Thermochemistry

1
Chapter 5

• Thermochemistry

2
Heat and work

• Heat flows from system to surroundings (q -ve)
• Heat flows from surroundings to system (q ve)
• Work done by system on surroundings (w -ve)
• Work done on system by surroundings (w ve)
• Since DE q w,
• If work is done by the system and heat flows from
  the system, DE lt 0
• If work done on system and heat flows to system,
  DE gt 0

3
State functions

• Figure 5.9 the internal energy of 50 g of water
  at 25oC is the same regardless of how the sample
  of water is obtained
• By warming 50g of ice to 25oC
• By cooling 50g hot water to 25oC

4
State functions

• Energy of the 50g of H2O is dependent only on the
  conditions of the experiment (e.g. room pressure)
• Internal energy is a state function a property
  of a system that is determined by specifying its
  condition or state
• The value of a state function depends only on its
  present condition, not on the particular history
  of the sample

5
State functions

• Energy (and DE) are state functions
• q and w are not state functions (although their
  sum, DE q w, is)

6
Enthalpy

• Many chemical reactions result in mechanical work
  being done (e.g., the reaction between an zinc
  metal and hydrochloric acid, Fig. 5.11)
• Zn(s) 2HCl(aq) ? ZnCl2(aq) H2(g)

7
Enthalpy

• As the piston in the experiment is pushed upwards
  against atmospheric pressure (by expansion of gas
  inside container) work is done (P-V work) w
  -PDV
• Expansion allows pressure inside container to be
  equal to external pressure
• The heat flow that occurs in a reaction at
  constant pressure is called enthalpy, H

8
Enthalpy

• H E PV (and DH DE PDV)
• 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

9
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

10
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

11
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

12
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)

13
Calorimetry

• Enthalpy changes (DH) for reactions can be
  measured experimentally at constant pressure,
  since heat flow will be accompanied by a
  temperature change.
• Measurement of heat flow is called calorimetry
• These experiments are carried out in calorimeters

14
Calorimetry

• The temperature change experienced by an object
  when it absorbs a certain amount of energy (heat)
  is determined by its heat capacity
• Heat capacity is defined as the amount of heat
  energy required to raise the temperature of a
  substance by 1oC (or K)
• Greater heat capacity means more energy required
  to increase the temperature
• Heat capacity for one mole of a substance is
  called molar heat capactiy
• Heat capacity of 1 g of a substance given another
  name (specific heat capacity or specific heat)

15
Calorimetry

• We measure the specific heat of a substance by
  measuring the temperature change (DT) that a
  known mass (m) of a substance undergoes when it
  gains or loses a specific quantity of heat

16
Calorimetry

• Example 209 J of heat is required to increase
  the temperature of 50.0 g of water by 1.00 K.
  What is the specific heat of water?

17
Calorimetry

• Q How much heat is required to warm about 250 g
  of water from 22oC to 98oC? (specific heat of
  water 4.18 J/g.K)
• Change in temp 98oC 22oC 76oC 76K

18
Calorimetry

• Q what is the molar heat capacity of water?
• If one g of water requires 4.18 J of energy to
  increase in temperature by one degree K, how much
  energy does one mole of water require?
• One mole of water weighs 18.02 g (MM H2O 18.02
  g/mol) so

19
Constant pressure calorimetry (coffee-cup
calorimetry)

• If we conduct an aqueous reaction in a styrofoam
  cup, following the temperature change with a
  thermometer, we can determine the heat flow of
  the reaction at constant pressure (qp DH)

20
Coffee-cup calorimetry

• For example
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• Reaction involves the addition of HCl to NaOH
• The temperature is observed to increase. What is
  the sign of q?
• The reaction produces heat. This heat is
  absorbed by the water molecules of the solvent,
  and transferred to the thermometer. Thus, heat
  flow is from system (rxn) to surroundings
  (solution) qp lt 0 (exothermic reaction)

21
Coffee-cup calorimetry

• Assumptions
• All heat is generated by the reaction itself
• No heat is lost to the calorimeter (styrofoam has
  a high heat capacity) no heat is lost by the
  solution to the surroundings
• Thus, all heat lost by the system (reaction) is
  gained by the surroundings (solution)
• qsoln -qrxn

22
A coffee-cup calorimetry problem

• Example when a student mixes 50 mL of 1.0 M HCl
  with 50 mL of 1.0 M NaOH in a coffee-cup
  calorimeter, the temperature of the resultant
  solution increases from 21.0oC to 27.5oC.
  Calculate the enthalpy change for the reaction,
  assuming the final solution has a density of 1.0
  g/mL and its specific heat is 4.18 J/gK
• Strategy
• Write the balanced chemical reaction
• Figure out the temperature change
• Figure out the mass of the solution
• Solve the specific heat equation using
  appropriate data

23
A coffee-cup calorimetry problem

• The temperature change for this reaction is
• Tfinal Tinitial 27.5oC 21.0oC 6.5oC (
  6.5 K)
• If the final volume is 50mL 50 mL 100 mL, and
  the solutions density 1.0 g/mL, then the
  mixture weighs 100 g
• The solution is obtained by solving for q

24
A coffee-cup calorimetry problem

• If qsoln 2.7 kJ, then 2.7 kJ of heat are have
  been gained by the solution from the system.
  Thus, qrxn for the reaction of 0.050 mol of HCl
  with 0.050 mol NaOH is -2.7 kJ
• Enthalpy is reported in terms of J/mol (or
  kJ/mol), so the enthalpy change (DHrxn) for this
  reaction would be

25
Bomb calorimetry

• A calorimetric technique that permits the study
  of combustion reactions

26
Bomb calorimetry

• Reaction is carried out inside a small chamber
  (bomb) that is capable of withstanding high
  pressures
• Bomb is contained in a water-filled chamber (mass
  of water known accurately) inside a calorimeter.
  Walls of calorimeter are insulated
• Bomb is saturated with O2(g) to permit complete
  combustion of the compound being studied
• Ignition is accomplished by passing a current
  into the sample to heat it up until combustion
  begins
• Bomb is contained in water. As the reaction
  occurs, heat generated increases the bomb
  casings temperature. Temperature increase of
  water is recorded.
• Volume of the bomb is constant

27
Bomb calorimetry

• Bomb calorimeter calculations are actually easier
  than coffee-cup calorimetry experiments.
• Once the bomb calorimeter has been calibrated vs.
  a known reaction, the temperature increase that
  accompanies a combustion is all that is needed to
  calculate the heat flow for a process
• Ccal (heat capacity of the calorimeter) is
  determined by combusting a substance that
  releases a known quantity of heat
• Ccal (units of J/oC or J/K)

28
Determining Ccal

• Example combustion of exactly 1g of benzoic acid
  is known to release 26.38 kJ of heat. If 1.000 g
  of benzoic acid is combusted in a bomb
  calorimeter, the temperature is found to increase
  by 4.857oC. What is Ccal?
• Ccal is the heat flow divided by the temperature
  change

29
Example bomb calorimetry

• The combustion of 1.000g of benzoic acid in a
  bomb calorimeter produces a temperature change of
  3.384oC. In a separate experiment, CH6N2 (4.00
  g) is combusted in the same bomb calorimeter.
  The temperature in this experiment was found to
  increase from 25.00oC to 39.50oC. What is the
  heat of reaction for the combustion per mole of
  CH6N2(l)?
• 2CH6N2(l) 5O2(g) ? 2N2(g) 2CO2(g) 6H2O(l)

30
Example - strategy

• First, determine Ccal
• Then, use Ccal to determine the heat flow
  associated with the 4.00 g of CH6N2(l)

31
Example - strategy

• This heat flow (-113 kJ) is the heat that is
  produced by combusting 4.00 g of CH6N2(l) (
  0.08676 mol)
• Enthalpy is an extensive property, so we expect
  that combustion of 1 mol of CH6N2(l) will produce
  more heat

32
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

33
Hesss Law

• For example, the combustion of methane
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890 kJ
• 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

34
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 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

35
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

36
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)

37
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

38
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)

39
Using enthalpies of formation to calculate
enthalpies of reaction

• 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

40
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

41
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, in terms of the last reaction,
• DHorxn 3(-393.5 kJ) 4(-285.8 kJ) - (-103.85
  kJ) (0 kJ)
• DHorxn -2220 kJ

42
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

43
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

44
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) DH
  -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

45
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)

46
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

47
Another midterm question

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

48
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

49
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? (ans -924.8 kJ)

50
Look at

• 5.59 (Hess)
• 5.77
• 5.51 (bomb calorimetry)
• 5.49 (coffee cup calorimetry)