Energetics HL only

15.1 Standard Enthalpy Changes Standard Enthalpy of Formation,  H Ϧ f The enthalpy change when 1 mole of a compound is produced from its elements under standard conditions, all reactants and products being in their standard states. e.g. Na (s) + ½Cl 2(g)  NaCl (s)  H Ϧ f = kJmol -1 By definition  H Ϧ f for all elements in their standard states = 0 kJmol -1 Write equations for the standard enthalpy of formation of i)Sodium carbonateii) Magnesium chloride iii) Ammoniaiv) Ethane

Standard Enthalpy of Combustion,  H Ϧ c The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, all reactants and products being in their standard states. e.g. CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)  H Ϧ c = kJmol -1 Write equations for the standard enthalpy of combustion of i)Hydrogenii) Graphite iii) Propaneiv) Ethanol

The equation below shows that when 1 mole of carbon burns completely 393 kJ of heat are evolved. C (s) + O 2(g)  CO 2(g) ;  H = kJ/mol How much heat energy would be released on complete combustion of a) 12 g of carbonb) 0.25 moles of carbon c) 10 moles of carbond) 18 g of carbon? What mass of carbon must be burnt to produce e) kJf) 786 kJ? g) What is the value of standard enthalpy of formation of carbon dioxide?

15.2 Born Haber Cycles Some important definitions: IONISATION ENTHALPY (energy),  H  i Enthalpy change occurring when 1 mole of electrons is removed from 1 mole of gaseous atoms to form 1 mole of ions with a 1+ charge e.g. Na (g)  Na + (g) + e-  H  i = +496 kJmol -1 Generally know as 1 st I.E., see earlier for other I.Es

ELECTRON AFFINITY,  H  ea Enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of ions with a 1- charge. e.g. Cl (g) + e-  Cl - (g)  H  ea = -364 kJmol -1 Can also have second E.A. This is always positive. Why? BOND DISSOCIATION ENTHALPY,  H  diss This is the standard molar enthalpy change which accompanies the breaking of a covalent bond in a gaseous molecule to form two gaseous free radicals e.g. Cl 2(g)  2Cl (g)  H  diss = +242 kJmol -1 CH 4(g)  CH 3(g) + H (g)  H  diss = +435 kJmol -1

ENTHALPY OF ATOMISATION,  H  at Standard enthalpy change which accompanies the formation of 1 mole of gaseous atoms e.g. ½Cl 2(g)  Cl (g)  H  at = +121 kJmol -1 n.b. this is ½  H  diss in this instance Na (s)  Na (g)  H  at = +107 kJmol -1 n.b. this is also  H  sub

LATTICE ENTHALPY,  H  L The enthalpy of lattice dissociation is the standard enthalpy change which accompanies the separation of 1 mole of solid ionic lattice into its constituent gaseous ions. e.g. NaCl (s)  Na + (g) + Cl - (g)  H  L = +771 kJmol -1 n.b. this is always positive but can also have lattice enthalpy of formation. e.g Na + (g) + Cl - (g)  NaCl (s)  H  L = -771 kJmol -1 So beware!

Born-Haber Cycles and Lattice Enthalpy The enthalpy of formation of an ionic solid can be split up into a number of steps which are then arranged into an energy cycle called a Born–Haber cycle. They are often used to calculate values for the lattice enthalpy. Beware sometimes FORMATION and DISSOCIATION. Guaranteed the biggest energy cycle that you will come across. Example 1: sodium chloridesodium chloride Example 2: magnesium chloridemagnesium chloride Both show dissociation but can simply reverse signs calculation for formation.

Try this: Name each of the changes A to E below, then construct a Born-Haber cycle for sodium bromide and calculate a value for the lattice enthalpy of formation for sodium bromide. A.Na (s) + ½Br 2(l)  NaBr (s)  H = -361 kJmol -1 B.Br (g) + e-  Br - (g)  H = -325 kJmol -1 C.Na (s)  Na (g)  H = +107 kJmol -1 D.Na (g)  Na + (g) + e-  H = +496 kJmol -1 E.½Br 2(l)  Br (g)  H = +112 kJmol -1

and more …… Construct a cycle for magnesium oxide and calculate a value for the lattice enthalpy of dissociation from the cycle. Click here to see answers By choosing suitable examples, investigate how ionisation enthalpy changes with: i)Size of ions ii)Charge on ions. Briefly explain your findings.

Theoretical I.E values from Born Haber Cycles vs Experimental (real) Values Look at the table below: NaClAgClAgI Theoretical value in kJmol Experimental value in kJmol Why are the experimental values higher than the theoretical values for AgCl and AgI? Do electronegativity differences give any clues?

15.3 and 15.4 Entropy and Spontaneity A spontaneous change is one which can occur in one particular direction but not in reverse (unless conditions such as temperature are changed). Many exothermic reactions are spontaneous. The natural direction of spontaneous change is from higher to lower enthalpy but there are some reactions that are spontaneous even though they are endothermic e.g. reaction of hydrochloric acid with potassium hydrogencarbonate. So there must be another factor to consider to decide whether a chemical reaction is spontaneous or not.

This other factor is called entropy, which is given the symbol S and measured in JK -1 mol -1. Entropy can be considered a measure of disorder. So an increase in entropy is an increase in disorder. So:solid  liquid  gas The particles get less ordered as we change from solid to liquid to gas so the entropy of a solid < a liquid << a gas. S   H 2 O (s) ) = 48 JK -1 mol -1 S   H 2 O (l) ) = 70 JK -1 mol -1 S   H 2 O (g) ) = 189 JK -1 mol -1 Simple substances have lower entropy values than more complex substances so S   CH 3 OH (l) ) = 127 JK -1 mol -1.

Entropy varies with temperature:

Entropy changes,  S The entropy change for a reaction can be calculated using the equation:  S   S  products -  S  reactants A spontaneous change involves an increase in entropy, i.e. an increase in disorder accompanies a spontaneous change. This is just like your file which starts off nice and neat but becomes disordered as the term goes by. Energy has to be put in to keep it ordered!

i) Calculate the entropy change that accompanies the reaction: C (s) + O 2(g)  CO 2(g) Comment on the value obtained. ii) Calculate the entropy change that accompanies the reaction: 2NaHCO 3(s)  Na 2 CO 3(s) + H 2 O (g) + CO 2(g) Comment on the value obtained.

Gibbs free energy,  G ,and spontaneity The key equation below links both the entropy and enthalpy factors mentioned so far:  G   H   S   G  must be negative (or zero) for a feasible change. Spontaneous and feasible mean much the same thing but use of the word feasible implies that it may not necessarily occur rapidly.

Calculating  G  i) Calculate the standard free energy change that accompanies the reaction: C (s) + O 2(g)  CO 2(g) Solution:  G   H   S   H  can be calculated  using Hess’s Law but in this case  H  =  H  c (graphite) = -394 kJmol -1. From earlier  S  = +3.3 JK -1 mol -1 So  G  = -394 – 298 x 3.3 / 1000 Where does the 1000 come from?  G  = -395 kJmol -1

ii) Calculate the free energy change that accompanies the reaction: 2NaHCO 3(s)  Na 2 CO 3(s) + H 2 O (g) + CO 2(g) What happens if the temperature is increased to 450 K? iii) Calculate the free energy change that accompanies the rusting of iron: 2Fe (s) + 1½O 2(g)  Fe 2 O 3(s) Is rusting feasible at 4000 K? Below what temperature is the rusting of iron feasible?

Calculating the Temperature at which a reaction becomes spontaneous A reaction becomes feasible when  G = 0. So  H = T  S And  T =  H /  S Beware of the units. Make sure you convert  S into kJ! At what temperature does the thermal decomposition of sodium hydrogencarbonate become feasible?

Entropy in Physical Changes e.g melting and boiling At the melting point and boiling point of a substance an equilibrium is established so there is no tendency for the system to move spontaneously in either direction. For a system at equilibrium,  G = 0. So  G   H   S  leads to  H   S  and   S   H  

Calculate the entropy change which accompanies the melting of ice. The enthalpy of fusion of ice is 6.0 kJmol -1. Calculate the entropy change which accompanies the boiling of water. The enthalpy of vaporisation of water is 44.0 kJmol -1.

From this data,  H L.E. = kJmol -1 Click above to return to notes