# Y13 Revision. Rates of Reaction explain and use the terms: rate of reaction order rate constant half-life Rate determining step.

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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

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

Rates of Reaction deduce, from a rate– concentration graph, the order (0, 1 or 2) with respect to a reactant;

Rates of Reaction (e) determine, using the initial rates method, the order (0, 1 or 2) with respect to a reactant; ABCRATE 1110.5 2111 2214 112 ABCRATE 1110.0045 2110.036 2210.0045 1120.009 ABCRATE 1110.01 211 2210.02 112

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  N 2 + 2CO 2 Rate = K[NO] 2 [CO] 2. SLOW: H 2  2H FAST: 2H + N 2  N 2 H 2 FAST: N 2 H 2 + 2H 2  2NH 3

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, MgCl 2 ) and relevant energy terms to: (i) construct Born–Haber cycles, (ii) carry out related calculations;

Element1 st IE2 nd IEAtomisation1 st EA 2 nd EA LE Sodium496NA107.3NANA Magnesium7381451147.7NANA ChlorineNANA121.7 -348 NA FluorineNANA79-328NA OxygenNANA249.2-141.1798 LEFormation NaCl-780------ NaF----573.6 MgCl 2 ----641.3 MgO----601.7 Na 2 O---- -414.2

Born Haber Cycles (MX) M (s) + ½ X 2 (g)

Born Haber Cycles (MX 2 ) M (s) + X 2(g)

Born Haber Cycles (MX) M (s) + ½ X 2(g)

Calculation of enthalpy of solution experimentally 1.Known volume of Water 2.Measure the temperature 3.Add known mass of solid 4.Stir to dissolve 5.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 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 known You can work out Enthalpy of solution

Entropy (a) explain that entropy is a measure of the ‘disorder’ of a system, and that a system becomes 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 =  S prod -  S react

Entropy Values C 3 H 6 (g) 266.9 J/K/mol H 2 (g) 65.3 J/K/mol C 3 H 8 (g) 269.9 J/K/mol  S =  S prod -  S react C 3 H 6 (g) + H 2 (g) C 3 H 8 (g)  S =  S prod -  S react 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.

Calculations  S =  S prod -  S react  H =  H f prod -  H f react HfHf -1206.9 -635.1 -393.5 CaCO 3  CaO + CO 2 CaCO 3 CaO CO 2 178.3 KJ/mol

Calculations  G =  H - T  S CaCO 3  CaO + CO 2 +130.5KJ/mol Will this reaction proceed spontaneously under standard conditions?  H = +ve,  S = +ve HENCE  G could be –ve or +ve  G =  H - T  S = 178.3 – (160.4*298/1000)

Entropy Values CaCO 3 92.9 J/K/mol CaO 39.7 J/K/mol CO 2 213.6 J/K/mol  S =  S prod -  S react CaCO 3(s) CaO (s) + CO 2(g)  S =  S prod -  S react Would we expect  S to be positive or negative? 160.4J/K/mol

Calculations  G =  H - T  S MgCO 3  MgO + CO 2  H = 100.6 KJ/mol  S = 174.8 J/K/mol  G = 48.5KJ/mol T>575.5K (or 302.5 celcius) Does MgCO 3 decompose at 298K What temp will it decompose?  H f  S KJ/mol J/K/mol MgCO 3 -1095.8 65.7 MgO -601.7 26.9 CO 2 -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-equations CO 2 + 2H + + 2e -  CO + H 2 O; Al  Al 3+ + 3e - F 2 O + 2H + + 4e -  2F - + H 2 O; IO 3 - + 3H 2 O  H 5 IO 6 + H + + 2e - oxidation numbers; H 4 XeO 6 + Cl -  XeO 3 + Cl 2 MnO 2 + HNO 2  Mn 2+ + NO 3 - NO 3 - + Cu  Cu 2+ + NO 2

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 with their 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 potentials describe how to measure, using a hydrogen electrode, 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 potentials describe how to measure, using a hydrogen electrode, 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 = E 0 cathode – E 0 anode

Feasibility of the reaction predict, using standard cell potentials, the feasibility of a reaction 3CO 2 + 6H + + 2Al  3CO + 3H 2 O + 2Al 3+ F 2 O + + 2IO 3 - + 5H 2 O  2F - + 2H 5 IO 6 Mn 2+ + 2NO 3 -  MnO 2 + 2NO 2 2NO 3 - + Cu +4H +  Cu 2+ + 2NO 2 + 2H 2 O Anode (where oxidation takes place) // Cathode (where reduction takes place): Eg Al/Al 3+ // CO 2,2H + /CO,H 2 O Standard cell potential = E 0 cathode – E 0 anode consider the limitations of predictions made using standard cell potentials, in terms of kinetics and concentration Does it state HOW FAST? Does it state concentration (if standard)?

Fuel Cells apply principles of electrode potentials to modern storage cells All energy obtained from liquid/solid fuels involve a redox reaction (except Nuclear) Eg H 2 + ½O 2  H 2 O, CH 4 + 2O 2  CO 2 + 2H 2 O Oxidation: ½H 2  H + + e - (½H 2 + OH -  H 2 O + e - ), CH 4  CO 2 + 4H + + 4e - Reduction: ½O 2 + 2e - + 2H +  H 2 O 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 ½H 2 /H + //½O 2,2H + /H 2 O or CH 4 /CO 2,4H + //½O 2,2H + /H 2 O 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 CO 2 (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 H 2 (Limited life of adsorption method, high pressure liquid / large fuel tanks) (ii) Hard to manufacture H 2 (Need electrolysis or need to convert fossil fuels to H 2 )

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