Y13 Revision
Apr 01, 2015
Y13
Revision
Rates of Reaction
• explain and use the terms:
• rate of reaction
• order
• rate constant
• half-life
• Rate determining step
Rates of Reaction
• deduce, from a concentration–time graph, the rate of a reaction and the half-life of a first order reaction;
• Understand how half-life is affected by Concentration of reactant
0
20
40
60
80
100
120
0 10 20 30 40 50
0
20
40
60
80
100
120
0 10 20 30 40 50
0
20
40
60
80
100
120
0 10 20 30 40 50
Rates of Reaction
• How to measure the rate of reaction from a concentration-time graph
• How to plot a rate-concentration graph from a concentration-time graph
0
20
40
60
80
100
120
0 10 20 30 40 50
0
20
40
60
80
100
120
0 10 20 30 40 50
0
20
40
60
80
100
120
0 10 20 30 40 50
Rates of Reaction
• deduce, from a rate–concentration graph, the order (0, 1 or 2) with respect to a reactant;
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
-20
0
20
40
60
80
100
120
140
0 0.5 1 1.5
Series1
Rates of Reaction
• (e) determine, using the initial rates method, the order
(0, 1 or 2) with respect to a reactant;
A B C RATE
1 1 1 0.5
2 1 1 1
2 2 1 4
1 1 2 0.5
A B C RATE
1 1 1 0.0045
2 1 1 0.036
2 2 1 0.0045
1 1 2 0.009
A B C RATE
1 1 1 0.01
2 1 1 0.01
2 2 1 0.02
1 1 2 0.02
Rates of Reaction
• deduce, from orders, a rate equation of the form: rate = k[A]m[B]n, for which m and n are 0, 1 or 2;
• calculate the rate constant, k, from a rate equation;
• explain qualitatively the effect of temperature change on a rate constant and hence the rate of a reaction;
Rates of Reaction• Use of rate equations to predict and propose a
reaction mechanism.• (i) for a multi-step reaction:• (ii) propose a rate equation that is consistent with the rate-
determining step,• (iii) propose steps in a reaction mechanism from the rate
equation and the balanced
1.
2NO + 2CO N2 + 2CO2
Rate = K[NO]2[CO]
2.
SLOW: H2 2H
FAST: 2H + N2 N2H2
FAST: N2H2 + 2H2 2NH3
Born Haber cycle• explain and use the terms
lattice enthalpy
enthalpy change of solution
enthalpy change of hydration;• explain, in qualitative terms, the effect of
ionic charge and ionic radius on the exothermic value of
a lattice enthalpy
an enthalpy change of hydration
Born Haber cycle
• Born–Haber cycle as a model for determination of lattice enthalpies and in testing the ionic model of bonding.
• use the lattice enthalpy of a simple ionic solid (ie NaCl, MgCl2) and relevant energy terms to:
• (i) construct Born–Haber cycles,
• (ii) carry out related calculations;
Element 1st IE 2nd IE Atomisation 1st EA 2nd EA LE
Sodium 496 NA 107.3 NA NA
Magnesium 738 1451 147.7 NA NA
Chlorine NA NA 121.7 -348 NA
Fluorine NA NA 79 -328 NA
Oxygen NA NA 249.2 -141.1 798
LE Formation
NaCl -780 ------
NaF --- -573.6
MgCl2 --- -641.3
MgO --- -601.7
Na2O ---- -414.2
Born Haber Cycles (MX)
M(s) + ½ X2 (g)
Born Haber Cycles (MX2)
M(s) + X2(g)
Born Haber Cycles (MX)
M(s) + ½ X2(g)
Calculation of enthalpy of solution experimentally
1. Known volume of Water2. Measure the temperature3. Add known mass of solid4. Stir to dissolve5. Measure the temperature every 10 secs
until the temperature drops (if exothermic) or increases (if endothermic)
6. Plot a temperature time graph and extrapolate to determine greatest temperature change (see graph)
7. Use E=mc t to determine energy change
8. Calculate number of moles of solid dissolved
9. Calculate H(soln) = -E/moles dissolved
Max T
Max T
Temp
Time
Theory behind dissolving
NaCl(s) NaCl(aq)
NaCl(aq) = Na+(aq) + Cl-(aq)
Hence the following process must have occurred
NaCl
Na+(g) Cl-(g)
Na+(aq) Cl-(aq)
HENCE if LE and HE knownYou can work out Enthalpy of solution
Entropy(a) explain that entropy is a measure of the ‘disorder’ of a system, and that a systembecomes energetically more stable when it becomes more disordered;
(b) explain the difference in magnitude of entropy:(i)of a solid and a gas,
(ii) when a solid lattice dissolves,
(iii) for a reaction in which there is a change in the number of gaseous molecules;
Entropy(c) calculate the entropy change for a reaction
given the entropies of the reactants and products;
S = Sprod - Sreact
Entropy Values
C3H6 (g) 266.9 J/K/mol
H2 (g) 65.3 J/K/mol
C3H8 (g) 269.9 J/K/mol
S = Sprod - Sreact
C3H6 (g) + H2 (g) C3H8 (g)
S = Sprod - Sreact
Would we expect S to be positive or negative?
-63.3 J/K/mol
• explain that the tendency of a process to
take place depends on temperature, T, the
entropy change in the system, ΔS, and the
enthalpy change, ΔH, with the surroundings;
• explain that the balance between entropy and enthalpy changes is the free energy change,
• Explain how ΔG, determines the feasibility of a reaction;
• state and use the relationship
ΔG = ΔH – TΔS;
• explain, in terms of enthalpy and entropy,how endothermic reactions are able to take place spontaneously.
CalculationsS = Sprod - Sreact
H = Hfprod - Hfreact
Hf
-1206.9
-635.1
-393.5
CaCO3 CaO + CO2
CaCO3
CaO
CO2 178.3 KJ/mol
CalculationsG = H - TS
CaCO3 CaO + CO2
+130.5KJ/mol
Will this reaction proceed spontaneously under standard conditions?H = +ve, S = +ve HENCE G could be –ve or +ve
G = H - TS = 178.3 – (160.4*298/1000)
Entropy Values
CaCO3 92.9 J/K/mol
CaO 39.7 J/K/mol
CO2 213.6 J/K/mol
S = Sprod - Sreact
CaCO3(s) CaO(s) + CO2(g)
S = Sprod - Sreact
Would we expect S to be positive or negative?
160.4J/K/mol
CalculationsG = H - TSMgCO3 MgO + CO2
H = 100.6 KJ/mol S = 174.8 J/K/mol G = 48.5KJ/molT>575.5K (or 302.5 celcius)
Does MgCO3 decompose at 298K
What temp will it decompose?
Hf S KJ/mol J/K/mol
MgCO3 -1095.8 65.7 MgO -601.7 26.9CO2 -393.5 213.6
REDOX• explain, for simple redox reactions, the terms
redox,
oxidation number,
half-reaction,
oxidising agent
reducing agent
REDOX• construct redox equations using
relevant half-equationsCO2 + 2H+ + 2e- CO + H2O; Al Al3+ + 3e-
F2O + 2H+ + 4e- 2F- + H2O; IO3- + 3H2O H5IO6 + H+ + 2e-
oxidation numbers;H4XeO6 + Cl- XeO3 + Cl2MnO2 + HNO2 Mn2+ + NO3
-
NO3- + Cu Cu2+ + NO2
Standard electrode potential• define standard electrode (redox) potential,E o • describe how to measure, using a hydrogen electrode,
standard electrode potentials of metals in contact withtheir ions in aqueous solution
V
RHS = Cathode (where reduction takes place)LHS = Anode (where oxidation takes place)Anode process // Cathode process
Standard electrode potential measured with Hydrogen cell placed at the anode
Measuring Standard potentialsdescribe how to measure, using a hydrogenelectrode, standard electrode potentials of non-metals in contact with their ions in aqueous solution
V
RHS = Cathode (where reduction takes place)LHS = Anode (where oxidation takes place)Cell notation: Anode process // Cathode process
Standard electrode potential measured with Hydrogen cell placed at the anode
Measuring Standard potentialsdescribe how to measure, using a hydrogenelectrode, standard electrode potentials of ions of different oxidation states in aqueous solution
V
RHS = Cathode (where reduction takes place)LHS = Anode (where oxidation takes place)Anode process // Cathode process
Standard electrode potential measured with Hydrogen cell placed at the anode
Standard electrode potential• calculate a standard cell potential by combining two
standard electrode potentials;V
RHS = Cathode (where reduction takes place)LHS = Anode (where oxidation takes place)Anode process // Cathode process
Standard cell potential measured as above practically
Standard cell potential = E0cathode – E0 anode
Feasibility of the reaction• predict, using standard cell potentials, the
feasibility of a reaction3CO2 + 6H+ + 2Al 3CO + 3H2O + 2Al3+
F2O + + 2IO3- + 5H2O 2F- + 2H5IO6
Mn2+ + 2NO3- MnO2 + 2NO2
2NO3- + Cu +4H+ Cu2+ + 2NO2 + 2H2O
Anode (where oxidation takes place) // Cathode (where reduction takes place):
Eg Al/Al3+ // CO2,2H+/CO,H2O
Standard cell potential = E0cathode – E0 anode
• consider the limitations of predictions made using standard cell
potentials, in terms of kinetics and concentrationDoes it state HOW FAST?
Does it state concentration (if standard)?
Fuel Cells• apply principles of electrode potentials to modern storage cellsAll energy obtained from liquid/solid fuels involve a redox reaction (except Nuclear)
Eg
H2+ ½O2 H2O , CH4 + 2O2 CO2 + 2H2O
Oxidation: ½H2 H+ + e- (½H2 + OH- H2O + e-), CH4 CO2+ 4H+ + 4e-
Reduction: ½O2 + 2e- + 2H+ H2O
We can create half cell for each process and hence create a battery
• explain that a fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage
½H2/H+//½O2,2H+/H2O or CH4/CO2,4H+//½O2,2H+/H2O
• explain the changes that take place at each electrode in a hydrogen–oxygen fuel cell;
V
Fuel Cells• outline that scientists in the car industry are developing fuel cell vehicles (FCVs),
fuelled by:
(i) hydrogen gas,
(ii) hydrogen-rich fuels (Ethanol/Methanol)Storage of fuel (using absorption/adsorption of gases on solid matix)
• State advantages of FCVs over conventional petrol or diesel-powered vehicles, in terms of:
(i) less pollution and less CO2 (only true for hydrogen gas)
(ii) greater efficiency (Only complete combustion occurs so maximum transfer of energy)
• State disadvantages of FCVs over conventional petrol or diesel-powered vehicles, in terms of:
(i) Storage of H2 (Limited life of adsorption method, high pressure liquid / large fuel tanks)
(ii) Hard to manufacture H2 (Need electrolysis or need to convert fossil fuels to H2)
Exam Questions