Edexcel Unit 4 Chem Notes

7/29/2019 Edexcel Unit 4 Chem Notes

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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-3

s-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 thegradient.

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

You can only find the order of a reaction *experimentally* there is NO theoretical order system.


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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 youre given the rate constant

(see below) and the order you can work out half life (you dont 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]n

Where:

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


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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-3

s-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 outthe 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 zerofor 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

Reaction rates wont be exactly double or treble due to experimental errors etc.


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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 > ZOR

2) X > Y + ZFrom the rate equation, we know that X and Y MUST be in the rate determining step, therefore, its

mechanism 1 which is the right one.


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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-

). In this case the Carbon (C+

) to Halogen (X-

) 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:

C+

X-

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

Rate = k[X] = 1 molecule in rate determining step = SN1 = tertiary/secondary halogenoalkane


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Activation Energy

We can calculate the activation energy using the Arrhenius equation:

Where; (you dont have to learn this, just understand the relationship)

k = rate constant EA = activation energy (J)

T = temperature (K) R = gas constant (8.31 JK-1

mol-1

)

A = another constant

Some relationships to note:

1) As EA increases, k will get smaller. Therefore large activation energy, means a slow rate thismakes sense!

2)

As T increases, k increases. Therefore at high temperatures, rate will be quicker this makessense too!

If we ln both sides of Arrhenius equation, we get;

ln k = EA/RT + ln A

(dont 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 beEA/R (m). Then R is just a number

that we know (8.31 JK-1

mol-1

) we can rearrange and find the activation energy.

EXAMPLE:

Iodine clock reaction

S2O82-

(aq) + 2I-(aq) > 2SO4

2-(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 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.


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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 productsGOOD

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 poisonBADSolid 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


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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 its 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. S

(H2O(s)) = 7.4 JK-1

mol-1

(see below)

Some randomness, some

entropy.

e.g. S

(H2O(l)) = 70 JK-1

mol-1

(see below)

Most randomness, highest

entropy.

e.g. S

(H2O(g)) = 189 JK-1

mol-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 entropy2. 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 entropyE.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, S, is the entropy of one mole of a substance under standard

conditions of298K and 1atm. The units are JK-1

mol-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 H see

below)


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LEARN THESE:

Where;

Ssys = Entropy change of a system, the entropy change between the reactants and the products

Ssurr = Entropy change of a surrounding

Stotal = Total entropy change, the sum of the entropy changes of the system and the surroundings

EXAMPLE:

NH3(g) + HCl(g) > NH4Cl(s)

Info: H = -315kJmol-1

S

(NH3(g)) = 192.3 JK-1

mol-1

S

(HCl(g)) = 186.8 JK-1

mol-1

S

(NH4Cl(s)) = 94.6 JK-1

mol-1

1) Find entropy of the systemSsys = Sproducts - Sreactants

= 94.6 (192.3 + 186.8)

= - 284.5 JK-1

mol-1

2) Find entropy of surroundings

= - (-315000)/298 [Note: H = -315kJmol-1

is in KILOJOULES, therefore x1000]

= + 1057 JK-1

mol-1

3) Find total entropyStotal = Ssys + Ssurr

= -284.5 + 1057

= + 772.5 JK-1

mol-1

[Note: must include sign (and units) with final answer]

Stotal = Ssys Ssurr

Ssys = Sproducts - Sreactants


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When will a reaction be spontaneous?

Total entropy must increase +Stotal = kinetically favourable (wants to react; spontaneous) Stotal = kinetically stable (will not react on its own; not spontaneous)

* You can predict ionic compound solubility using the same idea; ifStotal 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 0C2) 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 Ssurr is negative (because if

H is positive for endothermic reactions the equation of Ssurr means the overall Ssurr will benegative see above equation) the Ssys is GREAT ENOUGH to overcome it, meaning Stotal will be

positive still.

DEFINITIONS:

Thermodynamic stability where the Stotal is negative, at RTP, the reaction will simply not occur.

E.G. limestone> CaO + CO2

Kinetic inertness when the Stotal 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, Hhyd the enthalpy change when 1 mole of aqueous ions is

formed from gaseous ions. E.G. Na+(g) > Na

+(aq)

The standard lattice enthalpy, H

latt 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, Hsol the enthalpy change when 1 mole of solute is dissolved in

sufficient solvent, so no further enthalpy change occurs on further dilution. E.G. NaCl(s) > NaCl(aq)


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Factors affecting H

lattAND Hhyd include;

1) Ionic charge; = larger charge= more exothermic lattice energy

= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION

E.G. NaCl has H

latt = -780kJmol-1

whereas MgCl2has H

latt = -2526kJmol-1

because

magnesium has a charge of 2 which is greater than sodiums 1

2) Ionic radii; = smaller ionic radii= more exothermic lattice enthalpy

= higher charge density

= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION

E.G. Sodiums ionic radius is bigger than magnesiums (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;

H1= H2 H3REMEMBER (for Hsol ): GASEOUS IONS DOWN, AQUEOUS IONS UP

Guns InDetroit, Apples In Ukraine


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4.3 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 processmaking sulphuric acid

2SO2(g) + O2(g) 2SO3(g)

USES = fertilisers, dyes, medicines, batteries

2) Haber processmaking ammoniaN2(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 CHATELIERoppose

the motion!

Where does equilibrium move and why?

Increase Temperature Toward reactants, therefore less products; Move to the endothermicside. Higher kinetic energy so more chance of successful collision

LOW temp = high yield = but slow process...

Increase Pressure Toward side with less molecules ofgas (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)


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Kp / Kc

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 are2

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) Fe

3+(aq) + Ag(s)

1) Add 500cm3 of 0.1moldm-3 silver nitrate solution to 500cm3 of 0.1 moldm-3 of iron (II) sulfatesolution

2) Leave mixture in stoppered flask at 298K, it will reach equilibrium3) Take samples and titrate


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CALCULATION:

Reactant/Product Fe2+

(aq) Ag+(aq) Fe

3+(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;

Partial pressure of gas = 714 x

= 187.9kPa


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Equilibrium and Entropy are related

Stotal = 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

K = 1010

= reaction complete


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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)

BrnstedLowry 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 HYDROXONIUMIONS 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

mol2dm

-6

(We often define this with its own notationKw)

Kw = Kc x [H2O] = the ionic product of water = [H+][OH

-] with UNITS: mol

2dm

-6


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pHpower 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 isnt 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

pH = - log [H+]


18/27

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

-13mol 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 of3.75.

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 burette2) rough titration; swirl conical flask for approximate end point*3) accurate titration; drop by drop4) record amount of base needed to neutralise the acid5) 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 scale


19/27

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 orangeindicator

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.


20/27

Buffers

- RESIST changes in pH when small amounts of acid/alkali are added- Doesnt 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 (dont need to learn but be aware of)

Example Cells need constantpH for biochemical

reactions to take place

Blood need to be kept at pH 7.4 Food products changes in pHoccur due to fungi and

bacteria

Buffer Controlled by the

equilibrium between

dihydrogen phosphate

and hydrogen

phosphate

H2PO4-

H+

+ HPO42-

Carbonic acid (H2CO3)

H2CO3 H2O + CO2Lungs - 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

.

What is the pH of the buffer? 3. Find pH


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4.8 Further Organic Chemistry

Isomers

Structural: 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. CISsame side TRANSopposite 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 (C+

X-

) moveheterolytically to the X-

(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 (C+

X-

) move heterolytically to the X- (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


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Aldehydes and Ketones

They do not hydrogen bond with themselves as they dont have a polar O-

H+

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 C+

=O- bond. Oxygen uses its lone pair to

form hydrogen bonds with H+ 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 moveto 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

Tollens 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: Bradys: you can identify carbonyl compounds by the melting point of the orange precipitate

against known values


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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 O-

H+ bond. For this reason,

carboxylic acids have very high boiling points.

They can hydrogen bond with water due to their polar C+

=O- bond. Oxygen uses its lone pair to

form hydrogen bonds with H+

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 acid2) 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

ethanoic acid ethyl chloride


24/27

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 80C, 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 theester 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

Dicarboxylic acid + Diol > Polyester + Water


25/27

4.9 Spectroscopy and Chromatography

EM Radiation: Microwaves Ultraviolet

Wavelength: 1mm-1m 400nm-10nm

Why: Heating Initiating reactionsHow: 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 O-

H+

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

- Cl2UV> 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 abundancewhich 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.


26/27

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 theyre 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 shieldedby 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 = 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)


27/27

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 industryADV: non invasive, X-ray would be harmful

Infrared Spectroscopy

1) IR beam goes through sample2) IR energy is absorbed by the bonds, increasing their energy

(vibrational)

3) Different bonds in different environments absorb differentwavelengths

4) Any wavelengths that you need to know will be in the data bookUSES: 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 cant 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. silicaTube 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

ADV of HPLC over GC: HPLC can be used if sample is heat sensitive or has a high boiling point

Edexcel Unit 4 Chem Notes

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