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Chemistry A2 Notes

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1 Kinetics

KINETICS THE RATE OF CHEMICAL REACTIONS Rate of Reactions is affected by:

Surface Area of a Solid Reactant Temperature Pressure of gaseous reactants Catalyst Concentration of a reactant in Solution

DEFINITION OF A RATE OF A CHEMICAL REACTION; change in concentration of a substance time
-3
-1 Unit: moldm s NB: the use of square brackets denotes concentration e.g. [HCl]

Measuring Rate of Reactions Some techniques for doing this are:

1.Gas syringe.

2.Volumetric analysis - samples are removed at regular intervals.

3.Measuring changes in pressure (for gas reactions)

4.Colorimeter

5.A conductivity meter The rate at a particular instant in a reaction can be determined by finding the gradient of a tangent on a concentration versus time curve at that time. The gradient at t=0 is called the initial rate and is used to calculate k.

THE RATE EXPRESSION The rate equation, the rate constant and the orders of reaction can only be worked out experimentally. Consider the chemical equation A + B → products Rate expression: rate α [A]m[B]n GENERAL RATE EQUATION: rate = k [A]m[B]n The individual order is the power to which a concentration is raised in the rate The overall order of a reaction is the sum of all the individual orders in the rate equation eg = m + n

k is the rate constant - it's unique to each reaction and conditions, units depends on order of reaction. The rate constant, k, increases exponentially with increasing temperature.

1

1 Kinetics

ORDER OF REACTION

Zero Order wrt A rate = k, i.e rate is always constant rate = k (as [A]0 = 1 ) eg if conc of A doubles, the rate stays the same,

First Order wrt A (as [A]1 = [A] ) eg as the conc of A doubles, the rate doubles; as [A] triples, the rate triples ss

Second Order wrt A eg as [A] doubles, the rate will be 22= 4 times faster; if [A] triples, the rate will be 32
= 9 times faster Draw second graph, rate is directly proportional to (conc)2

DETERMINATION OF RATE EQUATION BY INSPECTION: General Rate equation = k[A]m[B]n When [A] is doubled, the rate doubles. When [A] is tripled, the rate tripled The reaction is 1st order with respect to [A]
When [B] is doubled, the rate is 22= 4x faster. When [B] is tripled, the rate is 32= 9x faster The reaction is 2nd order with respect to [B]

Rate = k[A][B]2 2

1 Kinetics

RATE-DETERMINING STEP The rate determining step is the slowest step in a multi-step reaction

•If a reactant appears in the rate equation, it must affect the rate

•This reactant OR A SPECIES DERIVED from this reactant must be in the RDS.

•This means that any species involved in or before the RDS could affect the rate and therefore appear in the rate equation.

Reactants in the Rate Eq affect the rate

•If a reactant does not appear in the rate eq, it will not be involved in the RDS (nor will any species derived from this reactant) Example; A + B → D + E

Rate = k[A]

MECHANISM: Step 1: A → C + D Step 2: B + C → E

RDS - SLOW (contains A found in rate eq) fast (does not appear in rate equation)

Individual orders of the reaction and the RDS The order of a reaction with respect to a reactant shows the number of molecules of that reactant that are involved in the RDS.

3

2 Equilibria

EQUILIBRIA CHEMICAL EQUILIBRIUM In a homogenous system, all the reactants and products are in the same phase. In the case of equilibrium, this is usually the LIQUID PHASE.

The equilibrium constant, Kc, is calculated from KNOWN CONCENTRATIONS at CONSTANT TEMPERATURE of the species involved in the dynamic equilibrium. In general:

aA(aq) + bB(aq) ⇌ cC(aq) + dD(aq) Kc = [C]c[D]d
[A]a[B]b

e.g. 2A + B ⇌ 2C Units are:

therefore Kc =

(moldm-3)2
=
(moldm-3)2(moldm-3)

[C]2
[A]2[B]

1
= mol-1dm3 moldm-3

CALCULATIONS WITH KC EXPRESSIONS General steps: 1. Work out equilibrium moles

2. Convert moles to concentration by diving by total volume in dm3

3. Put concentration numbers into Kc expression

4. Substitute each conc term for moldm-3 and cancel down to work out units. e.g.

N2

Initial moles Given eqm moles Workout

10

+

10-5 5

3H2 ⇌ 2NH3 20 20-15 5

0 10 10

then sub into equation etc...

EFFECT OF CHANGING CONDITIONS ON EQUILIBRIA Type of Reaction

Temperature Change

Effect on Kc

Effect on products

Effect on reactants

Direction change of equilibrium

Endothermic

decrease

decrease

decrease

increase

moves left

Endothermic

increase

increase

increase

decrease

moves right

Exothermic

increase

decrease

decrease

increase

moves left

Exothermic

decrease

increase

increase

decrease

moves right

4

3 Acids, bases and buffers

ACIDS, BASES AND BUFFERS An acid is a proton donator. A base is a proton acceptor. Water-soluble bases are called alkalis and produce OH- ions in aqueous solution. NB: monoprotic acid H+ eg HCl diprotic acid 2H+ eg H2SO4 triprotic acid 3H+ eg H3PO4 When we have complete dissociation of the acid/base in water it is fully ionised: Strong Acid HCl HNO3 H2SO4

Weak Acid CH3COOH CH3CH2COOH Citric Acid

Strong Base NaOH KOH Ca(OH)2

Weak Base Ammonia

CALCULATING pH (given always to 2dp) pH = -log[H+]

therefore 10-pH = [H+]

NB: The smaller the pH, the greater the concentration of H+ ions. pH changes with temperature

FINDING THE pH OF STRONG ACIDS AND BASES The ionisation of water Water is slightly ionised: H2O(l) ⇌ H+(aq) + OH-(aq) This equilibrium is established in water and all aqueous solutions. Kc = [H+(aq)][OH-(aq)]
[H2O(l)]
The conc of water is constant and is incorporated into a modified equilibrium constant Kw

Kw = [H+(aq)][OH-(aq)] - ionic product is 1.0 s 10-14 mol2dm-6

EXAMPLE: Calculate the pH of a 0.100moldm-3 sodium hydroxide solution:
[OH-(aq)] = 1.00 x 10-1moldm-3 Kw = [H+(aq)][OH-(aq)] = 10-14 therefore [H+] = 10-14

0.1
= 10-13 pH = -log[H+]
= -log(10-13)
= 13.00 (2dp) 5

3 Acids, bases and buffers

FINDING THE pH OF WEAK ACIDS AND BASES Weak acids and bases do not fully dissociate when dissolved in water. A weak acid which dissociates: HA(aq) ⇌ H+(aq) + A-(aq) The equilibrium constant is called the acid dissociation constant:

Ka = [H+(aq)]eqm[A-(aq)]eqm
[HA(aq)]eqm The larger the value of Ka, the further the equilibrium is to the right and the more the acid is dissociated and the stronger it is. EXAMPLE: Calculate the pH of 1.00moldm-3 ethanoic acid:

CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq) Before dissociation At equilibrium

1.00

1.00 - [CH3COO-]

Ka = [H+][CH3COO-]
[CH3COOH]

For weak acids, use:

0
[CH3COO-]

0
[H+]

but [CH3COO-]:[H+] is 1:1 therefore

Ka =

Ka = [H+]2

1.00
-5

1.7 x 10 = [H+]2
[H+] = √(1.7 x 10-5) pH = -log√(1.7 x 10-5)
= 2.39 (2dp)

[H+(aq)]2
[HA(aq)]

NB: does not apply to buffers

Excess calculations: work out moles of OH- and HA. If HA in excess, find moles HA left A- moles = moles OH- added Use Ka = [H+(aq)][A-(aq]]
[HA(aq)]

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3 Acids, bases and buffers

ACID-BASE TITRATIONS In an acid-base titration, an acid of known concentration is added to a base until neutralisation. Curves

Equivalence point - where sufficient base has been added to neutralise the acid. METHOD: Work out concentration of unknown acid: HCl + NaOH → NaCl + H2O The equivalence point is where the number of moles of HCl is equal to the number of moles of NaOH.
-Take known volume of acid and known volume+conc of base and equate moles to work out unknown conc.

CHOICE OF INDICATORS FOR TITRATIONS END POINT - volume of alkali or acid added when the indicator just changes colour. A suitable indicator needs the following properties:

1. colour change must be sharp rather than gradual

2. end point of titration given by indicator must be the same as the equivalence point

3. the indicator should give a distinct colour change.

An appropriate indicator must change within the vertical portion of the pH curve. Here methyl orange is acceptable but phenolphthalein isn't

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