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Doc Brown's Chemistry
Advanced Level Chemistry -
Kinetics
Part 2
Revision notes for GCE Advanced
Subsidiary Level AS Advanced Level A2 IB
Revise AQA OCR Edexcel Salters CIE revising courses for pre-university students
(equal to US grade 11 and grade 12 and Honours/honors level courses)
GCSE links GCSE
reaction rates notes *
GCSE rates coursework ideas
GCSE
notes on energy changes - reaction profiles *
Some basic
GCSE notes on enzyme activity
You must know the
basic GCSE stuff and most is NOT repeated here because this page is directed
at advanced level written papers but should be useful for kinetics study
projects.
Kinetics
Part 1 index:
1.
advanced
particle theory: 1a. distribution of particle energies :
1b.
effect of increasing temperature : 1c. effect of a catalyst
* 2. catalyst mechanisms : 2a. introduction
: 2b. heterogeneous examples-theory : 2c. homogeneous
examples-theory * 3a.
Examples of obtaining rate data : 3b. rate expressions
and orders of reactions :
3c.
deducing orders of reaction : 3d. Simple
exemplar rates Q's * PAGE 2 Kinetics Part 2 index: 4. case studies in kinetics * 5. Arrhenius equation for calculating activation energy * 6.
Kinetic versus thermodynamic stability : 6a. Introductory
points : 6b. Kinetic stability/instability examples : 6c.
Thermodynamic 'instability'
Also Computer simulations of kinetic particle theory -
Maxwell Boltzman Distribution of particle speeds/KE's
4.
Selected case studies of kinetics
4.1
Hydrolysis of halogenoalkanes : 4.2
Oxidation of iodide ions by
peroxodisulphate : 4.3 Enzyme kinetics : 4.4
The H2/I2/HI equilibrium : 4.5
Ester hydrolysis : 4.6 Acid catalyzed iodination
of propanone : Case study 4.7 Acid- thiosulphate
reaction :
Case study
4.1
The kinetics of the hydrolysis of halogenoalkanes
There
are a series of pages devoted organic reaction mechanisms
-
Reaction kinetics:
The possibility of two reaction mechanisms for the hydrolysis of
halogenoalkanes (RX) with sodium hydroxide or water has consequences for
the rate expressions.
-
The SN1
mechanism is sometimes described as 'unimolecular' because the rate
only depends on the concentration of the halogenoalkane.
-
The mechanism
for (i) has three steps
of bimolecular collisions. Here the rate is only
dependent on one reactant, the halogenoalkane, R3C-X, shown in step
(1), (but it still has to collide with the solvent, which never
seems to be shown at AS-A2 level and whose concentration is
essentially constant!)
-
Experimental results
produce the overall 1st order rate expression: rate = k1[RX]
-
1st order
kinetics suggests there is a rate determining step involving one of
the reactants, irrespective of the total number of steps, which in
this case is three.
-
This is because the
activation energy of the 1st step, forming the carbocation
intermediate by
heterolytic bond fission, is so high, that the speed is relatively
low. Therefore step (1)
alone determines the speed of the reaction. This is referred to as
the rate determining step (or rds in shorthand!). Steps
(2)/(3) have much lower
activation energies and are much faster. You would register zero order for the
order of reaction with respect to e.g. any hydroxide ion present or it might even
hydrolyse just in water!
-
Note the
simplicity of the rate expression, despite the complexity of the
mechanism!
-
The 1st
diagram (mechanism 10) shows the full reaction mechanism.
-
The 2nd
diagram (mechanism 42) shows the reaction profile with step (1)
having much the bigger activation energy and hence acting as the rate
determining step. The two 'troughs represent the formation of
the intermediates or transition states whatever their lifetime
maybe!
-
The SN2
mechanism below for reaction (ii), is referred to as 'bimolecular' because the rate depends
on the concentrations of both reactants.
-
Experimental results
produce the overall 2nd order rate expression: rate = k2[RX][OH],
-
2nd order
kinetics suggests there is a rate determining step involving both of
the reactants, irrespective of the total number of steps, though
in this case its just one step.
-
This is because it is a
simple single step mechanism involving
a bimolecular collision of the two
reactant molecules/ions. The rate depends on both the
halogenoalkane and hydroxide ion concentrations (1st
order with respect to both reactants).
-
The 1st
diagram above (mechanism 2) shows the 'simple' mechanism.
-
The 2nd
diagram (mechanism 33) shows the 'activated complex' or 'transition
state'* which is the peak of the potential energy of the system
(see diagram 41 below it) where the 'incoming' hydroxide ion is
half-bonded to the carbon and the 'outgoing' chloride ion is still
'half-bonded' to the carbon. No
intermediate is formed.
-
The 3rd
diagram (mechanism 41) shows the energy changes as a reaction
profile.
-
*
Note that
an 'activated complex' or 'transition state' is not the
same as an intermediate like a carbocation which is a definite
entity in its own right, however short its lifetime.
-
For full
details and discussion of all halogenoalkane mechanisms see
Mechanisms Part II.
Case
study 4.2 Oxidation of iodide ions by
peroxodisulphate
described
in section 2.
Case study
4.3 Enzyme
kinetics Please note
that this section is currently being re-edited.
-
KEY
in sequence: E
= free enzyme, S
= free substrate reactant molecule, ES
= enzyme-reactant complex, EP
= enzyme-product complex, E
= free enzyme, P
= free product
-
The mechanism
of enzyme catalysed:
-
Step (1)
E + S ==> ES
-
Step (2)
ES ==> EP
-
The conversion of the substrate-enzyme
complex ES into the enzyme-product EP, is
the rate determining step. The docking in of the substrate to
be held e.g.
by inter molecular forces is likely to have a low activation
energy and a relatively high probability initially, so E + S ==> ES cannot be the rds.
However, ES ==> EP
involves bond breaking and will have a much higher
activation energy giving a much lower
probability of ES ==> EP transformation.
-
However,
[ES] cannot be measured directly, so a simplified
kinetics approach is to treat the formation and transformation of
ES/EP through to P as
a function of the concentrations of E and S (see the rate
expressions below, which is a big bad fudge seen here and in most textbooks!).
The situation should be dealt with via the Michaelis-Menten
equation, but this goes way above
AS-A2 chemistry demands.
-
Step (3)
EP ==> E + P
A
possible detailed reaction
profile is shown below.
The
uncatalysed profile would be a single, and much higher 'hump' (see 2nd
simplified diagram below).
-
Here Ea1
is small for the formation of the enzyme-substrate complex or
intermediate ES. The energy of ES is lower than the reactants
because it involves an energy releasing interaction whether it be via
intermolecular forces or chemical bonding.
-
Ea2
is much larger for the transformation of the ES intermediate
into the EP intermediate because it involves bond breaking in
the substrate molecule.
-
Ea2
is still much smaller than Ea for the uncatalysed
reaction, which is not shown here, but is shown in the simplified
diagram below e.g. the uncatalysed decomposition of hydrogen
peroxide solution has an Ea of 76kJmol-1,
but with an enzyme like catalase/peroxidase, the Ea is
reduced to 30kJmol-1 AND the rate of reaction is
increased by a factor of 108, impressive
biochemistry!!!
-
Ea3
is relatively small for the breakdown of the EP intermediate into the
free product and free enzyme.
-
However,
the simplified reaction profile diagram below should be good enough in an exam and shows the
relevant activation
energies.
-
Here, Ea1 is for E+S ==> ES, Ea2
is for ES ==> EP ==> E + P
-
and Ea3 for the uncatalysed
reaction, which with no enzyme is significantly greater.
-
Treating step (2) as the rds, gives the rate expression ...
-
rate = kES[ES],
-
that is the rate
of reaction = rate of transformation of ES into EP .
-
However, since
you cannot measure the concentration of the enzyme-substrate
complex, the
rate is treated as a function of the concentrations of the enzyme and substrate, [E] and [S],
which form the ES intermediate, assuming a steady state
situation, giving
-
So we are
treating the reaction as 1st order with respect to both
enzyme and substrate, 2nd order
overall.
-
For a set of
experiments, using a fixed amount of enzyme, the initial rate should be
proportional to the concentration of substrate at relatively low
concentrations, so the
effective rate expression is pseudo 1st order ...
-
If you keep
the substrate concentration constant and vary the quantity of enzyme, the effective
rate expression is also pseudo 1st order ...
-
However, if the
substrate concentration is very high, the maximum number of active
centres on the enzymes are
occupied and the rate is completely controlled by the transformation
of the ES complex at its maximum possible
concentration and independently of the high substrate concentration. This means you are now dealing with the maximum possible
rate, which only depends on the amount of enzyme present,
-
so the rate
expression becomes zero
order with respect to the substrate,
-
and for a
constant [E], and varying [S] at high concentrations of S the rate
= k4,
-
The two situations are summed up by
graph (1) below for a constant amount
of enzyme. The graph moves from 1st order kinetics to zero
order.
-
Graph (2)
is a 'simple' comparison of the effects of no inhibitor, a
constant amount of a competitive or a
non-competitive inhibitor for a constant amount of enzyme.
-
Vo
is the initial reaction rate for a constant amount of enzyme
and varying the substrate concentration.
-
Vmax
is the maximum possible rate when the maximum number of active centres are
occupied with S or P.
-
Competitive
inhibition: Although the inhibitor initially decreases the
reaction rate, the substrate can increasingly compete with the
inhibitor to occupy the 'active sites' as the substrate
concentration increases (Le Chatelier's equilibrium concentration
principle) so
that the maximum reaction rate
Vmax
can eventually equal
Vmax for
the non inhibited enzyme.
-
Non-competitive
inhibition: Since the inhibitor molecule does not bind to the
'active site' of the enzyme it cannot be 'replaced' by increasing
the substrate concentration. However, because it constantly
interferes with a fraction of the enzymes, the reaction rate will be
reduced giving the lower the
Vmax
which can never reach
Vmax
however high the substrate concentration.
-
See also
enzyme structure and
function notes for more details on inhibition.
(1)  ,
(2)
Case study
4.4 The H2/I2/HI equilibrium
-
The gaseous phase equilibrium
and kinetics involving hydrogen, iodine and hydrogen iodide has been very well studied
quantitatively at temperatures of 250-500oC.
-
The reaction is: H2(g)
+ I2(g) 
2HI(g)
-
The reaction mechanism,
in either direction, is controlled by an initial bimolecular collision
(rds) with a 'transition state' or 'activated complex' consisting of two hydrogen atoms
and two iodine atoms. The structure of the 'transition state' is not known
and there are two possible mechanisms of either 2 or 4 steps. This proposed
initial rate determining step (rds) of I2 + H2
==>, for the forward reaction, and HI + HI ==> for the
backward reaction, is supported by the kinetics data which shows that
...
-
the rate expression for
the forward reaction at equilibrium is:
-
and the rate expression
for the backward reaction at equilibrium is:
-
Now the equilibrium
expression for the reaction is ...
-
Kc
= [HI(g]2/[H2(g][I2(g)],
the equilibrium constant Kc has no units (dimensionless),
-
but since the rate
expressions involve the same concentration expressions as the
equilibrium expression and the rates of
the forward and backward reaction are the same at equilibrium, we can rearrange
the rate expressions so that
-
[HI(g]2
= rateb/kb and [H2(g][I2(g)] =
ratef/kf
-
therefore we can
write: Kc
= (rateb/kb)/(ratef/kf)
= kf/kb
-
and the 'rates cancel out', so the equilibrium constant is the ratio
of the two rate constants for the forward and backward reactions.
-
This a nice simple
example to combine the concept areas of equilibrium and rates of reaction,
but many other equilibrium reactions are not so simple to analyse!
-
When a system is a
dynamic equilibrium the rate of the forward reaction = rate of the
backward reaction, so here the H2/I2/HI concentrations
remain constant, but two reactions are simultaneously occurring.
Case
study 4.5 Ester hydrolysis
-
Esters can be
hydrolysed by (i) water alone (but very slow), but is (II)
catalysed by acids (H+(aq) specifically) and
(iii) alkalis (OH-(aq) specifically). The
latter hydrolysis (iii), is sometimes called a saponification reaction.
-
Hydrolysis with water
alone is usually too slow to obtain meaningful rate data.
-
However, with a fixed
amount of acid catalyst e.g. HCl(aq), it is possible to follow
the reaction by alkali titration and show that ...
-
rate = k1[RCOOR'(aq)],
that is 1st order with respect to ester, however,
-
if the concentration
of acid catalyst is varied, things are more complicated,
-
and the 2nd
order expression, rate = k2[RCOOR'(aq)][H+(aq)]
then applies.
-
With alkali, 2nd
order kinetics are observed overall, and the reaction can be followed by
titrating the remaining alkali with acid. Phenolphthalein (Pk(ind)
= 9.3), would be a suitable indicator because one of the reaction products
is the salt of a weak carboxylic acid (approx. pH 9). The product of the
titration is neutral sodium chloride.
Case
study 4.6 Acid catalyzed iodination of propanone
-
Propanone readily forms
1-iodopropanone on reaction with acidified iodine solution, as do all 2-ones
('methyl ketones') I assume?
-
However, the rate
expression is: rate = k2[CH3COCH3(aq)][H+(aq)]
-
This suggests there is
a slow rate determining step involving the ketone and the hydrogen ion in
the mechanism and what ever happens next e.g. involving the iodine, is
much faster. A proposed mechanism is shown below, where R = CH3 if it was
propanone. Note that very little can be
absolutely proved in mechanistic detail and you will find variations of this
diagram on the web.
Case
study 4.7 The acid catalyzed decomposition of sodium
thiosulphate
-
The reaction
between e.g. dilute hydrochloric acid and sodium thiosulphate is a
redox reaction catalyzed by hydrogen ions. I do not know of any other
catalysts?
-
Its a typical
'rates' reaction at GCSE level to illustrate temperature and
concentration factors or used as a coursework investigation. Its
followed by the time it takes to form enough sulphur to obscure a
black X marked on white paper. The method described in more detail on
the GCSE rates page.
It is possible to follow the reaction with a colorimeter due to the
light scattering effect of the colloidal sulphur particles but the
absorbance does not follow Beers Law and processing results is
apparently difficult!
-
The reaction is
...
-
Na2S2O3(aq)
+ 2HCl(aq) ==> 2NaCl(aq) + SO2(aq)
+ S(s) + H2O(l)
-
which for advanced
level is much more appropriately written in the ionic form ...
-
S2O32-(aq)
+ 2H+(aq) ==> SO2(aq) + S(s)
+ H2O(l)
-
The thiosulphate
ion on face value has an S=S bond, one S=O and two S-O bonds, but all
four are 'merged' in the same delocalised pi bonding system.
-
You
could say that one sulphur is in the zero oxidation state and one
in the +4 ox. state,
-
OR, perhaps
'safer' to argue, that on average each sulphur is in the +2
state (oxygen is -2, overall charge on ion 2-).
-
In the products,
the oxidation states of sulphur are much clearer to define, +4 in SO2(aq)
and 0 in S(s).
-
On the basis
that two S's start off in the +2 state, you could argue that this
is a disproportionation reaction, in which an element in an
ion/compound/compound is simultaneously oxidised (+2 to +4) and
reduced (+2 to 0).
-
You would expect
that the rate might be controlled by the interaction of the negative
thiosulphate ion and a positive hydrogen ion. You would expect the
interaction of oppositely charged ions to have a relatively low
activation energy, so in the rate expression:
-
rate =
k[S2O32-(aq)]t[H+(aq]h,
you might expect the order t and h to be 1, t=1 is quoted on the
web.
-
but its likely to
be at least a two step mechanism, so whether h is 1 or 2 or ?, I don't
know? Whatever, the orders t and h can only be found by experiment.
5.
Deriving the activation energy Ea from kinetics-rates data
-
The Arrhenius
equation quantitatively describes the relationship between the
rate constant k, temperature and the activation energy. The rate
constant value increases with increase in temperature and nothing
else varies it!
-
k = A exp(-Ea/RT)
-
k = rate
constant (from the rate expression)
-
A = a
constant for a given reaction, sometimes called the
'frequency' or 'pre-exponential' factor and it seems to be
linked to stereochemical factors i.e. the spatial aspects of
reactant particle collision.
-
Ea
= activation energy in Jmol-1
-
R = ideal
gas constant = 8.314 Jmol-1K-1
-
T =
temperature in K (Kelvin = oC + 273)
-
Rewriting the
Arrhenius equation in logarithmic form gives:
-
From accurate rate
data at different temperatures (e.g. 5 or 10o
intervals and a minimum of four results) you
can calculate the values of k OR more simply, for a fixed 'recipe', a set
of 'rate' results.
-
You then plot the value of ln(k or relative rate) versus the reciprocal temperature in Kelvin.
-
The negative
gradient of the graph is equal to -Ea/R,
-
so, Ea = -R x
gradient (in J, and /1000 => kJ).
-
In terms of y =
mx + c, m = gradient = dy/dx = Ea/R, c = a constant =
ln(A),
-
so, in terms of the
Arrhenius equation, the algebra equates to
-
ln(k) =
- Ea/(RT) + ln(A)
-
Example
calculation 5.1
-
Some accurate rate constant
data for various temperatures is tabulated for the reaction between hydrogen and iodine to form
hydrogen iodide is presented on a
sub-web
page.
-
H2(g)
+ I2(g) ==> 2HI(g)
-
rate = k2
[H2(g)] [I2(g)]
-
The Arrhenius plot
of ln(k) versus 1/T is also shown.
-
The Excel plot routine
conveniently provides the best line fit.
-
y = -19867 + 25.651
-
therefore gradient =
-19867 = -Ea/R
-
so, Ea =
19867 x 8.314 = 165174 J mol-1,
-
activation
energy, Ea = 165 kJ
mol-1 (3 sf, 165.2 4sf)
-
The constant A can
be calculated as follows:
-
The constant 25.651
= ln(A), therefore A = e25.651 , so A = 1.325 x 1011,
-
so the full
Arrhenius expression for the hydrogen iodine reaction is
-
k = Ae(Ea/RT)
= 1.325 x 1011 x e(165174/8.314T)
The thermochemistry &
thermodynamics dealt with in section 6. below are being re-written and extended
via
Part 1: Thermochemistry -
Enthalpy changes etc.
*
Part 2: Born-Haber cycle *
Part
3: Entropy & Free Energy
6.
Kinetic versus thermodynamic stability
6a.
Introductory points
-
Even before rates
factors are considered, the feasibility of a reaction is governed
thermodynamically, that is in order for a reaction to be able to occur the
energy changes must be favourable.
-
In order for a reaction
to feasible the overall entropy change ...
-
For a reaction the overall
entropy change is given by:
-
Total entropy change =
entropy change of system + entropy change of surroundings
-
ΔSθtot
= ΔSθsys + ΔSθsurr
-
ΔSθsys
= = ΣSθproducts
- ΣSθreactants
-
and ΔSθsurr
= -ΔHθsys-reaction/T(K)
-
Note the connection
between ΔS and ΔG.
-
From above and
substituting ...
-
ΔSθtot
= ΔSθsys -ΔHθsys/T
-
multiplying through by
T gives
-
TΔSθtot
= TΔSθsys -ΔHθsys
-
changing signs
throughout gives
-
-TΔSθtot
= -TΔSθsys + ΔHθsys
-
-TΔSθtot
is called the Gibbs Free Energy change symbol G,
-
so giving the familiar
free energy equation ...
-
ΔGθsys
= ΔHθsys
- TΔSθsys
-
The free energy can be
thought of as heat energy that is available to do work.
-
OR in the case of
cells, where ΔGθsys
= -neEθF,
-
this gives the
electrical energy available to do useful work.
-
PLEASE NOTE you do
not have to derive the Gibbs Free Energy equation but some syllabuses
require you to be able to use when supplied with enthalpy and entropy
data.
-
The free
energy change of the reaction, ΔGθ
must be <=0 to be feasible.
-
If ΔSθtot
or ΔGθsys
is close to zero, then you are likely to be dealing with equilibrium
situation.
-
BUT, however feasible a
reaction might be thermodynamically, it does not necessarily mean it will
happen spontaneously because of kinetic limitations.
-
The speed at which a
reaction takes place depends on the many factors described here or on the GCSE
rates notes page. However, a feasible reaction that you might expect to take
place, may not happen because the activation energy is so high that
virtually no molecules have enough kinetic energy to change on collision.
This would be described as a kinetically stable mixture but thermodynamically
unstable.
The thermochemistry &
thermodynamics dealt with in section 6. below are being re-written and extended
via
Part 1: Thermochemistry -
Enthalpy changes etc.
*
Part 2: Born-Haber cycle *
Part
3: Entropy & Free Energy
6b.
Kinetic stability/instability examples
-
A mixture of hydrogen
and oxygen (e.g. in air at room temperature) is perfectly stable until a
means of ignition, e.g. a lit splint, match or spark etc., is applied. Thermodynamically
the mixture is highly unstable with a very negative free energy
ΔGθ
and shouldn't exist! However, the activation energy to break the strong H-H
or O=O bonds is so high, that they 'happily' co-exist without reacting,
because the particle collisions are not energetic enough to cause a
reaction. Therefore the
mixture is kinetically stable. A high temperature from a match or spark etc., gives
enough of the reactant molecules sufficient kinetic energy to overcome the
activation energy on collision*.
-
2H2(g) +
2O2(g) ==> 2H2O(l), ΔGθ
= -237 kJmol-1 or ΔHθ
= -286 kJmol-1
-
Note: A
very negative ΔHθ
is usually, but not necessarily, indicative of a negative free energy
change.
-
*The
transition metal palladium can reduce the activation energy so much that
it catalyses the spontaneous combustion/combination of hydrogen and
oxygen at room temperature!
-
Like the
methane-chlorine reaction below, it is free radical chain reaction. The
initiating energy produces the first free radicals.
-
A mixture of hydrogen/methane
and chlorine is stable in the dark, but exposed to light (particularly
ultra-violet), the mixture explodes to form hydrogen chloride/chloromethane gas. The
strong H-H/C-H and Cl-Cl bonds ensure a high activation energy, but the
absorption by chlorine (with the weakest bond) of a photon of light energy,
to start the Cl-Cl bond breaking and so initiating the fast and
exothermic free radical chain reaction (explosive!).
-
H2(g) +
Cl2(g) ==> 2HCl(g), ΔGθ
= -191 kJmol-1 or ΔHθ
= -185 kJmol-1
-
CH4(g) +
Cl2(g) ==> CH3Cl(g) + HCl(g),
ΔGθ
= -103
kJmol-1 or
ΔHθ
= -99
kJmol-1
-
Hydrogen peroxide is
thermally unstable, its aqueous solution is kept in a brown bottle to avoid
decomposition initiated by light. It has a decent shelf-life of a few weeks,
i.e., reasonably kinetically stable in the short term, until manganese(IV)
oxide powder is added and then the rapid exothermic reaction takes place!
The thermochemistry &
thermodynamics dealt with in section 6. below are being re-written and extended
via
Part 1: Thermochemistry -
Enthalpy changes etc.
*
Part 2: Born-Haber cycle *
Part
3: Entropy & Free Energy
6c.
Thermodynamic 'instability' - reaction feasibility
-
Energy distributed not just in translational
KE, but also in rotation, vibration and also distributed
in electronic energy levels (if input great enough, bond breaks).
-
All four forms of energy
are quantised and
the quanta ‘gap’ differences increases from trans. KE ==> electronic.
-
Entropy (S) and
energy distribution: The energy is distributed amongst the energy levels in
the particles to maximise their entropy.
-
Entropy is a measure of both
the way the particles are arranged AND the ways the quanta of energy can be
arranged.
-
We can apply ΔSθsys/surr/tot
ideas to chemical changes to test feasibility of a reaction:
-
ΔSθtot
= ΔSθsys + ΔSθsurr
-
ΔSθtot
must be >=0 for a chemical change to be feasible.
-
For example: CaCO3(s)
==> CaO(s) + CO2(g)
-
ΔSθsys
=
ΣSθproducts
- ΣSθreactants
-
ΔSθsys
=
SθCaO(s)
+ SθCO2(g) - SθCaCO3(s)
-
ΔSθsurr
is -ΔHθ/T(K)
and ΔH is very endothermic (very +ve),
-
Now ΔSθsys
is approximately constant with temperature
and at room temperature the
ΔSθsurr term is too negative for ΔSθtot
to be plus overall.
-
But, as the temperature is
raised, the ΔSθsurr term becomes less negative and eventually
at about 800oC
ΔSθtot
becomes plus overall (and
ΔGθ becomes negative), so
the decomposition is now chemically, and 'commercially' feasible in a
lime kiln.
The
thermochemistry & thermodynamics dealt with in section 6. above are
being re-written and extended via
Part 1: Thermochemistry -
Enthalpy changes etc.
*
Part 2: Born-Haber cycle *
Part
3: Entropy & Free Energy
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