= \frac{1}{j_3-k_4\conc{O}}\left( k_2\conc{O}\conc{O_2}\conc{M} - 2j_1\conc{O_2} \right) \\
&= \frac{1}{j_3-\left( \frac{j_1\conc{O_2}}{\conc{O_3}} \right)} \left( \frac{k_2j_1\conc{O_2}^2\conc{M}}{k_4\conc{O_3}}-2j_1\conc{O_2} \right)
\end{align} $$
First Analysis Concluded
We can multiply through by \(\conc{O_3}\) to remove it from the denominators
This allows us to find that
$$ \conc{O_3}=1.8\times 10^{13}\,\mathrm{cm}^{-3} $$
This can be used to find that
$$ \conc{O}=3.6\times 10^{8}\,\mathrm{cm}^{-3} $$
The actual \(\chem{O_3}\) density is lower by a factor of 2 to 4 due to nitrogen, hydrogen, and chlorine cycles that also participate
The \(\chem{HO_x}\) Cycle
The \(\chem{HO_x}\) cycle is dominated by two systems
Below an altitude of 30 km
$$ \begin{align}
\chem{OH+O_3} &\rightarrow \chem{HO_2+O_2} \\
\chem{HO_2+O_3} &\rightarrow \chem{OH+O_2+O_2}
\end{align} $$
$$\mathrm{Net:}\;\chem{2O_3\rightarrow 3O_2} $$
Above an altitude of 40 km
$$ \begin{align}
\chem{OH+O} &\rightarrow \chem{H+O_1} \\
\chem{H+O_2+M} &\rightarrow \chem{HO_2+M} \\
\chem{HO_2+O} &\rightarrow \chem{OH+O_2}
\end{align} $$
$$\mathrm{Net:}\;\chem{2O\rightarrow O_2} $$
Use the concentration of atomic oxygen and ozone as determined by the Chapman mechanism to find an expression for the steady-state concentration of \(\chem{NO}\), which appears as an intermediate in the \(\chem{NO_x}\) cycle, in terms of \(\conc{NO_2}\), \(\conc{O_2}\), \(\conc{O_3}\), and relevant rate constants. Assume for simplicity that no other reactions contribute significantly to the \(\chem{NO}\) concentrations.
Enzyme Catalysis
Enzymes are macromolecules that catalyze biochemical reactions
The often increase reaction rate by over eight orders of magnitude
Most enzymes are made up over 1000 atoms
Enzymes may be pure proteins or part protein (the apoenzyme) and part non-protein (the coenzyme)
In many cases, enzymes are much larger than the principal reactants, called substrates
Like all catalysts, the reactants form a complex with the catalyst and in this complex they react more quickly
Central Questions of Enzyme-Substrate Systems
How can we describe the structure of molecules with so many stable conformers?
How does this structure influence the enzyme's biochemical function?
What are the thermodynamics and kinetics governing the enzyme's biochemical function?
Biomolecular Structure
Typically to understand structure we would use infrared spectroscopy and microwave spectroscopy
Very large molecules rarely have a single geometry
As such, their vibrational spectrum is nearly a continuous series of absorptions throughout the entire range, \(<3000\, \mathrm{cm}^{-1}\)
Biomolecules are often found in water which has very strong IR absorptions
So how do we analyze their structure
electronic spectroscopy
NMR spectroscopy
optical rotation studies
x-ray diffraction of crystalized or fibrous biochemicals
Electronic Spectroscopy of Biomolecules
The spectroscopy of electronic transitions is most easily carried out on those molecules with relatively low-lying excited states.
Usually, ultraviolet and visible wavelength spectroscopy of biomolecules is usually sensitive to those regions of the molecule with a metal or sulfur atom, or by some other aromatic functional group
These sites on molecules are called chromophores
Chromophores offer the opportunity to probe large macromolecules at a specific location
It is more common to measure emitted fluorescence from chromophores than measuring absorption
Enzyme concentration is less that 1% the substrate concentrations so \(\chem{ES}\) is essentially a steady state
The third step is extremely fast compared the first and second so we can treat this as a two step mechanism
$$ \chem{E+S} \xrightleftharpoons{k_1}{k_{-1}} \chem{ES} \xrightleftharpoons{k_2}{k_{-2}} \chem{E+P} $$
If we focus on the early reaction, \(\conc{P}\) is small so the second step is essentially irreversible
$$ \chem{E+S} \xrightleftharpoons{k_1}{k_{-1}} \chem{ES} \xrightarrow{k_2} \chem{E+P} $$
Starting Work on the Rate Law
$$ \chem{E+S} \xrightleftharpoons{k_1}{k_{-1}} \chem{ES} \xrightarrow{k_2} \chem{E+P} $$
$$ \begin{align}
\frac{d\conc{P}}{dt} &= k_2\conc{ES} \\
\frac{d\conc{ES}}{dt} &= k_1\conc{E}\conc{S}-k_{-1}\conc{ES}-k_2\conc{ES} \\
&= k_1\conc{E}\conc{S}-k_{-1}\conc{ES}_{ss}-k_2\conc{ES}_{ss}\overset{ss}{=}0 \\
\conc{ES}_{ss} &= \frac{k_1\conc{E}\conc{S}}{k_{-1}+k_2} = \frac{k_1}{k_{-1}+k_2}\conc{E}\conc{S}
\end{align} $$
We can simplify this equation by defining the Michaelis-Menton constant, \( K_M\equiv \frac{k_{-1}+k_2}{k_1} \), therefore \( \conc{ES}_{ss} = \frac{\conc{E}\conc{S}}{K_M} \)
$$ \chem{E+S} \xrightleftharpoons{k_1}{k_{-1}} \chem{ES} \xrightarrow{k_2} \chem{E+P} $$
$$ \frac{d\conc{P}}{dt} = k_2\conc{ES} $$
$$ \conc{ES}_{ss} = \frac{\conc{E}_0\conc{S}}{K_M+\conc{S}} $$
Combining the last two equatons we find that
$$ \frac{d\conc{P}}{dt} = k_2\conc{ES}_{ss} = k_2 \frac{\conc{E}_0\conc{S}}{K_M+\conc{S}} $$
This is one form of the Michaelis-Menton equation. All the values on the right-hand side (RHS) are constant except for \(\conc{S}\).
Let's look at two limits of the Michaelis-Menton equation.
Let's look at the low limit
$$ \lim_{\conc{S}\rightarrow 0} \frac{d\conc{P}}{dt} = \frac{k_2\conc{E}_0\conc{S}}{K_M} $$
Let's look at the high limit
$$ \lim_{\conc{S}\rightarrow \infty} \frac{d\conc{P}}{dt} = \frac{k_2\conc{E}_0\conc{S}}{\conc{S}} = k_2\conc{E}_0 $$
Modification of the Michaelis-Menton Equation
Considering the high limit on \(\conc{S}\) we can define the maximum velocity
$$ v_{max}=k_2 \conc{E}_0 $$
Using this equation we can modify the Michaelis-Menton equation
$$ \frac{d\conc{P}}{dt} = k_2 \frac{\conc{E}_0\conc{S}}{K_M+\conc{S}} = k_2 \conc{E}_0 \frac{\conc{S}}{K_M+\conc{S}} = v_{max} \frac{\conc{S}}{K_M+\conc{S}} $$
Note that the rate constant \(k_2\) is also known as the enzymatic turnover number
the number of substrate molecules converted to product per unit time
it is often written as \(k_{cat}\)
The Michaelis-Menton Reaction Velocity
Uses of Michaelis-Menton Equation
The fraction of occupied reactive sites can be calculated from \(K_M\)
$$ f_{ES}=\frac{\conc{ES}}{\conc{E}_0}=\frac{\conc{S}}{\conc{S}+K_M} $$
One common way of finding \(K_M\) is from a Lineweaver-Burk plot which is based on
$$ \frac{1}{\bfrac{d\conc{P}}{dt}} = \frac{1}{v_{max}} + \frac{K_M}{v_{max}\conc{S}} $$
Complication
A complication may arise when an inhibitor is present
Inhibitor is a substance that blocks the catalytic activity of the enzyme
In the simpliest case, the inhibitor binds at the same enzyme site that the substrate uses
This binding can be written as
$$ \chem{E+I}\xrightleftharpoons{k_3}{k_{-3}} \chem{EI} $$
with an equilibrium constant of
$$ K_3=\frac{k_3}{k_{-3}}=\frac{\conc{EI}}{\conc{E}\conc{I}} $$
Applying the Complication
Applying our inhibitor equations to our enzyme concentrations we find that
$$ \begin{align}
\conc{E}_0 &= \conc{E}+\conc{ES}+\conc{EI}=\conc{E}+\conc{ES}+K_3\conc{E}\conc{I} \\
\conc{E} &= \frac{\conc{E}_0-\conc{ES}}{1+K_3\conc{I}}
\end{align} $$
Applying this change to \(\conc{ES}_{ss}\)
$$ \conc{ES}_{ss} = \frac{\conc{E}\conc{S}}{K_M} = \frac{\left( \conc{E}_0 - \conc{ES}_{ss} \right) \conc{S}}{K_M \left( 1+K_3\conc{I} \right)} $$
Doing a lot of algebra we can arrive at
$$ \conc{ES}_{ss} = \frac{\conc{E}_0\conc{S}}{K_M+K_MK_3\conc{I}+\conc{S}} $$
The Inhibited Michaelis-Menton Equation
Applying our inhibited steady-state concentration to our production rate we arrive at
$$ \frac{d\conc{P}}{dt} = \frac{k_2\conc{E}_0\conc{S}}{K_M+K_MK_3\conc{I}+\conc{S}} $$
More Enzyme Information
The location on the enzyme surface where the substrate binds is called the active site
The ratio of the number of active sites occupied to the total number of available active site is called the coverage, \(\theta\)
$$ \begin{align}
\theta &= \frac{\conc{ES}}{\conc{E}_0} = \frac{k_1\conc{S}}{k_{-1}+k_2+k_1\conc{S}} \\
&= \frac{\left( \frac{k_1}{k_{-1}+k_2} \right)\conc{S}}{1+\left( \frac{k_1}{k_{-1}+k_2} \right) \conc{S}} \\
&= \frac{K_{ads}\conc{S}}{1+K_{ads}\conc{S}}
\end{align} $$
where
$$ K_{ads}\equiv \frac{k_1}{k_{-1}+k_2} $$
Combustion Chemistry
Combustion is rapid and exothermic oxidation
We can make distinctions by comparing the system before and after combusion
If the final state is lower pressure and density than the original state we have deflagration
Flames are usually weak deflagrations
If the combustion is extrememly rapid, the pressure and density skyrocket then we have detonation
the product gases do note have time to flow away from the reaction before the fuel is consumed
Enormous pressure is generated in a small volume
This pressure is released as a shock wave through the surrounding medium
Laminar Flow
Laminar flow is fluid motion in a single direction without turbulance
turbulance is any discontinuity in the density or flow velocity
An ideal Bunsen burner would provide a laminar flame in which the \(\chem{CH_4}\) entering the conbusion region is completely oxidized to \(\chem{H_2O}\) and \(\chem{CO_2}\)
The principle elementary reactions relevant to this process are given in the following tables - there are other secondary elementary reactions possible
The shear number of elementary reactions makes analyzing the kinetics daunting so we need a way of simplifying things
Of all the reactants and products, only seven are chemically stable: \( \chem{CH_4,\,O_2,\,H_2O,\,CO_2,\,CO,\,H_2,\,CH_2O} \)
Important Reactions in the Combusion of \(\chem{CH_4}\) in \(\chem{O_2}\)
These reaction rates are determined at 2000 K in units of \(\left( \chem{cm^{3}\,mol^{-1}} \right)^{n-1}\,\chem{s}^{-1}\), when \(n\) is the molecularity of the reaction. When \(k_r\) is not given, the reverse reaction is negligible.
reaction
\(k_f\)
\(k_r\)
1.
\( \chem{H+O_2 \rightleftharpoons OH+O} \)
\( 2.92 \times 10^{11} \)
\( 1.15 \times 10^{13} \)
2.
\( \chem{H+O_2+M \rightleftharpoons HO_2+M} \)
\( 5.26 \times 10^{15} \)
\( 2.17 \times 10^{10} \)
3.
\( \chem{H_2+O \rightleftharpoons H+OH} \)
\( 8.97 \times 10^{12} \)
\( 7.02 \times 10^{12} \)
4.
\( \chem{H_2+OH \rightleftharpoons H+H_2O} \)
\( 8.34 \times 10^{12} \)
\( 8.36 \times 10^{11} \)
5.
\( \chem{2OH \rightleftharpoons H_2O+O} \)
\( 8.48 \times 10^{12} \)
\( 1.20 \times 10^{12} \)
6.
\( \chem{HO_2+H \rightleftharpoons 2OH} \)
\( 1.17 \times 10^{14} \)
\( 6.00 \times 10^{8} \)
Important Reactions in the Combusion of \(\chem{CH_4}\) in \(\chem{O_2}\) (cont.)
reaction
\(k_f\)
\(k_r\)
7.
\( \chem{HO_2+H \rightleftharpoons H_2+O_2} \)
\( 2.10 \times 10^{13} \)
8.
\( \chem{HO_2+H \rightleftharpoons H_2O+O} \)
\( 1.95 \times 10^{13} \)
9.
\( \chem{HO_2+OH \rightleftharpoons H_2O+O_2} \)
\( 1.30 \times 10^{13} \)
10.
\( \chem{CO+OH \rightleftharpoons CO_2+H} \)
\( 4.74 \times 10^{11} \)
\( 2.01 \times 10^{11} \)
11.
\( \chem{CH_4+H \rightleftharpoons H_2+CH_3} \)
\( 1.95 \times 10^{13} \)
\( 9.40 \times 10^{11} \)
12.
\( \chem{CH_4+OH \rightleftharpoons H_2O+CH_3} \)
\( 7.37 \times 10^{12} \)
\( 1.05 \times 10^{11} \)
13.
\( \chem{CH_3+O \rightleftharpoons CH_2O+H} \)
\( 7.00 \times 10^{13} \)
\( 5.30 \times 10^{7} \)
14.
\( \chem{CH_3+OH \rightleftharpoons CH_2O+2H} \)
\( 1.83 \times 10^{13} \)
Important Reactions in the Combusion of \(\chem{CH_4}\) in \(\chem{O_2}\) (cont. more)
reaction
\(k_f\)
\(k_r\)
15.
\( \chem{CH_3+OH \rightleftharpoons CH_2O+H_2} \)
\( 8.00 \times 10^{12} \)
16.
\( \chem{CH_3+OH \rightleftharpoons CH_2+H_2O} \)
\( 4.26 \times 10^{12} \)
17.
\( \chem{CH_3+H \rightleftharpoons CH_2+H_2} \)
\( 4.07 \times 10^{12} \)
18.
\( \chem{CH_3+H+M \rightleftharpoons CH_4+M} \)
\( 1.11 \times 10^{17} \)
\( 4.30 \times 10^{7} \)
19.
\( \chem{CH_2+H \rightleftharpoons CH+H_2} \)
\( 4.00 \times 10^{13} \)
\( 1.31 \times 10^{13} \)
20.
\( \chem{CH_2+OH \rightleftharpoons CH_2O} \)
\( 2.50 \times 10^{13} \)
21.
\( \chem{CH_2+OH \rightleftharpoons CH+H_2O} \)
\( 2.11 \times 10^{13} \)
22.
\( \chem{CH_2+O_2 \rightleftharpoons CO_2+2H} \)
\( 4.45 \times 10^{12} \)
Important Reactions in the Combusion of \(\chem{CH_4}\) in \(\chem{O_2}\) (completed)
reaction
\(k_f\)
\(k_r\)
23.
\( \chem{CH_2+O_2 \rightleftharpoons CO+OH+H} \)
\( 4.45 \times 10^{12} \)
24.
\( \chem{CH+O_2 \rightleftharpoons CHO+O} \)
\( 3.00 \times 10^{13} \)
25.
\( \chem{CH+OH \rightleftharpoons CHO+H} \)
\( 3.00 \times 10^{13} \)
26.
\( \chem{CH_2O+H \rightleftharpoons CHO+H_2} \)
\( 9.16 \times 10^{12} \)
\( 4.20 \times 10^{9} \)
27.
\( \chem{CH_2O+OH \rightleftharpoons CHO+H_2O} \)
\( 2.22 \times 10^{13} \)
\( 8.60 \times 10^{9} \)
28.
\( \chem{CHO+H \rightleftharpoons CO+H_2} \)
\( 2.00 \times 10^{14} \)
\( 2.80 \times 10^{5} \)
29.
\( \chem{CHO+OH \rightleftharpoons CO+H_2O} \)
\( 1.00 \times 10^{14} \)
\( 1.30 \times 10^{4} \)
30.
\( \chem{CHO+M \rightleftharpoons CO+H+M} \)
\( 1.04 \times 10^{13} \)
\( 3.63 \times 10^{12} \)
31.
\( \chem{CHO+O_2 \rightleftharpoons CO+HO_2} \)
\( 3.00 \times 10^{12} \)
Simplifying the System
We cannot simply neglect steps based on rate constant
Reaction rates depend on reactant concentrations
A reaction involving \(\chem{CH_4}\) will be much more important initially than a much faster reaction whose reactants are intermediates formed later in the reaction
An analysis of the numerical results can identify the most important series of reactions and break them up into different categories
The dominant chemistry of the methane flame can be described by the reduced set of multi-step reactions given on the following slides
These different steps occur at different points in the flame (upcoming figure)
Note: It is possible to decrease the amount of \(\chem{CO}\) in the exhaust gases by increasing the initial \(\chem{O_2}\) concentration or by seeding the fuel mix with other sources of \(\chem{OH}\).
Astrophysics and Spectroscopy
Physical chemistry and astrophysics share many common interest
The greatest similarities rest with the coincidence of observational methods
In both fields, the object of study is difficult to probe directly
in physical chemistry it is often a chemical sample too small or too reactive to isolate
in astrophysics it may be a star one thousand light years away
Spectroscopy has worked well both both
Stellar Emission
A star is a cloud of gas so massive and so dense that its own gravity compresses the matter at its core with enough force to enable nuclear fusion
The core is the hottest and densest region of the star
The star become cooler and more diffuse as the distance from the core increases
The interior gas of a star is gas of ionized atoms, mainly hydrogen and helium, too hot and dense for the emitted radiation to show any individual spectroscopic transitions
The Parts of a Stellar Atmosphere
In the inner part of the star, the matter density is so high over such a large distance that photons cannot leave without first being scattered or absorbed and re-emitted by these ions
These regions can't be penetrated by radiation without interaction with the matter
This type of region is called optically thick
Photons emitted in the outer shell of a star find the matter density low enough that they can leave the star without more scattering or absorption
This region of the star is essentially transparent to radiation
This type of region is called the photosphere of the star and is optically thin
Optical Thickness of a Stellar Interior
Example 14.4
The sun's emission spectrum is characteristic of a blackbody at a temperature of 5800 K. find the peak emission wavelength. The blackbody radiaton spectrum is given by
$$ \rho(\nu)\,d\nu = \frac{8\pi h\nu^3\,d\nu}{c^3\left( e^{\bfrac{h\nu}{k_BT}}-1 \right)} $$
Example 14.5
Heat is carried to the optically thin layer of a star primarily by convection. Radiation is just too inefficient a process in the dense core. Use Einstein's diffusion equation
$$ t\approx \frac{R^2}{6D} $$
to estimate the time required for a photon generated at the sun's center to diffuse out of the sun's core, a distance of \( 1.7\times 10^8\,\mathrm{m} \). Although the actual number density on the core approaches \(10^{26}\,\mathrm{cm}^{-3}\), the photons interact weakly with the matter, and have a mean free path of about \(0.1\,\mathrm{cm}\). Use the mean free path to estimate the diffusion constant
$$ D=\frac{\lambda^2\gamma}{2} $$
where \(\lambda\) is mean free path, \(\gamma\) is the collision frequency.
The Interstellar Medium
Most of the observable universe is believed to be interstellar gas and dust
Within our galaxy, the number density of matter between stars varies between \(10^3\) and \(10^6\,\mathrm{cm}^{-3}\)
Most of this matter is either \(\chem{H^+}\), \(\chem{H_2}\), or \(\chem{H}\)
This matter is largely known as the interstellar medium
The ionized gas corresponds to the largest, hotest, and sparsest regions of interstellar space
The molecular hydrogen exists only in compact, cold, relatively dense clumps known as molecular clouds
Molecular clouds are about \(10^{13}\) times less dense than our atmosphere
Energy Transfer in Molecular Clouds
Cosmic Abundances of the Dominant Elements, Relative to Hydrogen
Interstellar Chemical Synthesis: Detected Organic Molecular
\(\chem{CH}\)
\(\chem{C_2}\)
\(\chem{CN}\)
\(\chem{CO}\)
\(\chem{CS}\)
\(\chem{CH^+}\)
\(\chem{HOC^+}\)
\(\chem{HCS^+}\)
\(\chem{HCNH^+}\)
\(\chem{HCO^+}\)
\(\chem{HCN}\)
\(\chem{HNC}\)
\(\chem{HNCO}\)
\(\chem{HNCS}\)
\(\chem{HC_3N}\)
\(\chem{HC_5N}\)
\(\chem{HC_7N}\)
\(\chem{HC_9N}\)
\(\chem{HC_{11}N}\)
\(\chem{CH_3CN}\)
\(\chem{CH_3NC}\)
\(\chem{C_2H_3CN}\)
\(\chem{C_3H_5CN}\)
\(\chem{CH_3CHO}\)
\(\chem{HC_2CHO}\)
\(\chem{HC_3HOH}\)
\(\chem{CH_3CH_2OH}\)
\(\chem{CH_3OH}\)
\(\chem{C_3O}\)
\(\chem{HC_2O}\)
\(\chem{H_2CS}\)
\(\chem{HCOOH}\)
\(\chem{CH_3COOH}\)
\(\chem{CH_3SH}\)
\(\chem{NH_2CHO}\)
\(\chem{NH_2CN}\)
\(\chem{c-SiC_2}\)
\(\chem{c-C_3H_2}\)
\(\chem{C_3N}\)
\(\chem{CH_3C_3N}\)
\(\chem{CH_3C_4H}\)
\(\chem{CH_3CCH}\)
\(\chem{C_2H}\)
\(\chem{C_3H}\)
\(\chem{C_4H}\)
\(\chem{C_5H}\)
Note: "c-" prefix indicates a cyclic molecule.
Example 14.9
Based on the parameters given in the following table, estime the rates of formation of \(\chem{HCl^+}\) and \(\chem{HCl}\) at 80 K in the Orion Molecular Cloud from the reactions given.
$$ \begin{align}
\chem{Cl^++H_2} &\rightarrow^{k_1} \chem{HCl^++H} \\
\chem{Cl+H_2} &\rightarrow^{k_2} \chem{HCl+H}
\end{align} $$
Contraryto the apparent results of the previous example, the \(\bfrac{\conc{HCl}}{\conc{HCl^+}}\) ratio in the Orion Molecular Cloud is estimated at \( 2 \times 10^9 \). There is a more efficient way to produce the \(\chem{HCl}\) than by reaction 2 in the previous mechanism, and the \(\chem{HCl^+}\) is itself consumed in one of those reactions:
$$ \begin{align}
\chem{Cl+H_3^+ \xrightarrow{k_3} HCl^++H_2} &~~~~~~~~& k_3 &= 1.0 \times 10^{-9}\,\mathrm{cm}^3\,\mathrm{s}^{-1} \\
\chem{HCl^++H_2 \xrightarrow{k_4} H_2Cl^++H} &~~~~~~~~& k_4 &= 1.3 \times 10^{-9}\,\mathrm{cm}^3\,\mathrm{s}^{-1} \\
\chem{H_2Cl^++e^- \xrightarrow{k_5} HCl+H} &~~~~~~~~& k_5 &= 1.5 \times 10^{-7}\,\mathrm{cm}^3\,\mathrm{s}^{-1}
\end{align} $$
Reaction 3 is actually about four times more efficient at forming \(\chem{HCl^+}\) than reaction 1 (but doesn't offer such a straightfoward comparison with neutral-neutral kinetics). The net reaction is
$$ \chem{Cl+H_3^++e^- \rightarrow HCl+2H} $$
Find the steady-state number density of \(\chem{HCl^+}\) in this mechanism, using the parameters from our previous example.