Enthalpy

1
Enthalpy

• Enthalpy is heat flow under constant pressure
  (qp)
• Enthalpy is a state function because E, P, and V
  are all state functions
• If DH gt 0, heat has been gained by the system
  from its surroundings endothermic reaction
• If DH lt 0, heat has been lost by the system to
  the surroundings exothermic reaction
• DH is more useful in studying reaction
  thermochemistry because
• Easy to measure heat flow at constant pressure
• The difference between DH and DE is usually very
  small

2
Enthalpies of reaction

• Just like DE, we calculate DH for a chemical
  reaction as DHrxn H(products) H(reactants)
• Enthalpy change that accompanies a chemical
  reaction is called the enthalpy of reaction,
  DHrxn
• For the combustion of hydrogen
• 2H2(g) O2(g) ? 2H2O(g) DHrxn -483.6 kJ
• The thermochemical equation implies that when 2
  mol of H2(g) are combusted at constant pressure,
  483.6 kJ of heat are lost by the system to the
  surroundings

3
Guidelines for thermochemical equations (3)

• Enthalpy is an extensive property. The size of
  DH will depend on the amount of matter involved
  in the chemical reaction
• For
• 2H2(g) O2(g) ? 2H2O(g)
• If 2 mol H2(g) are combusted, DHrxn -483.6 kJ
• If 4 mol H2(g) are combusted, DHrxn -967.2 kJ

4
Guidelines for thermochemical equations

• The enthalpy change for a reaction is equal in
  magnitude, but opposite in sign, to DH for the
  reverse reaction
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g) DH 890 kJ

5
Guidelines for thermochemical equations

• The enthalpy change for a reaction depends on the
  state of the reactants and products
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) DH -802
  kJ
• (because 2H2O(l) ? 2H2O(g) DH 88 kJ)

6
Hesss Law

• Enthalpy is a state function. DH for a reaction
  depends only on
• The amount of matter involved in the reaction
• The initial state of the reactants and the final
  state of the products
• Thus the enthalpy change for a reaction is the
  same whether the reaction proceeds in one step or
  in a series of steps

7
Hesss Law

• For example, the combustion of methane
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
• This same reaction can be obtained by summing the
  following two reactions
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) DH -802kJ
• 2H2O(g) ? 2H2O(l)
  DH -88kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890 kJ

8
Hesss Law

• If a reaction is carried out in a series of
  steps, DH for the reaction will equal the sum of
  the enthalpy changes for the individual steps
• Overall enthalpy change is independent of the
  number of steps or the particular nature of the
  path by which the reaction is carried out
• Useful for calculating enthalpy changes for
  reactions without actually carrying out the
  experiment

9
Enthalpies of formation

• The heat flow that accompanies the formation of
  one mole of a compound from its constituent
  elements is termed the enthalpy of formation, DHf
• If the conditions are as follows
• 25oC
• 1 atm pressure
• Then we are dealing with the standard enthalpy of
  formation, DHof
• The symbol o indicates standard conditions

10
Enthalpies of formation

• Thus, the enthalpy change that accompanies the
  formation of one mole of a compound from its
  constituent elements, with all substances in
  their standard states, is DHof
• e.g.
• 2C(s) 3H2(g) ½O2(g) ? C2H5OH(l)
  DHof-277.7 kJ
• This reaction represents DHofC2H5OH(l)
• Notice that all the reactants are pure elements
  (this reaction shows the formation of 1 mol of
  ethanol from the elements that make up an ethanol
  molecule, C, H, and O. These elements are
  written as they appear in nature)

11
Elements in their standard states

• Some examples
• Carbon C(s) (graphite)
• Oxygen O2(g)
• Hydrogen H2(g)
• Sodium Na(s)
• Chlorine Cl2(g)
• Iodine I2(s)
• Phosphorus P4(s)
• Iron Fe(s)
• Trés important Enthalpy of formation of any
  element in its standard state is zero

12
Enthalpies of formation

• Which of the following represent standard
  enthalpy of formation reactions?
• 2K(l) Cl2(g) ? 2KCl(s)
• C6H12O6(s) ? 6C(diamond) 6H2(g) 3O2(g)
• 2Na(s) ½O2(g) ? Na2O(s)

13
Using enthalpies of formation to calculate
enthalpies of reaction

• Calculate the enthalpy change for the following
  reaction
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)

Appendix C has formation enthalpies for all of
these compounds. Can use Hesss Law to arrange
the formation reactions and then add them
together in a way that will yield the above
reaction
14
C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)

• Heres the data for the relevant compounds from
  Appendix C (formation enthalpies)
• 3C(s) 4H2(g) ? C3H8(g) DHofC3H8(g)
• C(s) O2(g) ? CO2(g) DHofCO2(g)
• H2(g) ½ O2(g) ? H2O(l) DHofH2O(l)
• Rearrange these, multiply where necessary, and
  then add them together to get the top equation.
• Add the corresponding DH values to get DHorxn for
  the top equation

15
C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)

• We can write the reaction as the sum of three
  formation reactions
• C3H8(g) ? 3C(s) 4H2(g) DH1 -DHofC3H8(g)
• 3C(s) 3O2(g) ? 3CO2(g) DH2 3DHofCO2(g)
• 4H2(g) 2O2(g) ? 4H2O(l) DH3 4DHofH2O(l)
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)
• DHorxn DH1 DH2 DH3

16
Using enthalpies of formation to calculate
enthalpies of reaction

• DHorxn DH1 DH2 DH3
• From Appendix C, we find the following
• DH1 -(DHofC3H8(g) 103.85 kJ
• DH2 3DHofCO2(g) -1186.5 kJ
• DH3 4DHofH2O(l) -1143.2 kJ
• DHorxn DH1 DH2 DH3 -2220 kJ

17
Using enthalpies of formation to calculate
enthalpies of reaction

• We can break any reaction into formation
  reactions, yielding the following general result
• DHorxn SnDHof, products SnDHof, reactants
• So, for the problem we just solved with Hesss
  Law,
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)
• DHorxn 3(-393.5 kJ) 4(-285.8 kJ) - (-103.85
  kJ) (0 kJ)
• products reactants
• DHorxn -2220 kJ

18
Using enthalpies of formation to calculate
enthalpies of reaction

• Calculate the standard enthalpy change for the
  combustion of 1 mol of benzene, C6H6(l), to
  CO2(g) and H2O(l)
• How to do
• C6H6(l) 15/2 O2(g) ? 6CO2(g) 3H2O(l)
• DHorxn SnDHof, products SnDHof, reactants
• DHorxn 3DHofH2O(l) 6DHofCO2(g)
  - DHofC6H6(l) (15/2)DHofO2(g)
• DHorxn 3(-285.8 kJ) 6(-393.5 kJ) - (49.0
  kJ) (0 kJ)
• DHorxn -3267 kJ

19
Using enthalpies of formation to calculate
enthalpies of reaction

• Compare the heat per gram of combustion of
  benzene with that of propane
• We know the heat of combustion per mole of each
  of these substances
• Figure out how many grams of each substance is
  equal to one mole then convert moles to grams
• C3H8(g) (-2220 kJ/mol)(1mol / 44.1 g C3H8)
  -50.3 kJ/g
• C6H6(l) (-3267 kJ/mol)(1 mol / 78.1 g C6H6)
  -41.8 kJ/g

20
A midterm question

• Complete combustion of 1 mol of acetone, C3H6O,
  results in the liberation of 1790 kJ of heat
• C3H6O(l) 4O2(g) ? 3CO2(g) 3H2O(l) DHrxn
  -1790 kJ
• Q Using this information together with the
  following enthalpies of formation, calculate the
  enthalpy of formation of acetone
• DHofH2O(l) -285.8 kJ
• DHofCO2(g) -393.5 kJ

21
A midterm question

• The equation wed need to use to calculate
  DHofC3H6O(l) would be
• DHorxn SnDHof, products SnDHof, reactants
• We know DHorxn already (-1790 kJ)
• What we dont know (and are looking to solve for)
  is DHofC3H6O(l)
• -1790 kJ (3DHofH2O(l)) (3DHofCO2(g))
  DHofC3H6O(l) DHofO2(g)

22
A midterm question

• Substituting in the appropriate numbers,
• -1790 kJ -1180.5 kJ -857.49 kJ -
  (DHofC3H6O(l) 0)
• DHofC3H6O(l) -247.99 kJ -248 kJ

23
Another midterm question

• From the following enthalpies of reaction,
• H2(g) F2(g) ? 2HF(g) DHo -537 kJ
• C(s) 2F2(g) ? CF4(g) DHo -680 kJ
• 2C(s) 2H2(g) ? C2H4(g) DHo 52.3 kJ
• calculate (using Hesss Law) DHo for the reaction
  of ethylene (C2H4) with F2
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)

24
Another midterm question

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• Where to start what molecules exist of the
  reactant side and what molecules exist as
  products in the reaction of interest?
• 2(H2(g) F2(g) ? 2HF(g)) 2(-537 kJ)
• 2(C(s) 2F2(g) ? CF4(g)) 2(-680 kJ)
• C2H4(g) ? 2C(s) 2H2(g) -(52.3 kJ)
• Adding together,
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g) -2486 kJ

25
Problem 5.75

• Calculate the standard enthalpy of formation of
  solid Mg(OH)2, given the following data
• 2Mg(s) O2(g) ? 2MgO(s) DHo -1203.6 kJ
• Mg(OH)2(s) ? MgO(s) H2O(l) DHo 37.1 kJ
• 2H2(g) O2(g) ? 2H2O(l) DHo -571.7
  kJ
• What to do? (answer DHof -924.8 kJ)