1.2 Thermochemistry - Hess's Law and Enthalpy Calculations

1.2a Hess's Law and its importance

Hess's Law is a version of the general law of conservation of energy i.e. the first law of thermodynamics which can be stated as energy cannot be destroyed or created but merely changed in form or distributed in different ways. Hess's Law states that the energy change from reactants A to products B is independent of the pathway taken no matter how many stages it involves. This is shown in the diagram of enthalpy cycles below.

ΔH1 = ΔH(A to B) Pathway 1 the most direct route with no intermediate stages,

 or  ΔH1 = ΔH2 + ΔH3 pathway 2 involving one set of intermediates C, or

or  ΔH1 = ΔH4 + ΔH5 + ΔH6 pathway 3 involving two sets of intermediates D and E

etc. etc. - there is no limit to the complexity of the Hess's Law Cycle as long as A and B are constant.

  

1.2b Using Hess's Law to perform enthalpy calculations

1.2b(i) Using known enthalpies of reaction or combustion etc.

Using a Hess's Law Cycle you can calculate enthalpies of formation which you could not determine by laboratory experiment. However, using these calculated enthalpies of formation and

experimentally determined enthalpies of combustion a huge variety of other enthalpies for other reactions can then be calculated.

 The 1st example of calculating the enthalpy of formation of methane illustrates the principles especially as this cannot be determined by direct laboratory experiment.

Given the following data below from text/data book calculate the enthalpy of formation of methane.

C(s) + O2(g) ==> CO2(g)   ΔHθc(C(s)) = ΔHθ

f(CO2(g)) = -393 kJmol-1

H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθc(H2(g)) = ΔHθ

f(H2O(l)) = -286 kJmol-1

CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)  ΔHθc(CH4(g)) = -890 kJmol-1

All of the three enthalpies above can be very accurately determined by direct experiment in a calorimeter.

 ΔHØ

f (methane) = ??? kJmol-1

C(s) + 2H2(g)   CH4(g)

ΔHθc(C(s)) + 2 x ΔHθ

c(H2(g))

= (-393) + (2 x -286)

C => CO2 and 2H2 => 2H2O +2O2(g)       

     

+2O2(g)

ΔHθc(CH4(g)) = -890

OR best to change the arrow round and change the signs as below, remember, change direction you change the sign BUT not the numerical energy value!

OR

ΔHθf(CO2(g)) + 2 x

ΔHθf(H2O(l))

due to coincidence of enthalpy names +2O2(g)  -2O2(g) 

= -ΔHθc(CH4(g))

= -(-890) = +890

CO2(g) + 2H2O(l)

The cycle for the standard enthalpy of formation of methane, ΔHθf

From Hess's Law, add up the sequence of enthalpy changes via the lower 'staged route' to get the overall enthalpy change for the enthalpy of formation of methane from its elements in their

normal stable states.

ΔHθf (methane) = (-393) + (2 x -286) + (+890) = -75 kJmol-1

  

2nd example of Hess's Law Cycle calculation

From the following thermochemical data

(1)  1/2H2(g) + 1/2Cl2(g) ==> HCl(g)   ΔHθf(hydrogen chloride) = -92.3 kJmol-1

(2)  2C(s) + 3H2(g) + 1/2O2(g) ==> CH3CH2OH(l)   ΔHθf(ethanol) = -278.0 kJmol-1

(3)  2C(s) + 11/2H2(g) + 1/2O2(g) + 1/2Cl2(g) ==> CH3COCl(l)   ΔHθf(ethanoyl chloride) = -275.0 kJmol-1

(4)  4C(s) + 31/2H2(g) + O2(g) ==> CH3COOCH2CH3(l)   ΔHθf(ethanol) = -481.0 kJmol-1

Calculate the enthalpy change for the reaction

(5)  CH3COCl(l) + CH3CH2OH(l) ==> CH3COOCH2CH3(l) +  HCl(g)   ΔHθr@(esterification) = ??? kJmol-1

ΔHØreaction(esterification) = ??? kJmol-1

CH3COCl(l) + CH3CH2OH(l)   CH3COOCH2CH3(l) + HCl(g)

-ΔHθf(CH3COCl(l))

+ {-ΔHθf(CH3CH2OH(l))}

= -(-275.0) + {-(-278.0)}

  

    

       

ΔHθf(HCl(g))

+ ΔHθf(CH3COOCH2CH3(l))

= (-92.3) + (-481.0)

4C(s) + 41/2H2(g) + O2(g) + 1/2Cl2(g)

from {2C(s) + 11/2H2(g) + 1/2O2(g) + 1/2Cl2(g)} + {2C(s) + 3H2(g) + 1/2O2(g)}

Adding up all the  ΔHθ's in the lower route of the cycle

ΔHθ298(esterification) = (+275.0) + (+278.0) + (-92.3) + (-481.0) = -20.3 kJmol-1

Note that the sign of enthalpy of formation of ethanoyl chloride and ethanol is reversed to fit in with the direction of change.

  

3rd example of Hess's Law Cycle calculation

From the following thermochemical data to do

 

1.2b(ii) Solving enthalpy problems using an 'algebraic-equation style' method

Given the following data

(1)  C(s) + O2(g) ==> CO2(g)   ΔHθ = -393 kJmol-1

(2)  H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθ = -286 kJmol-1

(3)  3C(s) + 4H2(g) ==> C3H8(g)  ΔHθ = -104 kJmol-1

calculate the standard enthalpy of combustion of propane

(4)  C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)  ΔHθc,298(propane) = ??? kJ mol-1

What you do is rearrange, if necessary, the data equations and add up the results - both equation components and delta H values, cancelling out the equation components should leave you with the

correct equation whose enthalpy value you require.

3C(s) + 3O2(g) ==> 3CO2(g) (1) add ΔHθ = 3 x -393 kJmol-1

4H2(g) + 2O2(g) ==> 4H2O(l) (2) plus ΔHθ = 4 x -286 kJmol-1

C3H8(g) ==> 3C(s) + 4H2(g) (3) plus ΔHθ = +104 kJmol-1  (equation and sign reversed)

C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l) (4) = ΔHθc,298(propane) = -2219 kJ mol-1

adding up (1) + (2) + (3) gives (4), cancelling out unwanted equation components, ==> treated as an = sign

 -

  

2nd example of 'algebraic' calculation style

Given the following thermochemical data:

(1)  1/2H2(g) + 1/2Cl2(g) ==> HCl(g) ΔHθ = -92.3 kJmol-1

(2)  2C(s) + 3H2(g) ==> CH3CH3(g) ΔHθ = -84.7 kJmol-1

(3)  2C(s) + 2H2(g) + Cl2(g) ==> ClCH2CH2Cl(l) ΔHθ = -166.0 kJmol-1

Calculate the enthalpy change for the reaction

(4)  2Cl2(g) + CH3CH3(g) ==> ClCH2CH2Cl(l) +  2HCl(g) ΔHθ = ??? kJmol-1

H2(g) + Cl2(g) ==> 2HCl(g) (1) add ΔHθ = 2 x -92.3 kJmol-1

CH3CH3(g) ==> 2C(s) + 3H2(g) (2)plus ΔHθ = +84.7 kJmol-1  (equation and sign reversed)

2C(s) + 2H2(g) + Cl2(g) ==> ClCH2CH2Cl(l) (3) plus ΔHθ = -166 kJmol-1

2Cl2(g) + CH3CH3(g) ==> ClCH2CH2Cl(l) +  2HCl(g) (4) = ΔHθ298(chlorination reaction) = -265.9 kJ mol-1

adding up (1) + (2) + (3) gives (4), cancelling out unwanted equation components, ==> treated as an = sign

-

  

1.2b(iii) A method using the summation of enthalpies of reactants and products

A very simple example of enthalpy summation method

molecule ethanoic acid + ethanol ==> ethyl ethanoate + waterequation CH3COOH(l) + CH3CH2OH(l) ==> CH3COOCH2CH3(l) + H2O(l)

ΔHθf,298/kJmol-1 -487 -278   -481 -286

ΔHreaction = ∑Hproducts - ∑Hreactants

ΔHθreaction,298 = ∑ΔHθ

f,298(products) - ∑ΔHθf,298(reactants)

ΔHθesterification = {ΔHθ

f(ethyl ethanoate) + ΔHθf(water)} - {ΔHθ

f(ethanoic acid) + ΔHθf(ethanol)}

ΔHθesterification = {-481 + -286} - {-487 + -278}

ΔHθ(esterification reaction) = (-767) - (-765) = -2 kJmol-1

What would be ΔHθ(hydrolysis)? Answer! just reverse the sign! i.e. +2 kJmol-1

2nd example of enthalpy summation method

A more complicated example where you need to think more about mole ratios in the equation.

Given the following data from laboratory measurements

(1)  C(s) + O2(g) ==> CO2(g)   ΔHθf,298(carbon dioxide) = -393 kJmol-1

(2)  H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθf,298(water) = -286 kJmol-1

(3)  C4H10(g) + 61/2O2(g) ==> 4CO2(g) + 5H2O(l)  ΔHθc,298(butane) = -2877 kJ mol-1

Calculate the standard enthalpy of formation of butane gas, which cannot be determined by experiment.

(4)  4C(s) + 5H2(g) ==> C4H10(g)  ΔHθ = ??? kJmol-1

To solve this you can use equation (3) and the data from equations (1) and (2) to obtain a value for equation (4)

ΔHreaction = ∑Hproducts - ∑Hreactants

for equation (3) ΔHcombustion(butane) = ∑ΔHθf(products) - ∑ΔHθ

f(reactants)

ΔHθc,298(butane) = {4 x ΔHθ

f(carbon dioxide) + 5 x ΔHθf,298(water)} - {ΔHθ

f(butane) + ΔHθf(oxygen)}

Since oxygen is an element, ΔHθf(oxygen) = 0, therefore after rearranging we get

ΔHθc,298(butane) = {4 x ΔHθ

f(carbon dioxide) + 5 x ΔHθf,298(water)} - {ΔHθ

f(butane)}

-2877 = {(4 x -393) + (5 x -286)} - ΔHθf(butane)

ΔHθf(butane) = 2877 - 1572 - 1430 = -125 kJmol-1

Question 1

(a) Define the standard enthalpy of (i) combustion; (ii) formation

(b) Given the following standard enthalpies of combustion (298K, 1 atmos.):

C(s) -393 kJ mol-1; H2(g) -285.6 kJ mol-1; C2H6(g) -1560 kJ mol-1

(note: the first two enthalpies correspond to the enthalpy of formation of carbon dioxide and water respectively) 

Calculate the standard enthalpy of formation of ethane, C2H6(g)

(c) Given the following bond enthalpies (bond energies) in kJ mol-1;

C-H 412; C-C 347; O-H 464; O=O 498; C=O 805 (for CO2); C-O 358

calculate the enthalpy of combustion of ethane. (you will not get -1560 kJ mol-1, see later)

 

Question 2

Given the following standard enthalpies of combustion (298K, 1 atmos.):

C(s) -393 kJ mol-1; H2(g) -285.6 kJ mol-1; C8H18(l) -5512 kJ mol-1;

Calculate the enthalpy of formation of octane, C8H18(l)

 

Question 3

(a) Given the following standard enthalpies of combustion (298K, 1 atmos.):

C(s) -393 kJ mol-1; H2(g) -285.6 kJ mol-1;

and the enthalpy of formation of cyclohexane, C6H12(l) -156 kJ mol-1;

Calculate the enthalpy of combustion of cyclohexane, C6H12(l)

(b) Using the appropriate bond energies from Q1c calculate the theoretical enthalpy of combustion of cyclohexane.

 

Question 4

(a) Given the following standard enthalpies of combustion (298K, 1 atmos.):

C(s) -393 kJ mol-1; H2(g) -285.6 kJ mol-1;

and the enthalpy of formation of ethanoic acid, CH3COOH(l) -487 kJ mol-1

Calculate the enthalpy of combustion of ethanoic acid, CH3COOH(l)  

(b) Using the bond enthalpies listed in Q1c calculate the enthalpy of combustion of ethanoic acid but noting that the bond enthalpy for C=O is +743 kJmol-1 when it is NOT in the carbon dioxide molecule.

 

Question 5

Given the following standard Enthalpies of Formation (298K, 1 atmos.):

NH3(g) -46.2 kJ mol-1; HCl(g) -92.3 kJ mol-1; NH4Cl(s) -315.0 kJ mol-1;

Calculate the enthalpy change for the reaction:

NH4Cl(s) ====> NH3(g) + HCl(g)

 

Question 6

Given the following standard Enthalpies of Formation in kJ mol-1 (298K, 1 atmos.):

CH4(g) -74.9; CH3Br(l) -36.0; HBr(g) -36.2

Calculate the enthalpy change for the reaction:

Br2(l) + CH4(g) ====> CH3Br(l) + HBr(g)

 

Question 7

(a) Define the term 'average bond enthalpy'.

(b) Given the following standard Enthalpies of Formation  in  kJ mol-1 (298K, 1 atmos.):

C3H8(g)   -104 ;      C3H7Cl(g)  -105 ;     HCl(g)  -92

Calculate the enthalpy change for the reaction:

  Cl2(g)   +   C3H8(g)    ====>    C3H7Cl(g)    +   HCl(g)

 

(c) Given the following average bond energies in  kJ mol-1:

      C-H  412 ;       Cl-Cl  242 ;       C-Cl  338 ;     H-Cl  431

      Calculate the enthalpy change for the same reaction as in (b).

(d) Explain which of the calculations in (b) or (c) will be the most accurate and why?

 

Question 8

(a) Given the following standard Enthalpies of Formation  in  kJ mol-1 (298K, 1 atmos.):

        I2(g) +7.9 ;     C2H6(g)   -84.7 ;      C2H5I(l)  -31.0 ;     HI(g)  +25.9

 

      Calculate the enthalpy change for the reaction:

      I2(g)   +   C2H6(g)    ====>    C2H5I(l)    +   HI(g)

 

(b) Given the following average bond energies in  kJ mol-1:

      C-H  412 ;       I-I  151 ;       C-I   238 ;     H-I  299

      Calculate the enthalpy change for the same reaction in (a).

(c) Suggest a reason why the calculations in (a) and (b) differ by a bigger % compared to similar calculations and data in Q's 1-4?

 

Question 9

Given the bond enthalpies in kJ mol-1: C-H 412, C-C 348 and H-H 436 (∆Hatom = 218 kJ mol-1).

Calculate the enthalpy of formation of propane if the enthalpy of atomisation of carbon (graphite) is 715kJ mol-1.

Q9   Calculating the enthalpy of formation of propane from bond energy data and ∆Hsub(carbon)

Don't forget to change the positive signs for bond enthalpies of dissociation into negative values of bond formation.

3C(s) + 4H2(g)   C3H8(g)

ΔHØsub(C(s)) + 4 x

ΔHØbond diss(H2(g))

= (3 x 715) + (4 x 436)

= 2145 +  1744 = +3889 kJ

2 x ΔHØbond form(C-

C(g)) + 8 x ΔHØbond

form(C-H(g))

(2 x -348) + (8 x -412) =

= -696 - 3296 = -3992 kJ

3C(g) + 8H(g)

∆Hform(C3H8) = +3889 + (-3992) =  -103 kJ mol-1 (data book value -104 kJ mol-1)

1.1 Advanced Introduction to Enthalpy (Energy) Changes in Chemical Reactions

1.1a I have ASSUMED you have studied the GCSE notes on the basics of chemical energy changes

in what you might call a lower level introduction which bridges GCSE and AS level and you are completely ok in interpreting enthalpy level and activation energy diagrams such as ...

 

... and can clearly distinguish between enthalpy change and activation energy and the activation energy change with a catalyst does NOT change the enthalpy value of the reaction (activation energy will be rarely mentioned until the end of Part 3), but all important ideas from the GCSE

page will be re-studied in their advanced level context, but I would still study the GCSE page first!

REMEMBER - all chemical changes are accompanied by energy changes or energy transfers, many of which can be directly measured or theoretically calculated from known values.

  

1.1b Enthalpy Changes and Thermochemistry

Some important initial definitions and examples:

The system: The reactants and products of the reaction being studied i.e. the contents of the calorimeter.

The surroundings: The means the rest of the 'world' including the i.e. a copper calorimeter, the surrounding air etc. etc.

Enthalpy H: The energy content of a substance. This cannot be determined absolutely but enthalpy changes for a chemical reaction can be measured directly or indirectly from theoretical calculations using known values.

Enthalpy change ΔH: The net heat energy transferred to a system from the surroundings or from the surroundings to a system at constant pressure. The Greek letter delta Δ in maths implies a change, in this case a net heat energy change.

ΔH = Hfinal - Hinitial  (the units of delta H are kJ mol-1)

or ΔH = ∑Hproducts - ∑Hreactants

or ΔHθf(reaction) = ∑ΔHθ

f(products) - ∑ΔHθf(reactants)

The Greek letter delta Δ implies 'change in' ....

The Greek letter ∑ implies 'sum of' ....'

ΔHθf denotes a standard enthalpy of formation - which is explained further down.

Exothermic reaction: A reaction in which heat energy is given out from the system to the surroundings i.e. the enthalpy of the reacting system decreases and the temperature of the system and surroundings rises.

This means Hreactants > Hproducts so that ΔH is negative (-ve).

The enthalpy of the reaction system is decreasing.

Example: All combustion reactions are exothermic

e.g. CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)  ΔH = -890 kJmol-1

i.e. the figure of 890 kJ released refers to the complete combustion of 1 mole of gaseous methane (24 dm3), using exactly 2 moles of gaseous oxygen (48 dm3) to form exactly 1 mole of gaseous carbon dioxide (24 dm3) and 2 moles of liquid  water. Note: (i) Not only the molar quantities must clearly indicated BUT the physical states of all the substances must be clearly stated too. (ii) This sort of combustion reaction can be measured in a calorimeter (seesection 1.3). All equations should be read in molar terms when dealing with enthalpy values i.e. a delta H value goes with a specific equation.

This is a convenient point to make the point about the importance of state symbols via the combustion of hydrogen. For ...

H2(g) + 1/2O2(g) ==> H2O(l)  ΔH = -285.9 kJ mol-1, but for

H2(g) + 1/2O2(g) ==> H2O(g)  ΔH = -241.8 kJ mol-1, but for

If the water forms remains as steam/vapour/gas, then 44.1 kJ less heat energy is released to the surroundings, because condensation is an exothermic process (g ==> l) and forming liquid water releases an extra 44.1 kJ. The -285.9 (~-286) kJ mol-1 is the usual value for the enthalpy of combustion of

hydrogen you will encounter in your studies because at the standard temperature of 298K water is a liquid in its normal stable state.

 

Endothermic reaction: A reaction in which the system takes in or absorbs heat energy from the surroundings i.e. the enthalpy of the system increases and the temperature of the system and surroundings falls OR the system must be heated to initiate the reaction and provide the heat absorbed.

This means Hproducts > Hreactants so that ΔH is positive (+ve).

The enthalpy of the reaction system is increasing.

Example: The thermal decomposition of calcium carbonate

CaCO3(s) ==> CaO(s) + CO2(g)  ΔH = +179 kJmol-1

i.e. 179 kJ of heat energy must be absorbed to decompose 1 mole of solid calcium carbonate into 1 mole of solid calcium oxide and 1 mole of gaseous carbon dioxide. Mr(CaCO3) = 100, so 17.9 kJ of heat energy is absorbed in decomposing 10g of limestone. This reaction requires an experimental temperature of 800-1000oC to achieve an appreciable rate of reaction and cannot be studied quantitatively in the laboratory. However it can be theoretically calculated from known enthalpy change values by means of a Hess's Law cycle calculation.

  

The two diagrams below illustrate how exothermic (left) and endothermic (right) reactions are specified on an enthalpy level diagram.

 

Standard conditions

This means the reactants/products start/finish at a specified temperature, pressure and concentration whatever the 'temporary' temperature change in the reaction - which is required to calculate the enthalpy change.

The net energy change is based on the products returning to the same temperature and pressure that the reactants started at. The most frequently used standard conditions are a temperature of 298 K/25oC (K = 273 + oC) and a pressure of 1 atm/101 kPa and a concentration of 1.00 mol dm-3.

The use of standard conditions enables a database of delta H change to be assembled from which you can do theoretical calculations (see section 1.2 using Hess's Law).

Strictly speaking the standard conditions should be indicated in terms of the standard temperature and the reactants involved and standard delta H values are denoted with the Greek letter theta (θ).

By using data based on standard. agreed and defined conditions, then the data can be used universally by any laboratory around the world and also allows scientists to check each others experimental results.

Standard Enthalpy of Reaction ΔHr/react/reaction is the enthalpy change (heat absorbed/released, endothermic/exothermic) when molar quantities of reactants as stated in an equation react under standard conditions (i.e. 298K/25oC, 1 atm/101kPa)

Examples

NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(aq)  (exothermic)

ΔHθr,298 = -57.1 kJ mol-1 (can also be described as an 'enthalpy of neutralisation')

CaCO3(s) ==> CaO(s) + CO2(g)  (endothermic)

ΔHθr,298 = +179 kJ mol-1 (can also be described as an 'enthalpy of thermal decomposition')

Standard Enthalpy of Formation ΔHf/form/formation is the enthalpy change when 1 mole of compound is formed from its constituent elements with both the compound and elements in their standard states ('normal stable states). It may be endothermic or exothermic.

The standard state is the most stable state at the standard temperature and pressure e.g. at 298K/25oC and 1 atm/101kPa

e.g.  H2(g)  H2O(l)  C(s)  O2(g), C3H8(g)  C8H18(l)  C24H50(s)  CO2(g)  CH3CH2OH(l)  etc.

Examples

C(s) + 2H2(g) ==> CH4(g)   ΔHθf,298(methane) = -74.9 kJ mol-1

2C(s) + 2H2(g) ==> C2H4(g)   ΔHθf,298(ethene) = +52.3 kJ mol-1

2C(s) + 3H2(g) + 1/2O2(g) ==> CH3CH2OH(l)   ΔHθf,298(ethanol) = -278 kJ mol-1

1/2N2(g) + O2(g) ==> NO2(g)   ΔHθf,298(nitrogen dioxide) = +33.9 kJ mol-1

Note (i) The values can be positive/endothermic or negative/exothermic.

(ii) The enthalpy of formation of elements in their standard stable states is arbitrarily assigned a value of zero. This definition, together with experimental values of enthalpy changes allows a body of enthalpy change data to be accumulated and extended via theoretical calculations.

Standard Enthalpy of Combustion ΔHc/comb/combustion is the enthalpy change when 1 mole of a fuel (or any combustible material) is completely burned in oxygen (or air containing oxygen). You should ensure just 1 mole of fuel appears in the equation to accompany the delta H value which is always negative i.e. always exothermic.

Examples

C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)  ΔHθc,298K(propane) = -2219 kJ mol-1

CH3COOH(l) + 2O2(g) ==> 2CO2(g) + 2H2O(l)  ΔHθc,298K(ethanoic acid) = -876 kJ mol-1

In the calculations explained below just the subscripted letters r/f/c will be used for brevity and a temperature of 298K and a constant pressure 1atm assumed unless otherwise stated.There is more the enthalpies of combustion of alkanes and alcohols in section 1.4a

Standard enthalpy of neutralisation is the energy released when unit molar quantities of acids and alkalis completely neutralise each other at 298K.

NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)  ΔHθneutralisation = -57.1 kJ mol-1

Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l)  ΔHθneutralisation = -116.4 kJ mol-1

More on enthalpies of neutralisation

Bond Enthalpy ('bond energy')

This is the average energy absorbed to break 1 mole of a specified bond when all species involved are in the gaseous state.

e.g. H2(g) ==> 2H(g)  ΔH = +436 kJ mol-1  or  CH3CH2Br ==> CH3CH2 + Br  ΔH = +276 kJ mol-1

It is always endothermic and the reverse process - bond formation, is always exothermic. In many cases the values are averaged from a variety of 'molecular' situations. More on this in the bond enthalpy section.

  

General points: Arrows pointing downwards represent exothermic changes and arrows pointing upwards represent endothermic changes

1. The energy released when 1 mole of aluminium oxide is formed. The ΔH value of -1669 kJ mol-1 corresponds to the very exothermic enthalpy of formation of Al2O3 or the enthalpy of the complete combustion of two moles of Al. The very exothermicity of the reaction suggests, and correctly, that aluminium oxide is a very stable compound - it is thermally stable to at least 2500oC.

2. The endothermic enthalpy of formation of gold(III) oxide. It is a compound not readily formed and it decomposes on heating at ~150oC - contrast this thermal instability with that of aluminium oxide.

3. This is a much more complex enthalpy level diagram. The +436 kJ represents the bond enthalpy for splitting hydrogen molecules into hydrogen atoms. The +242 kJ is the bond energy of chlorine molecules. The -184 kJ is the enthalpy of formation of hydrogen chloride gas. The very exothermic -862 kJ is the energy released theoretically when two moles of hydrogen chloride are formed directly from hydrogen and chlorine atoms.

Problems

1. Is it possible that graphite is thermodynamically stable and diamond is kinetically stable while they're practically the same thing? 

2. Explain how kinetics relate to thermodynamics. Use the terms 'energy of motion', 'energy of heat', and an example from the module in your answer. 

3. Why would it be beneficial for a thermodynamically-stable reaction to use an energy input in the form of an enzyme or a catalyst even if it doesn't require energy to proceed? 

4. How come gas doesn't spontaneously combust inside a fuel tank?5. How is the rate constant k related to equilibrium? How does the rate constant change if heat is added to the reaction? 6. If the difference in energy between the reactants and products is negative, is the reaction spontaneous/nonspontaneous?

Most likely kinetically-stable, or thermodynamically-stable? 

Answers

1. Yes. Their different structures will differentiate their polarity and charge, and will cause the two compounds to act differently. Thus, one can be thermodynamically stable, while the other can be kinetically-stable. 

2. The energy of motion is related to kinetics, which determines how fast the reaction will reach equilibrium, related to thermodynamics. The energy of motion (kinetics) added to a reaction, will cause the reaction to happen faster, using energy of heat as a way by which to accelerate the reaction. An example of this is the cup of water with the sugar while it is being heated. The heat energy converts into kinetic energy (energy of motion), accelerating the reaction between the water molecules and the sugar crystals. 

3. A catalyst or enzyme will still be beneficial in a thermodynamically-stable reaction because it will simply accelerate it. 4. Fuel is kinetically-stable, which means it'll require an energy input in order to react. The energy inputed is the spark

caused by the ignition of the car. 5. The rate constant k is related to equilibrium in that it tells us about how fast the reaction reaches equilibrium. If heat is

added to a reaction, its rate will increase due to increased kinetic energy. 6. The reaction will be spontaneous, thermodynamically-stable. This is because the energy is given-off, not consumed by the

reaction.