This is for the Unit 4 of Edexcel Chemistry A2 Level. Enjoy, and any feedback is very welcome.
4.1 Rates of Reactions
Reaction Rate = change in amount of reactants/products per unit time (units: mol dm-3s-1) Following a reaction;
gas volume produced (gas syringe) mass lost (balance)
colour change (colorimeter)
clock reaction (sudden change at particular time means specific concentration of product has been reached - the shorter the time taken, the faster the rate)
electrical conductivity (number of ions will change as reaction occurs) Concentration-Time Graph
Rate at any point can be found by drawing a tangent at that point on the graph and finding the gradient.
Orders of Reaction
The order of reaction = how the reactants concentration affects the rate INCREASE REACTANT – RATE STAYS THE SAME – ORDER OF 0
INCREASE REACTANT – RATE INCREASES BY 1 FACTOR – ORDER OF 1 INCREASE REACTANT – RATE INCREASES BY 2 FACTORS – ORDER OF 2
Shapes of Rate-Concentration Graphs tell you the order.
ZERO ORDER FIRST ORDER SECOND ORDER
*square brackets indicate concentration. For example [X] = concentration of X. Half-life = time taken for half the reactant to react
If the half life is constant = first order If the half life is doubling = second order
You can also calculate the half life using reaction rates. For example, if you’re given the rate constant (see below) and the order you can work out half life (you don’t need to know how, just to be aware of it)
Rate Equations
Rate equation = tell you how the rate is affected by the concentrations of reactants. E.G.
Rate = k[A]
m[B]
nWhere: m = order of A n = order of B n+m = overall order
k = rate constant (always the same for a reaction at specific temp and pressure, increase temp = increase k = bigger value of k = faster reaction)
0 2 4 6 time (s) [X] 0 2 4 6 time (s) [X] 0 2 4 6 time (s) [X] 0 1 2 3 4 5 [X] rate 0 2 4 6 [X] rate 0 2 4 6 8 [X] rate
EXAMPLE
Propanone + Iodine —> Iodopropanone + H+ + I- (reaction occurs in acid)
Info: First order with respect to propanone and H+ and zero order with respect to iodine Rate equation = k[propanone]1[H+]1[iodine]0
Simplify to;
Rate equation = k[propanone][H+] (because anything to the power of 0 is 1) How to calculate rate constant from the orders and rate?
Rearrange to make k the subject and calculate.
Units of k can be found as you know concentration is moldm-3 and rate is moldm-3s-1 using a normal “cancelling” method.
Using data to deduce the order
1) The experiment: titrate sample solutions against sodium thiosulfate and starch to work out the concentration of the iodine. Repeat experiment, changing only the concentration for ONE REACTANT at a time.
experiment 1 2 3 4 5 6 7
[propanone] 0.4 0.8 1.2 0.4 0.4 0.4 0.4
[iodine] 0.002 0.002 0.002 0.004 0.006 0.002 0.002
[H+] 0.4 0.4 0.4 0.4 0.4 0.8 1.2
Here, first we changed the concentration of propanone for experiments 1, 2 and 3. Then, we changed the concentration of iodine in experiments 4 and 5.
Lastly, we changed concentration of H+ in experiments 6 and 7.
2) From this table we can plot 7 Concentration-Time graphs. Finding the gradient at time zero for each of these plots will give us the INITIAL rate of each.
3) Compare the results e.g.
Experiment Change compared to experiment 1
Rate of reaction Change
1 --- 0.033 ---
2 [propanone] doubled 0.062 Rate doubled
3 [propanone] trebled 0.092 Rate trebled
4 [iodine] doubled 0.034 No change
5 [iodine] trebled 0.032 No change
6 [H+] doubled 0.058 Rate doubled
7 [H+] doubled 0.094 Rate trebled
4) Now we can work out the rate equation:
Rate is proportional to [propanone] so the reaction is of order 1 with respect to propanone. Rate does not change/is independent of [iodine] so the reaction is of order 0 with respect to
iodine.
Rate is proportional to [H+] so the reaction is of order 1 with respect to [H+]. Rate determining step = slowest step in a multi-step reaction
(if a reactant appears in the rate equation it MUST be a rate determining step including catalysts which may appear in a rate equation)
PREDICITIONS
The order of a reaction with respect to a reactant shows the number of molecules that the reactant is involved in with regard to the rate-determining step. EXAMPLE: rate = k[X][Y]2. Here, one molecule of X and 2 molecules of Y will be involved in the rate determining step.
Chlorine free radicals in the ozone consist of 2 steps:
Cl•(g) + O3(g) —> ClO•(g) + O2(g) slow rate determining step
ClO•(g) + O•(g) —> Cl•(g) + O2(g) fast reaction
Therefore, Cl• and O3 must be in the rate equation as they are the reactants from the slowest step.
RATE = k*Cl•+*O3]
Predicting Mechanisms:
Once you know what the rate determining reactants are, you can think about what reaction mechanism it follows.
EXAMPLE:
If the rate equation is: rate = k[X][Y] And the two different mechanisms are:
1) X + Y —> Z OR
2) X —> Y + Z
From the rate equation, we know that X and Y MUST be in the rate determining step, therefore, it’s mechanism 1 which is the right one.
Halogenoalkanes – Nucleophilic Substitution (SN)
Halogenoalkanes can be hydrolysed by OH- ions by nucleophilic substitution. This is where a
nucleophile (e.g. :OH-) attacks a molecule and is swapped/substituted for one of the attached groups (e.g. Br δ-). In this case the Carbon (Cδ+) to Halogen (Xδ-) bond is POLAR as halogens are much more electronegative than the carbon so they draw in electrons making the Carbon slightly/delta positive. The bond looks like this:
Cδ+ — X
δ-Thus, the carbon can be easily attacked by a nucleophile who likes positive areas. This mechanism occurs:
*C-Br bond breaks heterolytically (unevenly)
Primary – react by SN2 where 2 molecules/ions are involved in the rate determining step Secondary – react by SN1 and SN2
Tertiary – react by SN1 where 1 molecule/ion is involved in the rate determining step You can see by the rate equation if there are 1 or 2 molecules in the rate determining step, which in turn, tells you if the mechanism is SN1 or SN2.
EXAMPLE:
Rate = k[X][Y] = 2 molecules in rate determining step = SN2 = primary/secondary halogenoalkane OR
Activation Energy
We can calculate the activation energy using the Arrhenius equation:
Where; (you don’t have to learn this, just understand the relationship) k = rate constant EA = activation energy (J)
T = temperature (K) R = gas constant (8.31 JK-1mol-1) A = another constant
Some relationships to note:
1) As EA increases, k will get smaller. Therefore large activation energy, means a slow rate – this
makes sense!
2) As T increases, k increases. Therefore at high temperatures, rate will be quicker – this makes sense too!
If we “ln” both sides of Arrhenius’ equation, we get; ln k = – EA/RT + ln A
(don’t forget, ln A is just a constant, a number) This looks a bit like:
y = mx + c
If we plot ln k (y) against 1/T (x), the gradient we produce will be –EA/R (m). Then R is just a number
that we know (8.31 JK-1mol-1) we can rearrange and find the activation energy. EXAMPLE:
Iodine clock reaction
S2O82- (aq) + 2I- (aq) —> 2SO42- (aq) + I2 (aq)
Rate of reaction is inversely proportional to the time taken for the solution to change colour i.e. increased rate = decreased time taken
k α 1/t
We can say that 1/t is the same as k (rate constant) and we can substitute 1/t instead of k in Arrhenius’ equation and find the gradient again to find a value for EA.
Catalysts
Catalyst = increases rate of a reaction by providing an alternative reaction pathway with a LOWER activation energy (EA). A catalyst will be chemically unchanged at the end of a reaction.
Adv: Small amount needed to catalyse a lot of reactions, also they are remade, thus reusable. Disadv: High specificity to the reactions they catalyse.
There are two types of catalysts:
HOMOGENOUS CATALYSTS HETEROGENOUS CATALYSTS
These are catalysts in the same state as the reactants.
E.G. when enzymes catalyse reactions in your body, all reactants are aqueous, this is a homogenous catalysis.
These are catalysts in different physical states to the reactants.
They are easily separated from products – GOOD
Can be poisoned (i.e. a substance clings to a catalyst stronger than the reactant would, preventing reaction speeding up) example: sulphur in the Haber process is a “poison”– BAD Solid catalysts provide a large surface area for the reaction to occur e.g. mesh/powder
E.G. vanadium pentoxide in the contact process to make sulphuric acid
4.2 Entropy
Entropy = a measure of how much disorder there is in a substance, how many different ways particles can be arranged.
Systems are MORE energetically stable when disorder/entropy is HIGH.
EXAMPLE: A gas will want to escape its bottle because the room it’s in is much bigger and the particles can be arranged in lots of different ways.
SOLID LIQUID GAS
No randomness, therefore lowest entropy.
e.g. Sѳ (H2O(s)) = 7.4 JK-1mol-1
(see below)
Some randomness, some entropy.
e.g. Sѳ (H2O(l)) = 70 JK-1mol-1
(see below)
Most randomness, highest entropy.
e.g. Sѳ (H2O(g)) = 189 JK-1mol-1
(see below) *Note that zero entropy will only occur in a perfectly ordered crystal
Affecting Factors:
1. More quanta (packets of energy) = More ways to arrange particles = More entropy 2. More particles = More arrangements = More entropy.
E.G. X -> 2Y 2 moles of Y produced from 1 mole of X therefore entropy has increased 3. Increase in temperature = Increase in energy = More entropy
E.G. - from solid to liquid entropy has increased a bit - from liquid to gas entropy has increased a lot 4. Complicated/complex molecules = more entropy DEFINITIONS:
Standard entropy of a substance, Sѳ, is the entropy of one mole of a substance under standard conditions of 298K and 1atm. The units are JK-1mol-1.
We expect exothermic reactions to be the spontaneous ones; however some endothermic reactions are spontaneous too. This is to do with entropy. If entropy is high enough, the reaction will be spontaneous, whether the reaction is exo/endothermic.
EXAMPLE:
NaHCO3(s) + H+(aq) —> Na+(aq) + CO2(g) + H2O(l)
1 mole 1 mole 1 mole 1 mole 1 mole
Solid aqueous ions aqueous ions gas liquid
Here, the products have high entropy states (e.g. gas) and there are more moles (e.g. reactants to products = 2:3) And so, overall entropy has increased = SPONTANEOUS (also depends on ΔH – see below)
LEARN THESE:
Where;
ΔSsys = Entropy change of a system, the entropy change between the reactants and the products
ΔSsurr = Entropy change of a surrounding
ΔStotal = Total entropy change, the sum of the entropy changes of the system and the surroundings
EXAMPLE:
NH3(g) + HCl(g) —> NH4Cl(s)
Info: ΔH = -315kJmol-1
Sѳ (NH3(g)) = 192.3 JK-1mol-1 Sѳ (HCl(g)) = 186.8 JK-1mol-1 Sѳ (NH4Cl(s)) = 94.6 JK-1mol-1
1) Find entropy of the system
ΔSsys = Sproducts - Sreactants
= 94.6 – (192.3 + 186.8) = - 284.5 JK-1mol-1
2) Find entropy of surroundings
= - (-315000)/298 [Note: ΔH = -315kJmol-1 is in KILOJOULES, therefore x1000] = + 1057 JK-1mol-1
3) Find total entropy ΔStotal = ΔSsys + ΔSsurr
= -284.5 + 1057
= + 772.5 JK-1mol-1 [Note: must include sign (and units) with final answer]
ΔS
total
= ΔS
sys
+ ΔS
surr
When will a reaction be spontaneous? Total entropy must increase
+
ΔStotal = kinetically favourable (wants to react; spontaneous)
— ΔS
total = kinetically stable (will not react on its own; not spontaneous)* You can predict ionic compound solubility using the same idea; if ΔStotal is positive √, if negative X
ENDOTHERMIC experiments that are spontaneous: 1) Ba(OH)2(s) and NH2Cl(s)
Ba(OH)2.8H2O(s) + 2NH2Cl(s) —> BaCl2(s) + 10H2O(l) + 2NH3(g)
When you add barium hydroxide to ammonium chloride: Smell of ammonia gas
Temperature drops below 0˚C
2) Cold pack – NH4NO3(s) and H2O(l)
NH4NO3(s) —H2O(l)—> NH4+(aq) + NO3-(aq)
When you dissolve ammonium nitrate crystals in water:
Looking at the states in both these experiments, we have an INCREASE in entropy (from solids to liquids/aqueous). These reactions are spontaneous EVEN THOUGH the ΔSsurr is negative (because if
ΔH is positive for endothermic reactions the equation of ΔSsurr means the overall ΔSsurr will be
negative – see above equation) the ΔSsys is GREAT ENOUGH to overcome it, meaning ΔStotal will be
positive still. DEFINITIONS:
Thermodynamic stability – where the ΔStotal is negative, at RTP, the reaction will simply not occur.
E.G. limestone –> CaO + CO2
Kinetic inertness – when the ΔStotal of a reaction is positive, a reaction can happen spontaneously,
however the rate of reaction at RTP is so slow because the activation energy needed for it to start is so high. E.G. diamond –> graphite
The enthalpy change of hydration, ΔHhyd – the enthalpy change when 1 mole of aqueous ions is
formed from gaseous ions. E.G. Na+(g) —> Na+(aq)
The standard lattice enthalpy, ΔHѳlatt – the enthalpy change when 1 mole of a solid ionic compound is
formed from gaseous ions under standard conditions (298K and 1atm). E.G. Na+(g) + Cl-(g)—> NaCl(s) The enthalpy change of solution, ΔHsol – the enthalpy change when 1 mole of solute is dissolved in
Factors affecting ΔHѳlatt AND ΔHhyd include;
1) Ionic charge; = larger charge
= more exothermic lattice energy
= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION E.G. NaCl has ΔHѳlatt = -780kJmol-1 whereas MgCl2 has ΔHѳlatt = -2526kJmol-1 because
magnesium has a charge of 2+ which is greater than sodium’s 1+ 2) Ionic radii; = smaller ionic radii
= more exothermic lattice enthalpy = higher charge density
= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION E.G. Sodium’s ionic radius is bigger than magnesium’s (because Mg has one more proton which has a stronger positive nuclear attraction to its electrons – see unit 1/2) therefore magnesium will have a more negative lattice enthalpy/hydration enthalpy.
Finding the enthalpy of solution
where we use a similar principle to Hess’ Law;
ΔH
1
= ΔH
2
+ ΔH
3
REMEMBER (for ΔHsol ): GASEOUS IONS DOWN, AQUEOUS IONS UP4.3 Equilibria
RECAP: (for exothermic reaction)
At equilibrium the amount of reactants and products is the SAME.
Dynamic Equilibrium – a reaction that occurs in both ways at the same time (conditions; in a closed system at constant temperature)
Many industrial reactions are reversible; we use this sign for equilibria: E.G. Both these experiments are good economically
1) Contact process – making sulphuric acid
2SO2(g) + O2(g) 2SO3(g)
USES = fertilisers, dyes, medicines, batteries 2) Haber process – making ammonia
N2(g) + 3H2(g) 2NH3(g)
USES = fertilisers, producing nitrogen-based compounds EXPERIMENT: Hydrogen-Iodine Reaction (REVERSIBLE)
There is a relationship between the concentration of initial reactants/products and the equilibrium concentrations which are produced from them
E.G. H2(g) + I2(g) 2HI(g)
Initial concentration: H2 = 1.0moldm-3 I2 = 1.0moldm-3
Equilibrium concentration: H2 = 0.228moldm-3 I2 = 0.228moldm-3
From this we can see that the ratio has remained the same, i.e. 1:1 LE CHATELIER – oppose
the motion!
Where does equilibrium move and why?
Increase Temperature Toward reactants, therefore less products; Move to the endothermic side. Higher kinetic energy so more chance of successful collision LOW temp = high yield = but slow process...
Increase Pressure Toward side with less molecules of gas (only affects gases). Particles are pushed together, which increases chances of successful collision. HIGH pressure = high yield = expensive!
Introduce Catalyst NO EFFECT ON EQUILIBRIUM POSITION (will affect rate)
K
p/ K
c What is Kp / Kc?K p / Kc is the ratio of product concentration to reactant
concentration, and is commonly known as the equilibrium constant. For example, in the hydrogen-iodine reaction Kc will be;
*Note: products are 2 because in a balanced equation, there is a 2 in front – see below E.G. 4X 2Y + 3Z
*We can calculate Kp using partial pressures (see below)
As long as the equilibrium is HOMOGENOUS (all reactants/products in the same state) then we can use this general rule for finding Kc;
If the equilibrium is HETEROGENOUS (where reactants/products are in different states) then you must LEAVE OUT any concentrations that are
solid.
For Kp, HOMOGENOUS equilibriums can be calculated using;
If the equilibrium is HETEROGENOUS then you only take into account the gases.
*Note: we dont use square brackets for equilibrium partial pressures
EXPERIMENT: Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s)
1) Add 500cm3 of 0.1moldm-3 silver nitrate solution to 500cm3 of 0.1 moldm-3 of iron (II) sulfate solution
2) Leave mixture in stoppered flask at 298K, it will reach equilibrium 3) Take samples and titrate
CALCULATION:
Reactant/Product Fe2+(aq) Ag+(aq) Fe3+(aq) Ag(s)
Initial concentration (moldm-3)
0.05 0.05 0 0
Equilibrium concentration (from titre results)
0.0439 0.0439 (1:1 ratio) 0.0061 (0.05 – 0.0439) solid Equilibrium constant Units
Calculating partial pressures;
Minty FruitsTaste Minty
EXAMPLE:
When 3.0 moles of PCl5 is heated in a closed system, the equilibrium mixture has 1.75 moles of Cl. If
total pressure of the mixture is 714kPa, what is the partial pressure of PCl5?
Step 1) Find moles at equilibrium of all reactants and products;
We know 1.75 moles of Cl2, therefore we must also have 1.75 moles of PCl3 and so (3 - 1.75) will
leave us with the moles at equilibrium for PCl5 which is 1.25 moles. Adding these together we get
1.25+1.75+1.75 = 4.75 total moles at equilibrium. Step 2) Find the mole fraction;
Mole fraction of a gas in mixture = = = 1.66 Step 3) Find partial pressure;
Equilibrium and Entropy are related
∆S
total= R lnK
When the total entropy, ∆Stotal, increases, the equilibrium constant, K, will also increase.
If; K = 10-10 = reaction will not occur K = 10-5 = mostly reactants
K = 1 = balanced products and reactants K = 105 = mostly products
4.7 Acid/base Equilibria
From the timeline we can see the change in definition of acids through history. The main ones to know are:
Arrhenius definition – when acids/bases dissolve in water then completely/partially dissociate into charged particles (ions)
Brønsted–Lowry definition – an acid is a proton donor and a base is a proton acceptor; acids (proton donors) will never release a H+ on its own, it is always combined with H2O to form HYDROXONIUM
IONS – H3O+
*NOTE: Acid-base equilibria involves the transfer of protons, either donated or accepted.
example reason
Strong Acid Base
HCl(g) —> H+(aq) + Cl-(aq)
NaOH(s) + H2O(l) —> Na+(aq) + OH-(aq)
Strong acids and bases ionise almost completely in water.
*HCl has a pH of 0 = completely ionised Weak Acid
Base
CH3COOH(aq) CH3COO-(aq) + H+(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Weak acids and bases only slightly ionise. Equilibrium is set up with mostly reactants (to the left)
Conjugate acid base pairs
HA and A- are conjugate pairs H2O and H3O+ are conjugate pairs
WATER is special – it can behave as a base and an acid. You can work out the equilibrium constant in the same manner as we did before e.g.
However, the equilibrium is very far left and so the equilibrium constant for this reaction is said to have a constant value;
At 298K/1atm, the Kc of water is 1.0 x 10
-14
mol
2
dm
-6
(We often define this with its own notation – Kw)
pH – “power of hydrogen” - is a measure of the hydrogen ion concentration
CALCULATION: finding the pH of a strong acid 1) Calculate the pH of 0.05 moldm-3 of nitric acid. pH = - log[H+] pH = - log[0.05]
= 1.3 (pH value is small – expected for a strong acid) 2) An acid has a pH of 2.45, what is the hydrogen ion concentration?
pH = - log[H+] [H+] = 10-pH
= 3.55 x 10-3 moldm-3
*NOTE: H2SO4 dissociates to give 2[H+] and you will have to divide the final answer by 2 to find your
hydrogen ion concentration
CALCULATION: finding the pH of a weak acid
Weak acids do not fully dissociate so it isn’t as straight forward as above. Another constant called Ka
is introduced. There are some assumptions to make first:
a) Only a tiny amount of product dissociates so initial concentration of reactant = equilibrium concentration of reactant
b) All H+ ions come from the acid i.e. concentration of product 1 = concentration of product 2 1) Calculate the hydrogen ion concentration and the pH of a 0.02 moldm-3 solution of propanoic acid (CH3CH2COOH). The Ka of propanoic acid is 1.3 x 10-5 moldm-3.
Ka = [H+]2/[CH3CH2COOH]
[H+] = 5.09 x 10-4 pH = -log[5.09 x 10-4] pH = 3.29
CALCULATION: finding the pH of a strong base
One OH- ion fully dissociates per mole of base so the concentration of OH- ions and concentration of the base is the same. However to work out pH from the formula, we need [H+]. Therefore, we use our knowledge of (@ 298K), Kw = 1.0 x 10-14 mol2dm-6
1) Find the pH of 0.1 moldm-3 of NaOH at 298K. [H+] = [ ] = = 1.0 x 10-13 mol dm-3 Therefore,
pH = -log [1.0 x 10-13]
= 13.0 (ph value is large – expected for a strong alkali) CALCULATION: finding the pKa
1) Calculate the pH of 0.05moldm-3 of methanoic acid (HCOOH). Methanoic acid has a pKa of
3.75.
3.75 = -log[Ka] Ka = 1.78 x 10-4
1.78 x 10-4 = [H+]2/[0.05] [H+]= 2.98 x 10-3 pH = -log[2.98 x 10-3] pH = 2.53 Lastly, you should be aware that;
diluting a strong acid (e.g. HCl) by a factor of 10 increases the pH by 1 diluting a weak acid (e.g. CH3COOH)by a factor of 10 increases the pH by 0.5
1) add a measure of acid (with known concentration) to burette 2) rough titration; swirl conical flask for approximate end point* 3) accurate titration; drop by drop
4) record amount of base needed to neutralise the acid 5) Repeat for more accurate readings
*end point: when the solution changes colour (also known as equivalence point – see below)
INDICATORS Indicator Colour in acid pH when colour change Colour in alkali
Methyl orange red 3.1-4.4 yellow
phenolphthalein colourless 8.3-10 pink
acid alkali
pKa = - log [Ka]
pH Curves
Equivalence point = where a tiny amount of alkali causes a sudden big change in pH, where the acid is JUST neutralised. Equivalence point will vary depending on acid/alkali used. For the last graph between a weak acid and weak alkali, a pH meter is the best thing to use to find the equivalence point as the colour change is gradual and unclear.
CALCULATION: finding the Ka of a weak acid using a pH curve
At the half equivalence point [HA] = [A-] Therefore;
Thus, we can say that the half equivalence point is also the pKa of the weak acid, then we can use Ka = 10-pKa
Strong acid/Strong alkali Strong acid/Weak alkali
Weak acid/Strong alkali Weak acid/Weak alkali Phenolphthalein indicator
Phenolphthalein indicator
Methyl orange indicator
The half equivalence point is the point where half the acid has been neutralised, where half the volume of strong base has been added the weak acid before equivalence.
Buffers
- RESIST changes in pH when small amounts of acid/alkali are added - Doesn’t stop the pH from changing completely
- They only work when small amounts of acids/alkalis are added
ACIDIC BUFFERS ALKALINE BUFFERS
Weak acid + Salt Weak base + Salt
CH3COO-Na+(aq)—> CH3COO-(aq) + Na+(aq)
This fully dissociates; therefore mostly ethanoate ions
CH3COOH(aq) CH3COO-(aq) + H+(aq)
This only slightly dissociates; therefore mostly ethanoic acid
NH4Cl(aq)—> NH4+(aq) + Cl -(aq)
This fully dissociates; therefore mostly ammonium ions
NH4+(aq) H+(aq) + NH3(aq)
This only slightly dissociates; therefore mostly ammonium
ADDING ACID: (small amount)
[H+] increases which combines with the
CH3COO- to form CH3COOH so equilibrium shifts
to left, no change in pH.
ADDING ALKALI: (small amount)
[OH-] increases which combines with the H+ to form H2O which removes the H+ ions from
solution, so more H+ dissociate from CH3COOH
so equilibrium shifts to right, no change in pH.
ADDING ACID: (small amount)
[H+] increases which combines with the NH3 to
form NH4 so equilibrium shifts to left, no change
in pH.
ADDING ALKALI: (small amount)
[OH-] increases which combines with the H+ to form H2O which removes the H+ ions from
solution, so more H+ dissociate from NH4+ so
equilibrium shifts to right, no change in pH.
Biological environments (don’t need to learn but be aware of) Example Cells – need constant
pH for biochemical reactions to take place
Blood – need to be kept at pH 7.4 Food products – changes in pH occur due to fungi and
bacteria Buffer Controlled by the
equilibrium between dihydrogen phosphate and hydrogen phosphate H2PO4- H+ + HPO4 2-Carbonic acid (H2CO3) H2CO3 H2O + CO2
Lungs - by breathing out CO2,
levels of H2CO3 decrease and so
equilibrium moves to the right H2CO3 H+ + HCO3
-Kidneys control this equilibrium
Sodium citrate
Citric acid citrate ions Or
Phosphoric acid phosphate ions
Or
Benzoic acid benzoate ions CALCULATION:
A buffer solution of 0.4 moldm-3 of 1. Equation for Ka methanoic acid and 0.6 moldm-3 sodium
methanoate. 2. Rearranging for [H+]
For methanoic acid Ka=1.6 x 10-4 moldm-3.
4.8 Further Organic Chemistry
IsomersStructural: compounds with the same molecular formulae but different structural formulae
Stereoisomerism: Optical E/Z
mirror images of each other, only occurs in double bonds where they cannot be superimposed. rotation is restricted i.e. groups are
*known as fixed in position.
enantiomers
The chiral carbons has four different Z-ISOMER E-ISOMER groups attached to it. CIS “same” side TRANS “opposite sides” They are optically active (they rotate determined by how the heaviest plane-polarised light) – one will rotate molecules are distributed around clockwise and the other anticlockwise. the double bond.
A racemic mixture contains equal quantities of each enantiomer of an optically active compound (rotates plane polarised light).
Optical Activity can be used to work out a reaction mechanism. For example, nucleophillic substitution can occur in two different ways;
If a reaction is SN1 and you start with one enantiomer, the product will be a racemic mixture of two optical isomers. The electrons move in the polar bond (Cδ+ — Xδ-) move heterolytically to the Xδ- (1 stage)
If a reaction is SN2 and you start with one enantiomer, the product will be a single enantiomer which will rotate the polarised light. First the nucleophile attacks a carbon and then the electrons in the polar bond (Cδ+ — Xδ-) move heterolytically to the Xδ- (2 stages)
*Remember: from rates of reaction; SN1 means only 1 molecule will be involved in the rate determining step and SN2 means there are 2 molecules in the rate determining step
Aldehydes and Ketones
They do not hydrogen bond with themselves as they don’t have a polar Oδ-—Hδ+ bond. For this reason, aldehydes and ketones have lower boiling points than alcohols (which can hydrogen bond) They can hydrogen bond with water due to their polar Cδ+=Oδ- bond. Oxygen uses its lone pair to form hydrogen bonds with Hδ+ atoms on the water molecules.
Note: small ketones/Aldehydes will dissolve due to the polarity mentioned above, however large ketones/Aldehydes will have very strong intermolecular forces and will not dissolve.
NUCLEOPHILIC ADDITION
Hydrogen cyanide is a weak acid – it partially dissociates in water HCN H+ + CN-
CN- is a nucleophile and attacks the slightly positive carbon atom and donates its electrons to it. The electrons in the C=O bond move to the oxygen. H+ from water/hydrogen cyanide bond to the oxygen forming OH.
NOTE: HCN is a very toxic gas; acidified potassium cyanide is used to reduce the risk. Experiment must be conducted in fume cupboard.
Evidence of optical activity: carbonyl group is planar; nucleophile can attack from either side. Asymmetric (not symmetrical) ketone/aldehyde + CN- —> racemic mixture/two optical isomers. This is what you expect if the CN can attack either side, producing two different isomers.
Tests to identify
TEST Info Ketone Aldehyde
Bradys reagent 2,4-dinitropheylhydrazine Orange Orange
Tollen’s reagent
+ heat (water bath – not flame as flammable!)
Colourless solution of silver nitrate dissolved in ammonia which gets reduced and changes colour;
Ag(NH3)2+(aq) + e- —> Ag(s)
+2NH3 (aq)
No change Silver mirror (Ag(s))
Aldehyde oxidised
Fehlings/Benedicts solutions
Blue solution of copper(II) ions dissolved in NaOH(aq) become Cu+ ions;
Cu2+(aq) + e- —> Cu+(aq)
No change Brick red precipitate (Cu+ ions)
Aldehyde oxidised
Iodine in alkali + heat (tests for CH3 on carbon
attached to oxygen)
Positive test = yellow precipitate If aldehyde positive = ethanal If ketone positive = one end is CH3
*NOTE: Brady’s: you can identify carbonyl compounds by the melting point of the orange precipitate against known values
OXIDISING
Aldehyde —> Carboxylic acid √ [heat with acidified potassium dichromate (VI) ions (oxidising agent)]
colour change: ORANGEto GREEN
Ketone —> Carboxylic acidX [acidified dichromate (VI) ions are not a strong enough oxidising agent] REDUCING: [LiAlH4 in dry ether]
Aldehyde —> Primary alcohol Ketone —> Secondary alcohol Carboxylic Acids
They hydrogen bond with themselves as they do have a polar Oδ-—Hδ+ bond. For this reason, carboxylic acids have very high boiling points.
They can hydrogen bond with water due to their polar Cδ+=Oδ- bond. Oxygen uses its lone pair to form hydrogen bonds with Hδ+ atoms on the water molecules. Therefore, carboxylic acids are soluble, however as they get bigger they become less soluble as the intermolecular forces get too strong.
Dimer: when a molecule hydrogen bonds with just one other molecule, increasing the size and intermolecular forces of the molecule, meaning the boiling point is also higher.
Making a carboxylic acid:
1) Primary alcohol – oxidised – aldehyde – oxidised – carboxylic acid
2) Nitrile – hydrolysed (reflux with HCl then distil) – distilled product is carboxylic acid REACTIONS OF CARBOXYLIC ACIDS:
Neutralisation; 1) CH3COOH + NaOH —> CH3COONa + H2O
ethanoic acid sodium ethanoate
*NOTE: CO2 causes effervescence
2) 2CH3COOH + Na2CO3 —> 2CH3COONa + H2O + CO2
E.G. 12.5 ml of 0.1 moldm-3 of NaOH exactly neutralises 25ml of orange juice. What is the concentration of citric acid in the juice?
3NaOH + C6H8O7 —> Na3C6H5O7 + 3H2O
1) Find moles mols = conc x vol = 0.0125 x 0.1 = 0.00125mols 2) Find ratio/moles 3mol NaOH neutralised 1mol citric acid; 3:1
0.00125 ÷ 3 = 0.000417mol 3) Find concentration conc = mols ÷ vol
0.025 ÷ 0.000417 = 0.017 moldm-3
Reduction; 1) CH3COOH —LiAlH4 (in dry ether)—> 2CH3OH
2) CH3COOH + PCl5 —> CH3COCl + POCl3 + HCl
Making an ester:
Carboxylic acid + alcohol (heat/reflux/acid catalyst) ester
It is a reversible reaction so in order to get the ester you must distil off the liquid at 80˚C, and then mix with sodium hydrogen carbonate solution to remove any acid. Then separate the top layer (ester) using a funnel.
USES: ethyl ethanoate is used as a solvent in chromatography as well as pineapple flavouring. Naming; the alcohol that was added comes first i.e. ethanol + methanoic acid will produce an ester call ethyl methanoate
Acyl chlorides and Esters
REACTIONS OF ACYL CHLORIDES WATER (produce carboxylic acid) - Vigorous reaction with cold water acyl chloride + H2O —> COOH + HCl
ALCOHOL (produce ester) - Violent reaction @298K
acyl chloride + OH —> COOCH + HCl AMMONIA (produce amide)
- Violent reaction at 298K
acyl chloride + NH3 —> CONH2 + HCl
AMINE (produce Nsub-amide) Violent reaction at 298K
acyl chloride + CNH2 —> CONH2C + HCl
*NOTE: HCl gas is always given off (observation)
REACTIONS OF ESTERS
Acid hydrolysis – adding water so that the ester splits into an acid and an alcohol (reverse of making ester) using reflux/heat/acid catalyst. Base Hydrolysis
Reflux an ester with DILUTE ALKALI (e.g. NaOH) producing a carboxylate ion (H3COO-)
and an alcohol.
USES: making soaps; hydrolysing vegetable oils and animal fats (trimesters) and heating them with NaOH produces glycerol (tri-ol) and sodium salt (soap) that we use every day Trans-Esterification (TE)
Hydrogenation: adding hydrogen to remove the double bonds.
Use: making low fat spread from butter, biodiesel
Problem: some trans isomers have been linked to various diseases
Solution: to hydrogenation: trans-esterification;
Ester + Alcohol —> New ester Forming a polyester
4.9 Spectroscopy and Chromatography
EM Radiation: Microwaves Ultraviolet
Wavelength: 1mm-1m 400nm-10nm
Why: Heating Initiating reactions
How: Radiation causes electric field; food (also polar e.g. fats, sugars) rotate to line up with the field. Dryer food with less water content will take longer to cook as water has polar Oδ-—Hδ+ bonds.
Has enough energy to split molecules and produce free radicals
Example: Cooking – Microwave oven
Surgery – to kill cancer cells Chemical industry – heating
Initiating reactions such as substitution between halogen and alkane
- Cl2 —UV—> 2Cl•
- CF3Cl —UV—> CF3 + Cl•
Danger: n/a This initiation can cause one Cl• can
cause the destruction of two O3
molecules and another Cl• Massive chain reaction.
Mass Spectroscopy
Some common RFM of fragment ions: CH3+ 15 C2H6+ 29 C3H7+ 43 OH+ 17 CHO+ 29 COOH+ 45
The base peak is the 100% relative abundance which is used to find the RFM
M peak is caused by the whole molecular ion which breaks up into fragments of free radicals and positive ions, but only the positive ion shows up on a mass spectrometer.
The other peaks are fragment ions of a broken ethanol molecule. See below.
NMR Spectroscopy
This gives you information about the structure using the idea that every atomic nucleus (with an odd number of protons/neutrons) has a weak magnetic field due to its nuclear spin, and applying a strong magnetic field will display accordingly.
Hydrogen is a single proton and so we can use proton NMR to find how many hydrogens there are and how they’re arranged...
Normally protons are spinning randomly, however when you apply a STRONG EXTERNAL MAGNETIC FIELD all the protons line up. Some protons are aligned in the direction of the magnetic field and others are opposing it. Those opposing it are at a higher energy level and can emit a radiowave to move to the lower energy level. Those in the direction of the magnetic field are at a lower energy level and can absorb a radiowave and move to a higher radiowave.
NMR measures the absorption of energy.
Protons in different environments absorb different amounts of energy; due to them being shielded by electrons experiencing the effects of the strong magnetic force instead.
Examples of different environments:
2 environments: 4 environments:
Chemical shift – is the difference in absorption of a proton relative to TetraMethylSilane (Si(CH3)4).
Where δ = 0 is the value of TMS.
Each peak = one environment. In the graph opposite, there are two environments (2 peaks)
The less shielded a proton is, the further left the shift will be.
Spin-spin coupling – in high res, the peaks of an NMR usually split into smaller peaks, this is because the magnetic field of neighbouring protons interact. The peaks follow an n+1 rule whereby;
2 splits [doublet] = 1 neighbouring proton (or hydrogen) 3 splits [triplet] = 2 neighbouring protons (or hydrogens) 4 splits [quartet] = 3 neighbouring protons (or hydrogens)
Magnetic Resonance
- Patient is placed in a very large magnet and irradiated with radio waves - Hydrogen nuclei in the water in patients body interacts with the radiowaves - Different frequencies of wave are absorbed by different densities of tissue - A series of images can be produced by moving the beam to build a 3D image
USES: cancer/bone and joint treatment, brains studies, checking purity in pharmaceutical industry ADV: non invasive, X-ray would be harmful
Infrared Spectroscopy
1) IR beam goes through sample
2) IR energy is absorbed by the bonds, increasing their energy (vibrational)
3) Different bonds in different environments absorb different wavelengths
4) Any wavelengths that you need to know will be in the data book
USES: in the chemical industry to determine the extent of a reaction by seeing what bonds are present
Chromatography – good at separating and identifying things Mobile phase – where molecules can move i.e. liquid/gas Stationary phase – where molecules can’t move i.e. solid
Gas chromatography GC High performance liquid chromatography HPLC Stationary phase is a viscous liquid in a long
coiled tube e.g. oil
Stationary phase is small particles of a solid packed into a tube e.g. silica
Tube is built into an oven Tube is not heated
Sample injected and vaporised Sample forced through tube by high pressure Both rely on different amounts of the sample being moved from the top of the tube to the bottom known as the “retention time”