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Chemical thermodynamics answers questions about whether or not reactions
are favorable/spontaneous (can happen without “outside help”) given certain
conditions, and yields information about equilibrium concentrations.
–  Typical terminology: equilibrium constant K, free energy, enthalpy, entropy,
law of mass action.
Chemical kinetics answers questions about how fast reactions happen, and
how rapidly concentrations change.
–  Typical terminology: rate constant k, activation barrier, activation energy,
transition state, Arrhenius equation
The first thing to remember about thermodynamics and kinetics is…
Keep them separated!
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Whether or not a reaction is spontaneous tells you nothing at all about how
fast it is, and vice versa. For example, self-combustion of human beings is
spontaneous, but fortunately very, very slow (kinetically limited).
Only in special cases is there a direct link between the two fields.
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Epot
Reagents
Products
Ea
ΔG
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ΔG, the (Gibbs) free energy change, determines whether the reaction is
favorable (spontaneous) or not, and where the equilibrium lies. In this case
ΔG < 0, so the reaction is spontaneous ⇔ the equilibrium will lie on the side
of the products.
Ea, the activation energy (or activation barrier) tells you how fast the reaction
is. A high barrier means a slow reaction, a low barrier a fast reaction. Chemical reactions are generally reversible, at least in principle. Let
a A + b B → c C + d D
be a general chemical reaction in the “forward” direction. (That is, a
molecules of A and b molecules of B react to form c molecules of C and
d molecules of D.) The reverse reaction is then
cC+dD→aA+bB
We usually write:
aA+ bB ⇔ cC+dD
At equilibrium, the law of mass balance (a.k.a. the law of chemical
equilibrium) states that:
Where K is the equilibrium constant for the reaction.
Note that in this formulation K will be dimensionless if a+b = c+d, and otherwise have
dimensions of (concentration)(c+d-a+b)
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The value of the ratio
tells you how far you are from
chemical equilibrium, and on what side.
•  If the ratio is smaller than K, the reaction will tend to proceed in
the forward direction, with more reagents (A and B) being converted
into products (C and D) until the equilibrium is reached. • If the ratio is greater than K, the reaction will proceed in the
reverse direction, with products (C and D) being converted into
reagents (A and B).
Remember again that the equilibrium constant tells you nothing about
how fast this will happen! The system may sit far from equilibrium for a
billion years if the reactions are slow enough.
(Later on we will see that equilibrium constants can be expressed as a
ratio of the rate constant of the forward and reverse reactions…)
In the gas phase, it is more convenient to use partial pressures instead of
concentrations. The law of mass balance can then be reformulated to give:
In this formulation, K’ is now always dimensionless.
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The link between equilibrium constants and free energies is:
Where ΔG is the Gibbs free energy change for the reaction. (We are here assuming
reaction conditions of constant pressure and temperature. If some other set of variables is kept constant, other free
energies have to be used instead.)
Negative ΔG ⇒ equilibrium on the side of the products, favorable reaction
Positive ΔG ⇒ equilibrium on the side of the reactants, unfavorable reaction
It should be noted that ΔG usually depends on the conditions (temperature,
pressure etc), so be careful when using some standard-condition ΔG values.
E.g. ΔG for gas-phase reactions is often given at 298 K and 1 atm, if the
pressure of some reactant pX differs from 1 atm then:
Often ΔG is expressed as a sum of enthalpy (H) and entropy (S) terms:
ΔG = ΔH - TΔS.
The equilibrium of a species X between the gas phase and a liquid phase
(typically, an aqueous solution) can be represented as:
X(g) ⇔ X(aq)
where the solvent in the liquid phase (here, water) is not explicitly indicated.
X(aq) can be thought to correspond to X•(H2O); a molecule of X solvated by
(one or more) water molecules.
The expression for this equlibrium is usually given as:
[X(aq)] = HXpX
where HX is called the Henryʼs law constant for X and typically has units of
mol/(L×atm) or mol/(L×Pa). Note that mol/L is often denoted simply as “M”, not to be
confused with the symbol for molar mass…
Question: what is the SO2 concentration in a solution at equilibrium with air
containing 100 ppt of SO2 at 298K? (HSO2 at 298 K is 1.23 M atm-1)
Answer: [SO2(aq)] = 1.23 M atm-1 × 100 × 10-12 atm = 1.23 × 10-10 mol/L
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In acid – base reactions, equilibrium constans are called acidity (or basicity)
constants, Ka and Kb.
AH + H2O ⇔ A- + H3O+
B + H2O ⇔ BH+ + OH-
Where did [H2O] go?
It is assumed to remain constant (at about 55.5 mol/L), and is therefore
included in the acidity and basicity constants:
Ka = K×[H2O]; Kb = K× [H2O]. Normally, the constants are given in negative logarithimic form:
pKa = -log10(Ka), pKb = -log10(Kb)
This is stricty speaking a bit sloppy, as you can only take logarithms of dimensionless
numbers. We should probably write pKa = -log10{Ka /(1 M)} etc. to be exact. For the so-called autoprotolysis reaction
H2O + H2O ⇔ OH- + H3O+
(where again the water concentration is included in the constant Kw, and the constant
is commonly given in the form pKw = -log10(Kw / M2),
We can thus easily calculate that for neutral water,
Now, for any acid and its conjugate base: Ka × Kb = KW
Or equivalently:
pKa + pKb = pKw = 14
For this reason, “basicity constants” are almost never given – you just take
the acidity constant of the corresponding conjugate acid and use the
equations above.
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pH is the measure of the acidity of a water solution. It is defined as:
pH = -log10([H3O+]/M)
With [H3O+] given in mol/L. • Neutral solutions have a pH of 7.
• Acidic solutions have a pH less than 7
• Basic solutions have a pH greater than 7.
Typically, problems involving acid-base equilibria will almost always involve
computing the pH. Acid-base reactions tend to be very fast, so we can almost always assume
that the solution is at chemical equilibrium.
The pH can then be calculated by setting up the appropriate equilibrium
expression(s) and solving for [H3O+].
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Even the cleanest rainwater is slightly acidic due to the influence of
atmospheric carbon dioxide, which dissolves into rain droplets and forms
carbonic acid. There are three equilibrium expressions to consider:
CO2(g) ⇔ CO2•H2O
Note: CO2•H2O can also be written as H2CO3
CO2•H2O + H2O ⇔ HCO3- + H3O+
HCO3- + H2O ⇔ CO32- + H3O+
Given values of pCO2=350 ppm, HCO2=0.034 M atm-1, Kc1=4.3 × 10-7 M
and Kc2 = 4.7 × 10-11 M , how do we compute the pH?
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Answer: substitute the first expression into the second, and the second into
the third:
We now have expressions for all the concentrations in terms of H3O+ and a set of known constants. To calculate [H3O+], we still need one thing: the electro-
neutrality equation. This is a useful trick, based on the fact that the sum of positive
charges (concentration of positive ions multiplied by their charge) and the sum of
negative charges in a solution (as above) have to be equal. So we get:
Which can be solved (with a calculator) to give [H3O+] ≈ 2.5×10-6 M ⇒ pH ≈ 5.6 Chemical kinetics describe a reaction process of molecules, e.g. –  probability of reactions, –  concentration development of compounds, –  speed of change.
Important because of the understanding of the contribution of different
reactions to the overall processes.
- Example: chemical lifetime τchem
Often a so-called steady-state (or quasi steady-state/pseudo-steady
state) is assumed, which describes the equilibrium conditions, e.g.
equilibrium concentrations, if reactions are sufficiently fast that other
processes such as strongly time dependent emissions and atmospheric
transport do not affect the situation remarkably.
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Chemical reaction equations can be classified into two types.
1)Elementary reactions, where the described reaction happens in one step, i.e. the
reagents in the equation actually react directly with each other and form the
products 2)Composite reactions, where the reaction actually occurs via several steps.
E.g. the oxidation of methane in the troposphere can be written as CH4 + 2 O2 ⇒ CO2 + 2 H2O
but the reaction mechanism actually contains tens of different elementary reactions,
and involves several other compounds.
Elementary reactions can be either unimolecular (1-body), bimolecular (2-body) or
trimolecular (3-body). It should be noted that real elementary trimolecular
reactions are very rare (never in the gas phase at atmospheric pressure, some
examples in the liquid phase, triple-α fusion in the sun…).
A
B+C
k: reaction rate coefficient [s-1]
Example: Decomposition of
N2O5 (important storage for NOx
during night)
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Two molecules A and B (e.g. NO and ozone) react in one step:
A+B
C+D
kbim.: reaction rate coefficient
[cm3 molecule-1 s-1]
Special case: A = B
Note: The reaction rate coefficient includes both the collision cross-section
of A + B as well as the probability to react (fate) of the molecules!
If their concentration is the same, noble gas atoms and OH radicals will
collide with hydrocarbons at roughly the same frequency, but only the
latter collisions will ever lead to a reaction. (Weʼll see later that this is due
to a difference in activation energies.)
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If but only if reaction A + B ⇔ C + D is a reversible elementary reaction
–  A + B ⇒ C + D happens in one step, rate equation:
–  C + D ⇒ A + B happens in one step, rate equation:
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Then at equilibrium (since concentrations donʼt change) we can write:
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So the equilibrium constant is the ratio of the rate constants.
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A + B → AB*
(1)
AB* → A + B (2)
AB* + M → AB + M*
(3)
stabilization (removal of excess energy)
M* → M + heat. (4)
M is any inert molecule (typically O2 or N2) and * denotes excess energy, leading to instability.
Reaction (1) is the limiting (slowest) reaction and thus the 3 reactions can be
approximated as a single trimolecular reaction:
A + B + M → AB + M
k in cm6 molecule-2 s-1
How do we find a rate expression for a simple composite reaction like this?
Solving the system of differential equations is doable, but leads to unpleasant
mathematics.
Usually, you can avoid the math by using the Pseudo Steady State Approximation.
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Consider the previous
”trimolecular” reaction system.
AB* is quite reactive. Therefore,
itʼs reactions are probably fast
compared to changes in the
concentrations of A, B and AB. We can thus assume that the
reaction system reaches a pseudo
– steady state, where the
production and consumtion of
AB* are equal ⇒ d[AB*]/dt = 0.
From this we get an expression for
the steady-state concentration of
AB*. We can then obtain the rate of
formation of the final product AB in
terms of [A], [B], [M] and the rate
constants.
A + B → AB*
AB* → A + B
AB* + M → AB + M*
(M* → M + heat.)
(1)
(2)
(3) (4)
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Take a closer look at the
”trimolecular”
rate
expression
obtained.
If k3[M] >> k2, we can write the
”high-pressure” approximation:
–  The reaction is now effectively
”second
order”,
with
rate
constant k∞ = k1 • 
If k3[M] << k2, we can write the ”lowpressure” approximation:
–  The reaction is now effectively
”third order”, with rate constant
k0 = k1k3/k2 .
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The original rate expression can
now be written in terms of the
limiting rate constants k∞ and k0 :
The temperature dependence of an elementary reaction is often expressed
in terms of the Arrhenius equation:
where A is called the “pre-exponential factor” (it is related to the collision
cross-section, among other things) and EA is the activation energy.
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Naturally enough, reactions with low barriers are fast (high k) and
reactions with high barriers are slow (low k).
Also, elementary reactions tend to speed up with increasing
temperature (EA ≥ 0,almost always).
Sometimes the pre-exponential factor has to be replaced with a weakly
temperature – dependent expression A(T), e.g. a polynomial of T.
Arrhenius – like expressions are sometimes also used for composite reactions.
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Trace gases in the atmosphere are present at low concentrations.
→ no frequent reactions
Only exception are compounds reacting sufficiently fast: radicals
Radical: a compound with an unpaired electron in its valence shell. Radicals are
unstable, and their reactions thus usually have low activation energies.
e.g. OH (8 + 1 electron = 9 electrons)
radical
NO (7 + 8 electrons = 15 electrons)
radical
HNO3 (1 + 7 + 3x8 electr. = 32 electr.) non-radical (also called closed shell)
Note: some compounds (most notably O and O2 in their ground states) have two unpaired
electrons. They are then “biradicals” even though they have an even number of electrons
in total. They are often also quite reactive. (Determining which species are biradicals requires a
bit of quantum mechanics that is beyond this course.)
Excess energy (e.g. from sunlight) is needed to start reactions:
(1) non-radical + hν ⇒ radical + radical
(photolysis)
(2) radical + non-radical ⇒ radical + non-radical (reaction chain)
(3) radical + radical ⇒ non-radical + non-radical (termination)
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