1. Thermodynamics Change in Enthalpy, H
CHEM 3310
Thermodynamics Change in Enthalpy, H

2. Enthalpy, H For processes measured under constant pressure
condition, the heat of the reaction is qp.
The subscript p reminds us that the heat measured is under constant pressure condition.
E = q + w
= qp – Pext  V
Solve for qp
qp = E + Pext  V
The heat measured under constant pressure condition is the sum of the change in internal energy plus PV work.
CHEM 3310

3. H = E + PV H = qp Enthalpy, H
At this point, we define Enthalpy, H.
H = E + PV
H = E + (PV)
= E + PV + VP
Now we can relate H to E. The difference is PV work!
For a constant pressure process, VP = 0
H = E + PV
Recall this quantity is qp from the previous slide.
H = qp
CHEM 3310

4. Difference of 2 kJ per mole of propane!
Enthalpy, H
H = E + PV
How big is the difference between E and H?
Let’s compare the combustion of propane gas, C3H8.
C3H8(g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
6 moles of gas moles of gas
Constant Volume
Constant Pressure
No PV work
There is PV work
Energy released is kJ per mole of propane
Energy released is kJ per mole of propane
E = kJ
H = kJ
Difference of 2 kJ per mole of propane!
H = E + PV
-2043 = PV
PV = 2 kJ per mole of propane
The system does 2 kJ/mole of work in the constant pressure expansion from 6 moles of gas to 7 moles of gas.
CHEM 3310

5. Enthalpy is an extensive property.
Enthalpy, H
H = E + PV
H is a state function. This means H is path independent.
H represents the amount of heat released/absorbed at constant P for a reaction as written by the balanced chemical equation.
C3H8(g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) H = kJ
If the reaction is carried out with 0.5 mole of propane, then (0.5)(-2043 kJ) = kJ
or 1022 kJ of energy is released.
Enthalpy is an extensive property.
Physical states of reactants and products must be specified as solid (s), liquid (l), gas (g), aqueous (aq) when enthalpy changes are reported.
Enthalpy change for a reaction is equal in magnitude but opposite in sign to the H for the reverse reaction.
3 CO2 (g) + 4 H2O (g)  C3H8(g) + 5 O2 (g) H = kJ
CHEM 3310

6. More energy released for reaction (ii)
Enthalpy, H
H = E + PV
More energy released for reaction (ii)
H for the combustion of propane is H = kJ when H2O is produced as a gas H = kJ when H2O is produced and condensed to a liquid
(i) C3H8(g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) H = kJ
(ii) C3H8(g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) H = kJ
The difference between the two H for the above reactions is kJ of energy.
It requires 176 kJ of energy to convert 4 moles of water from a liquid to a gas.
Less energy released for reaction (i) because 176 kJ is absorbed to convert water to steam.
H2O (l)  H2O (g) H = 44.0 kJ H2O (l)  4 H2O (g) H = 4 (44.0) = 176 kJ
CHEM 3310

7. Enthalpy, H H = E + PV H = Hfinal – Hinitial
Enthalpy of Reaction:
H = Hfinal – Hinitial
= Hproducts – Hreactants
H > 0 when Hproducts > Hreactants
H < 0 when Hproducts < Hreactants
Endothermic
Exothermic
CHEM 3310

8. Get used to seeing reaction not written from left to right!
Enthalpy, H
H = E + PV
For many reactions, the H of the reaction cannot be measured directly.
We can go home, or we must think of using alternative path to attain the H of the reaction that is of interest.
Hess’ Law
When a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for each step.
Path 1 is the direct path.
Path 2 is the indirect path A  C Ha C  F Hb
H = Ha + Hb
Get used to seeing reaction not written from left to right!
CHEM 3310

9. Enthalpy, H Hess’ Law in action for Experiment 7
Hhydration
Na2CO3 (s) + 10 H2O (l)  Na2CO3 • 10 H2O (s)
Path 1
Find an alternative path!
Direct path cannot be done!
H’a
H’b
Path 2
Na2CO3 (aq) + 10 H2O (l)
Path 1 is the direct path.
Path 2 is the indirect path Na2CO3 (s) + 10 H2O (l)  Na2CO3 (aq) + 10 H2O (l) H’a Na2CO3 (aq) + 10 H2O (l)  Na2CO3 • 10 H2O (s) H’b
Hhydration = H’a + (-H’b)
CHEM 3310

10. How can H2 be determined?
Enthalpy, H
Example:
The complete combustion of octane in an automobile engine yields carbon dioxide and water.
(i) C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (l) H1 = kcal
Health officials are concerned about a side-reaction which puts a dangerous pollutant, carbon monoxide, into the automobile exhaust.
(ii) C8H18 (l) + 17/2 O2 (g)  8 CO (g) + 9 H2O (l) H2 = ?
H2 cannot be measured directly since reaction (i) will always occur simultaneously to some extent.
How can H2 be determined?
What might be also interesting is how much energy you are losing if the fuel in your car is not undergoing complete combustion.
CHEM 3310

11. Enthalpy, H H2 = H1 - (H3) = H1 - (8 H4) = -1302.7 - 8(-67.6)
Example:
What is the heat of reaction for
(ii) C8H18 (l) + 17/2 O2 (g)  8 CO (g) + 9 H2O (l) H2 = ?
H1 = kcal
Find an alternative path!
C8H18 (l) + 25/2 O2 (g)
 8 CO2 (g) + 9 H2O (l)
Direct path cannot be done!
H3
H2 = ?
8 CO (g) + 9 H2O (l) + 4 O2 (g)
H3 is the combustion of carbon monoxide – Doable!
CO (g) + ½ O2 (g)  CO2 (g) H4 = kcal
H2 = H1 - (H3) = H1 - (8 H4)
= (-67.6)
= kcal/mole
Energetically, for every mole of octane that undergoes incomplete combustion, (761.9/1302.7)100 = 58.5%
you only get 58.5% of the energy.
CHEM 3310

12. Example:
Calculate the heat of combustion of carbon to form carbon monoxide, CO
(i) C (s) + 1/2 O2 (g)  CO (g) H = ?
Given that
(ii) C (s) + O2 (g)  CO2 (g) H = kJ
(iii) CO (g) + 1/2 O2 (g)  CO2 (g) H = kJ
Answer:
Reaction (i) is not doable, but the alternate paths are reactions (ii) and (iii). Rearrange reactions (ii) and (iii) to give reaction (i).
Add reaction (ii) to the reverse of (iii)
(ii) C (s) + O2 (g)  CO2 (g) H = kJ
+ (iii) CO2 (g)  CO (g) + 1/2 O2 (g) H = kJ
C (s) + 1/2 O2 (g)  CO (g) H = kJ

13. q  nAT q = nACAT H and Heat Capacity
Recall from the discussion of heat capacity, in an isolated system containing Substance A
q  nAT
where nA is the number of moles of substance A T = T2 – T (T2 = final temperature, T1 = initial temperature)
Introduce a proportionality constant, CA, such that
q = nACAT
CA is the molar heat capacity; it is characteristic of the substance.
Molar heat is the quantity of heat necessary to raise the temperature of one mole of the material by one degree Celsius.
CHEM 3310

14. H and Heat Capacity qp = nCpT q = nCT
The subscript p reminds is that:
heat measured is under constant pressure condition
heat capacity used in the equation is the heat capacity determined under constant pressure condition.
For a constant pressure process,
qp = nCpT
Since H = qp,
For infinitesimal changes,
Integrate, and assume that Cp is constant over T1 to T2.
When Cp varies with temperature, it is expressed as a series expansion.
Cp = a + bT + cT2 + …
where a, b, and c are constants determined empirically.
CHEM 3310

15. H and Heat Capacity q = nCT
When Cp is not constant over T1 to T2, Cp cannot be taken out of the
Integral sign.
When Cp varies with temperature, it is expressed as a series expansion.
Cp = a + bT + cT2 + …
where a, b, and c are constants determined empirically.
Temperature is
in Kelvin!
For example, Cp of water vapour varies with temperature,
Cp (H2O, g) = x 10-3 T x 10-7 T2 cal mol-1 K-1
CHEM 3310

16. H and Heat Capacity: Relate Cp to Cv
Recall from the definition of enthalpy, H,
H = E + PV
For infinitesimal changes,
dH = dE + d(PV)
where
dH=nCpdT
dE=nCvdT
d(PV)=nRdT
Substitute dH, dE, and d(PV), we have
nCpdT = nCvdT + nRdT
Integrate, and assuming that Cp and Cv are independent of temperature over T1 and T2.
CpdT = CvdT + RdT
(Cp-Cv)dT = RdT
Cp is greater than Cv by the amount of R, the gas constant. This is the PV work associated with maintaining the system at constant pressure.
(Cp – Cv)T= R T
Cp - Cv = R
For monatomic ideal gases, the molar heat capacity
Cp = Cv + R = 3/2R + R = 5/2R = J deg-1 mole-1
CHEM 3310

17. Standard Enthalpy of Formation, Hfo
The enthalpy change when one mole of the compound is formed from the elements in their stable forms at 25oC and 1 atm.
Tables of Hfo are available for various compounds.
Example: The standard enthalpy of formation for CO2(g) is kcal/mole. What that means is,
C (s, graphite) + O2 (g)  CO2 (g) Hfo (CO2, g) = kcal/mole
In general, Ho means the change in enthalpy under standard conditions of 25oC and 1 atm.
By definition, the enthalpy of formation of any element in its most
thermodynamically stable form is zero.
Hfo (O2, g) = 0
Hfo (Na, s) = 0
Hfo (Br2, l) = 0
etc.
CHEM 3310

18. Standard Enthalpy of Formation, Hfo
Enthalpy of formation of a substance can be calculated from knowing the enthalpy of other reactions.
Example: Calculate the enthalpy of formation of ethanol. Given
(i) The complete combustion of ethanol is Ho = kJ/mole
(ii) The decomposition of water is Ho = kJ/mole (iii) The heat of formation of CO2 (g) is Ho = kJ/mole
The reactions corresponding to the above heats of reaction are:
(i) C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (l) Ho = kJ/mole
(ii) H2O (l)  H2 (g) + 1/2 O2 (g) Ho = kJ/mole (iii) C (graphite) + O2 (g )  CO2 (g) Ho = kJ/mole
The reaction that we are looking for is
2 C (graphite) + ½ O2 (g) + 3 H2 (g)  C2H5OH (l) Hfo = ?
Reverse (i) + 3(Reverse (ii)) + 2(iii)
Check that Hfo = kJ/mole
CHEM 3310

19. Use of Hess’ Law to calculate H
Hess' Law states that H for a reaction can be found indirectly by summing
H values for any set of reactions which sum to the desired reaction.
As a result, we can use the tabulated Heats of Formation to calculate H.
The enthalpy of a reaction is the sum of the enthalpy of formation of each product multiplied by its stoichiometric coefficient in the balanced equation subtracted from the enthalpy of formation of each reactant multiplied by its stoichiometric coefficient in the balanced equation.
CHEM 3310

20. Use of Hess’ Law to calculate H
Example
Determine the heat of reaction for
ZnO (s) + 2 HCl (g)  ZnCl2 (s) + H2O (l) H = ?
Given, Hfo (ZnO) = kcal/mole
Hfo (HCl) = kcal/mole
Hfo (ZnCl2) = kcal/mole
Hfo (H2O) = kcal/mole
H = Hfo (ZnCl2) + Hfo (H2O) – (Hfo (ZnO) + 2 (Hfo (HCl))
= – ( (-22.1))
= kcal/mole
CHEM 3310

21. Using Bond Enthalpy to determine H
The bond enthalpy (or bond energy) is the energy required to break
a chemical bond. It is usually expressed in units of kcal/mole or kJ/mole,
measured at 298 K.
The exact bond enthalpy of a particular chemical bond depends upon the
molecular environment in which the bond exists. Bond enthalpy values
given in chemical data books are averaged values.
CHEM 3310

22. Using Bond Enthalpy to determine H
The enthalpy of dissociation of the O-H bond in a water molecule in the
gas phase is
H2O (g)  H (g) + OH (g) H = 120. kcal
The enthalpy of dissociation of the O-H bond in the hydroxyl radical in the
gas phase is
OH (g)  O (g) + H (g) H = 101 kcal
The bond energy is defined as the average of the two values.
Bond energy (O-H) = ½ ( ) = 111 kcal/mole
CHEM 3310

23. Using Bond Enthalpy to determine H
H>0 for bond dissociation
For the diatomic element, H2, the bond energy is
H2 (g)  H (g) + H (g) H = 104 kcal
H<0 for bond formation
Reverse the process, we get the Heat of Bond Formation
H (g) + H (g)  H2 (g) H = -104 kcal
The Heat of Bond Formation is the enthalpy change when one mole of
a particular type of bond is formed from gaseous atoms.
Cl (g) + Cl (g)  Cl2 (g) H = -58 kcal
H (g) + Cl (g)  HCl (g) H = -103 kcal
CHEM 3310

24. Using Bond Enthalpy to determine H
The reaction between hydrogen gas, H2 (g), and chlorine gas, Cl2 (g),
forms HCl (g). The heat of reaction of
H2 (g) + Cl2 (g)  2 HCl (g) H = -44 kcal
Consider the above reaction from bond breaking and bond formation.
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
Bond breaking: H2 (g)  H (g) + H (g) H = 104 kcal
Cl2 (g)  Cl (g) + Cl (g) H = 58 kcal
Bond forming: H (g) + 2 Cl (g)  2 HCl (g) H = 2 (-103) = -206 kcal
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
ΔH = – 206
ΔH = -44 kcal
CHEM 3310

25. Using Bond Enthalpy to determine H
Mean (single) Bond Energies at 298K (kcal per mole of bonds) H C N O F S Cl Br I 71 57 - 48 50 43 36 88 66 58 51 52 46
103 79 49 61 81 62 68 63
135
116 65 44 37
111 82 33 93 70 38 99 83
104
Mean (multiple) Bond Energies at 298K (kcal per mole of bonds)
C=C; 147
C=N; 147
C=O; 192
N=N; 100
C C; 200
C N; 210
C O; 256
N N; 226
O=O; 119
S=S; 100
S=O; 125
CHEM 3310

26. Using Bond Enthalpy to determine H
Bond enthalpy values:
Single Bonds: generally, H < 100 kcal/mole eg – single carbon-carbon bond H = 83 kcal/mole
Double Bonds: generally, H < 200 kcal/mole eg – double carbon-carbon bond H = 147 kcal/mole A double bond is usually not as strong as 2 single bonds.
Triple Bonds: generally, H < 300 kcal/mole eg – triple carbon-carbon bond H = 200 kcal/mole A triple bond is usually not as strong as 3 single bonds.
Usually single bonds with different atoms are stronger than those between like atoms eg – X-Y bond is stronger than X-X and Y-Y bonds
CHEM 3310

27. Using Bond Enthalpy to determine H
For an reaction that is exothermic, H < 0
“weak” bonds  “strong” bonds
For an reaction that is endothermic, H > 0
“strong” bonds  “weak” bonds
For the following reaction,
H2 (g) + Cl2 (g)  2 HCl (g) H = -44 kcal
The reaction goes from reactants with weaker bonds to the product that is formed with a stronger bond.
CHEM 3310

28. Using Bond Enthalpy to determine H
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
Example Find H for the following reaction from bond energies
2 H2 (g) + O2 (g)  2 H2O (g) H = ?
Bond breaking: 2 H-H H = 2 (104) = 208 kcal
1 O=O H = 119 kcal
Bond forming: O-H H = 4 (-111) = -444 kcal
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
ΔH = – 444
If you use this formula, remove the signs associated with bond breaking and bond forming.
ΔH = -117 kcal
Or,
Keep the signs associated with bond breaking (+) and bond forming (-) and simply add them up to get ΔH.
Eg: ΔH= ( )+(-444)=-117 kcal
CHEM 3310

29. Compared to the measured heat of combustion of propane H = -2043 kJ
Using Bond Enthalpy to determine H
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
Example Find H for the combustion of propane from bond energies
CH3CH2CH3 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) H = ?
Bond breaking: 2 C-C H = 283 = 166 kcal
8 C-H H = 899 = 792 kcal
5 O=O H = 5119= 595 kcal
Bond forming: 8 O-H H = 8(-111) = -888 kcal
6 C=O H = 6(-192) = kcal
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
ΔH =
ΔH = -487 kcal (-2038 kJ)
Compared to the measured heat of combustion of propane H = kJ
CHEM 3310

30. Using Bond Enthalpy to determine H
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
Example Show, by bond energies, that
C (graphite) + ½ O2 (g)  CO (g) H = kcal
Heat of vaporization of carbon is kcal/mole
Bond forming: 1 C≡O H = = -256 kcal
Bond breaking: C (graphite)  C (g) H = kcal
½ O=O H = ½ 119= 59.5 kcal
ΔH = ∑ ΔH(bonds breaking) - ∑ ΔH(bonds forming)
ΔH =
Bond enthalpy calculation of the reaction is close to the reported H value.
ΔH = -25 kcal
CHEM 3310

31. Variation of H with Temperature
Suppose we want to determine the enthalpy change at non-standard
Temperature, T. (To is the standard temperature)
Reactants at T  Products at T
HT
H1
H2
Reactants at To  Products at To
Ho
If To > T, then
H1 is the heat necessary to warm the reactants from T to To at constant pressure.
H2 is the products undergoing the reverse temperature change.
HT = H1 + Ho + H2
HT = Ho +
HT = Ho +
CHEM 3310

32. Variation of H with Temperature
Example
(a) Calculate the enthalpy change in the formation of 1 mole of ethanol at 25oC
by the reaction shown. Thermodynamic data are given below for each substance.
C2H4 (g) H2O (l)  C2H5OH (l)
Ho
Hfo (kcal/mole)
12.50
-57.80
-66.36
Cp (cal/mole deg)
10.41
8.02
26.64
Answer:
(a)
Ho = Hfo (ethanol) - ( Hfo (ethene) + Hfo (water))
= – ( ( ))
= kcal/mole
CHEM 3310

33. Variation of H with Temperature
C2H4 (g) + H2O (l)  C2H5OH (l)
Hfo (kcal/mole)
12.50
-57.80
-66.36
Cp (cal/mole deg)
10.41
8.02
26.64
(b) What is the H when the reactants are initially at 15.0 °C and the product, at
the end of the reaction, is at 100.°C?
Reactants at 15.0oC  Products at 100oC
HT
H1
H2
Reactants at 25.0 °C  Products at 25.0 °C
Ho (from (a)) = kcal/mole
HT = H1 + H° + H2
HT =
HT = kcal
CHEM 3310

34. ln 𝑃 = − ∆ 𝐻 𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑅 1 𝑇 + ln(A)
The Clausius-Clapeyron Equation
The vaporization curves of most liquids have similar shape.
The vapor pressure increase as the temperature increases.
Experiments showed that the pressure P, enthalpy of vaporization, Hvaporization, and temperature T are related.
p = b eaT
This is the Clausius-Clapeyron equation.
where R is the gas constant and A is a constant for the liquid-gas system.
ln 𝑃 = − ∆ 𝐻 𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑅 1 𝑇 + ln(A)
CHEM 3310

35. Compare The Clausius-Clapeyron Equation p = b eaT k = b eaT CHEM 3310
Arrhenius Equation
The Clausius-Clapeyron equation allows us to estimate the vapor pressure (P2) at another temperature (T2), if the vapor pressure (P1) is known at some temperature (T1), and if the enthalpy of vaporization, Hvap, is known.
CHEM 3310

36. The Clausius-Clapeyron Equation
The boiling points of a pure compound was
measured at two different pressures. Estimate the heat of vaporization of the compound.
p = b eaT
T (K)
P (kPa)
260
3.3
300
325
Check your answer: Hvap = 74.4 kJ/mole
If the phase transition is from solid to gas, the enthalpy of sublimation. (Hsub) is the change in enthalpy when one mole of solid vaporizes to form one mole of gas.
CHEM 3310

37. Compare The Clausius-Clapeyron Equation
Slope = - Hvap / R
Slope = -Ea / R
ln 𝑃 = − ∆ 𝐻 𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑅 1 𝑇 + ln(A)
ln 𝑘 = − ∆ 𝐸 𝑎 𝑅 1 𝑇 + ln(A)
Clausius-Clapeyron Equation
Arrhenius Equation
The Clausius-Clapeyron equation allows us to estimate the vapor pressure (P2) at another temperature (T2), if the vapor pressure (P1) is known at some temperature (T1), and if the enthalpy of vaporization, Hvap, is known.
CHEM 3310

38. The Clausius-Clapeyron Equation
The vapor pressures of ice at two temperatures are summarized below. Estimate the heat of sublimation of ice.
T (K)
P (torr)
268
2.965
273
4.560
760 torr = 1 atm
2.965 torr = atm
4.560 torr = atm
Check your answer: Hsub = J mol-1
CHEM 3310

39. H = E + PV H = nCp(T2-T1) Summary:
Change in enthalpy of a system, H, can be determined by:
H = qp
The subscript p reminds us that the heat measured is under constant pressure condition.
Relating H to E,
H = E + PV
In terms of heat capacity, assuming that Cp is constant of T1 and T2.
H = nCp(T2-T1)
For a monatomic gas, Cv = 3/2R
Cp = R + Cv
Hess’ Law
Clausius-Clapeyron equation
CHEM 3310