1. 1ST LAW THERMODYNAMICS
2nd part
ROHAZITA BT BAHARI

2. First Law Thermodynamic Basic concept: Work, heat and energy
The internal energy
Expansion work
Heat transaction
Adiabatic changes
Thermochemistry:
Standard enthalpy changes
Standard enthalpies of formation
The temperature dependence of reaction enthalpies.
First Law
Thermodynamic
State of function and exact differentials:
Exact and inexact differentials
Changes in internal energy
The Joule – Thomson effect

3. THERMOCHEMISRTY

4. Thermochemistry: Basic Terms
Thermochemistry is the study of energy changes that occur during chemical reactions.
System: the part of the universe being studied.
Surroundings: the rest of the universe.

5. Types of systems: open (exchange of mass and energy)
closed (exchange of energy)
isolated (no exchange)
Open system
Close System
Isolated system

6. A closed system; energy (not matter) can be exchanged.
Types of Systems
Open: energy and matter can be exchanged with the surroundings.
Closed: energy can be exchanged with the surroundings, matter cannot.
Isolated: neither energy nor matter can be exchanged with the surroundings.
After the lid of the jar is unscrewed, which kind of system is it?
A closed system; energy (not matter) can be exchanged.

7. Enthalpy (H) Enthalpy can be defined as:
Enthalpy is a state function (values depend on current state of the system, not on HOW the system acquired that state, which is independent of path.)
Volume
Internal energy
Pressure

8. ΔH = Hproduct - Hreactant
ENTHALPY (H)
The energy possessed by a system is called the enthalpy or heat content of the system and is given the symbol H.
The change of the heat content is given by ΔH,
ΔH = Hfinal - Hinitial
ΔH = Hproduct - Hreactant

9. Heat (q) Heat is energy transfer resulting from thermal
differences between the system and surroundings
“flows” spontaneously from higher T  lower T
“flow” ceases at
thermal equilibrium
EOS

10. Enthalpy(H) vs heat (q)
When reactions take place at constant pressure;
heat change = enthalpy change
q = ΔH
The quantity of heat energy liberated or absorbed during a reaction taking place at constant pressure is the enthalpy change of the reaction.

11. Exothermic reaction Eg : C(s) + O2(g) CO(g)
A chemical reaction that releases/gives off heat
is called an exothermic reaction
During an exothermic process, heat flows out of the system and into the surrounding. Thus the energy of the system is negative.
ΔH = -ve
ΔH = kJ/mol
Eg : C(s) + O2(g) CO(g)

12. Maximum temperature rise.
Exothermic Reaction
Surroundings are at 25 °C
Typical situation: some heat is released
to the surroundings,
35.4 °C
32.2 °C
some heat is absorbed by the solution.
25 °C
Hypothetical situation: all heat is instantly released to the surroundings. Heat = qrxn
In an isolated system, all heat is
absorbed by the solution.
Maximum temperature rise.

13. Endothermic reaction
A chemical reaction that absorbs/takes in heat
is called an endothermic reaction.
During an endothermic process, heat flows into the system from the surrounding.
Thus the energy of the system is positive.
ΔH = +ve

14. Enthalpy (constant volume)
Consider a constant-volume process:
For an infinitesimal change,
Constant volume

15. Internal Energy Change at Constant Volume
For a system where the
reaction is carried out at
constant volume, V = 0 and U = qV.
All the thermal energy produced by conversion from chemical energy is released as heat; no P- V work is done.

16. Enthalpy (constant pressure)
Energy transferred as heat at constant pressure is equal to the change in enthalpy of the system.
HOW?

17. Enthalpy (constant pressure)
Rearranging,
For an infinitesimal change,
For a measurable change

18. internal Energy Change at Constant Pressure
For a system where the reaction is carried out at constant pressure,
U = qP – PV or
U + PV = qP
Most of the thermal energy is released as heat.
Some work is done to expand the system against the surroundings (push back the atmosphere).

19. Enthalpy and Enthalpy Change
The evolved H2 pushes back the atmosphere; work is done at constant pressure.
Enthalpy is the sum of the internal energy and the pressure-volume product of a system:
H = U + PV
For a process carried out at constant pressure,
qP = U + PV
Mg + 2 HCl  MgCl2 + H2
Most reactions occur at constant pressure, so for most reactions, the heat evolved equals the enthalpy change. so
qP = H

20. Enthalpy Diagrams Values of H are measured experimentally.
Negative values indicate exothermic reactions.
Positive values indicate endothermic reactions.
An increase in enthalpy during the reaction; H is positive.
A decrease in enthalpy during the reaction; H is negative.

21. Thermochemical Equations
A themochemical equation is an equation that represents a reaction and shows the overall enthalpy changes during the reaction.
The following are three important rules for writing thermochemical equations:
The physical states (s, l, g) of all reactants
and products must be specified.
2H2O(l) 2H2O(g) ΔH= 88.0 kJ/mol

22. the value of ΔH is reversed in sign
2. When both sides of a thermochemical equation is multiplied by a factor, then ΔH must also be multiplied by the same factor.
H2O(s)
2H2O(s)
H2O(l) ΔH= 6.01 kJ/mol
2H2O(l) ΔH= kJ/mol
3. When a chemical equation is reversed,
the value of ΔH is reversed in sign

23. Reversing a Reaction H changes sign when a process is reversed.
Same magnitude; different signs.

24. How to measure enthalpy change
1. DIRECT METHOD ( through experiment)
Calorimetry
Heat absorb or evolve is measured using a
calorimeter
Two types of calorimeter :
Coffee-cup calorimeter
Bomb calorimeter

25. Calorimetry Calorimetry is a technique used to measure heat
exchange in chemical reactions
Calorimetry – measuring the heat flow associated with a chemical reaction by measuring the temperature.
EOS

26. Calorimetry We measure heat flow using calorimetry.
A calorimeter is a device used to make this measurement.
A “coffee cup” calorimeter may be used for measuring heat involving solutions.
A “bomb” calorimeter is used to
find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned.

27. Calorimetry, Heat Capacity, Specific Heat
Heat evolved in a reaction is absorbed by the calorimeter and its contents.
In a calorimeter we measure the temperature change of water or a solution to determine the heat absorbed or evolved by a reaction.
The heat capacity (C) of a system is the quantity of heat required to change the temperature of the system by 1 °C.
C = q/T (units are J/°C)
Molar heat capacity is the heat capacity of one mole of a
substance.
The specific heat (s) is the heat capacity of one gram of a pure substance (or homogeneous mixture).
s = C/m = q/(mT)
q = s m T

28. Heat Capacity: A Thought Experiment
Place an empty iron pot weighing 5 lb on the
burner of a stove.
Place an iron pot weighing 1 lb and containing 4 lb water on a second identical burner (same total mass).
Turn on both burners. Wait five minutes.
Which pot handle can you grab with your bare hand?
Iron has a lower specific heat than does water. It takes less heat to “warm up” iron than it does water.

29. specific heat capacity
The specific heat capacity (or “specific heat”) is the heat required to raise the temperature of 1 gram of a substance by 1 oC.
Specific heat capacity of water :
4.184 J /(g oC).
4.184 J energy is needed to raise the
temperature of 1 g of water b 1 oC

30. How does a calorimeter measure heat (q)
Heat evolved in a reaction is absorbed by the calorimeter and its contents (water or solution).
In a calorimeter the temperature change of water or a solution is measured to determine the heat absorbed or evolved by a reaction.

31. TO CALCULATE q q = m c T q is calculated using
q = mass x specific heat capacity x T
q = m c T
If T is positive (temperature increases), q is positive and heat is gained by the system.
If T is negative (temperature decreases), q is negative and heat is lost by the system

32. Calculation of q - example
Calculate the heat absorbed when the temperature of
15.0 grams of water is raised
from 20.0 oC to 50.0 oC.
(The specific heat of
water is J/g.oC.)
using the equation q = mc T
T = (50oC – 20oC ) = 30oC, therefore q = x 15 x 30 = J
= 1.88 kJ

33. Calculating enthalpy using direct method
When 23.6 grams of calcium chloride, CaCl2, was dissolved in 500 mL water in a calorimeter, the temperature rose from 25.0 oC to 38.7 oC.
Given the heat capacity of water is 4.18 J/g oC, and density of water is 1 g/mL, what is the enthalpy change per mole of calcium chloride?

34. Heats of Reaction: Calorimetry
– First, calculate the heat absorbed by the
calorimeter.
q = mc T
q= 500 g x 4.18 J/g oC x ( ) oC
q= J = kJ
Now we must calculate the heat per mole of calcium chloride.

35. Heats of Reaction: Calorimetry
Calcium chloride has a molecular mass of
111.1 g, so
(1 mol CaCl2 )
(23.6 g CaCl2 ) 
 mol CaCl2
111.1 g
Now we can calculate the heat per mole of calcium chloride.
qrxn
H 
  kJ   kJ / mol
mol CaCl2
0.212 mol

36. Calculating ΔH θ by calorimetry n
solution = 4.18J/(K g). Calculate the standard molar
enthalpy of neutralization.
1. calculate heat change using
25.0 cm3 of 1.00 M HCl at 21.5oC were placed in a polystyrene cup and 25.0 cm3 of 1.00 M NaOH at 21.5oC was added. The mixture was stirred, and the
temperature rise to 28.2oC.
The density of each
q  mcT

37. q= mct = 50.0 g x 4.18 J/(K g) x 6.7 OC= 1400 J = 1.4 kJ
2. Calculate heat change per mole (ΔHnθ )
mole of H2O formed in the reaction=
25.0 x10-3 dm3 x 1.0 M = 2.5 x10-2 mol
2.5 x10-2 mol H2O formed = 1.4 kJ
1 mol H2O formed =
1 x 1.4
2.5 x10-2 mol
= 56 kJ/mol

38. 2. Calculating enthalpy using data from
Standard Enthalpies of Formation
The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.
Ho  nHo (products) 
 
mHo (reactants) f f
  is the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation.

40. Using ΔHf to calculate enthalpy of reactionθ
Using ΔHf to calculate enthalpy of reaction
Calculate the enthalpy of reaction for the following reaction:
2Al(s)
+ 2 Cr(s)
+ Cr2O3(s) Al2O3(s)
given ΔHfθ (Cr2O3(s)) = kJ/mol
ΔHfθ (Al2O3(s)) = kJ/mol

41. Using ΔHf to calculate ΔH rxn
θ θ
Using ΔHf to calculate ΔH rxn
2Al(s)
+ 2 Cr(s)
+ Cr2O3(s) Al2O3(s)
ΔHfθ = 0
ΔHfθ = ΔHfθ = ΔHfθ =0
θ = Σ of ΔH of products - Σ of ΔH θ of reactants ΔHrxn fθ f
= [2(0) + (-1128)] – [(-1669) + 2(0)]
= 541 kJ/mol

42. Hess’s Law
Hess’s law states the enthalpy change of a reaction is constant whether the reaction is carried out directly in one single step or indirectly through a number of steps.
ΔH1θ
System A
System B
ΔH4θ
ΔH2θ
ΔH3θ
System A’
System A’’

43. Example 1 : Hess’s law H2(g) I2(s) I2(s) H2(g) I2(g)
Hydrogen iodide can be prepared from hydrogen and
iodine using two separate routes.
Route I:
Eqn 1:
Route II: +
→2HI(g)
∆H = kJ
H2(g) I2(s)
eqn 2:
eqn 3: → +
∆H= kJ
∆H= -9.2 kJ
I2(s)
H2(g)
I2(g)
→ 2HI(g)

44. Eqn 1 can be obtained when eqn 2 is added to eqn 3;. eqn 1 = eqn 2
Eqn 1 can be obtained when eqn 2 is added to eqn 3; eqn 1 = eqn 2 + eqn 3
eqn 2:
eqn 3:
I2(s)
H2(g)
∆H= kJ
∆H= -9.2 kJ
→ I2(g) +
→ 2HI(g)
→ 2HI(g
I2(g)
I2(s)
Hence ∆H1 = ∆H2
+ ∆H3
= 61.3 – 9.2 = kJ
The total enthalpy change for the route I is the same as that for route II.

45. Hess’s Law S(s)  O (g)  SO (g); Ho  -297 kJ
Example 2; ΔH for formation of SO3 cannot be obtained directly but the enthalpy of these reactions are known:
S(s)  O (g)  SO (g); Ho  -297 kJ
2 2
2SO (g)  2SO (g)  O (g); Ho  198 kJ
3 2 2
The above data can be used to obtain the enthalpy change for the formation of SO3 according to the following reaction?
2S(s)  3O2 (g)  2SO3 (g);
Ho  ?

46. 2S(s)  2O2 (g)  2SO2 (g); H  (-297 kJ) (2)
The third equation can be obtained by multiplying the first equation by 2 and added to the reverse of the second equation, they will sum together to become the third. o
2S(s)  2O2 (g)  2SO2 (g); H  (-297 kJ) (2)
2SO2 (g)  O2 (g)  2SO3 (g); H  (198 kJ) (-1)
2S(s)  3O (g)  2SO (g); Ho  -792 kJ
2 3

47. Fe2O3(s) + FeO(s) Fe3O4(s) exercise Fe3O4(s)
Determine the heat of reaction;
Fe2O3(s) + FeO(s) Fe3O4(s)
Using the information below:
2Fe(s) + O2(g)
4Fe(s) + 3O2(g)
3Fe(s) + 2O2(g)
2FeO(s) ∆Ho = kJ
2Fe2O3(s) ∆Ho = kJ
∆Ho = kJ
Fe3O4(s)
(Answer : kJ)

48. Thermochemical cycle / energy diagram
Reactant ( A + B)
route 2
∆Ho 2
route 1 ∆Ho
∆Ho 3
route 3
Product (C + D )

49. Enthalpy diagram illustrating
Hess’s law.

50. Hess’s Law: An Enthalpy Diagram
We can find H(a) by
subtracting H(b) from H(c)

51. Born Harber cycle
Born Haber cycle is an energy cycle used to calculate the lattice energy which cannot be obtained by direct experimental method. They only can be obtained by applying Hess’s law in this cycle which involving breaking and forming bonds.

52. Na+(g) Al2O3(s) + 3O2 (g) Lattice energy
The enthalpy change when one mole of ionic compound (crystalline substance) is formed from its gaseous ions.
+ Cl-(g)
NaCl(s)
Al2O3(s)
∆Hlat = -788 kJ/mol
∆Hlat = kJ/mol
Na+(g)
2Al3+(g) -
+ 3O2 (g)

53. ii) Enthalpy of atomization•
The enthalpy change when one mole of gaseous atoms is formed from its elements under standard conditions •
Na(s)
½ O2(g)
Na(g)
O(g)
H = +108 kJ/mol
H = +247 kJ/mol
iii) Electron Affinity •
The enthalpy change when one mole of gaseous atom gains
one mole of electron to form anion.
+ e- O-(g)
H = kJ/mol
(1st electron affinity)
H = kJ/mol
(2nd electron affinity)
O(g)
+ e-
O-(g)
O2-(g)

54. Mg(g) Mg+(g) Mg+(g) iv) Ionization Energy
The standard ionization energy is the enthalpy required to remove one mole of electrons from one mole of gaseous metallic atoms to form one mole of positively charged ions. The process is endothermic because energy is absorbed to release electrons from an atom.
- e-
Mg(g)
Mg+(g)
Mg+(g)
Mg2
+(g)
(1st ionization energy)
(2nd ionization energy)

56. Born haber cycle for LiF

57. Calculate the lattice energy for NaCl(s) using the following data.
standard enthalpy of formation of NaCl(s) ∆Hf = -411kJ/mol
standard enthalpy of atomization of Na(s) ∆Ha = +108 kJ/mol
c. first ionization energy of Na(s)
∆HIE
= kJ/mol
d. standard enthalpy of atomization of Cl2(s) ∆Hf
= +121 kJ/mol
first electron affinity of Cl2(g)
Standard lattice enthalpy of NaCl(s)
∆HEA
∆Hlat
= -364 kJ/mol
= x kJ/mol

59. Measuring Enthalpy – Kirchhoff’s Law
The standard enthalpy of many important reactions have been measure at Different temperatures.
However, in the absence of this information standard reaction enthalpies at different temperature may be calculated from heat capacities and reaction enthalpy at some at some other temperature.
When temperature is increased, the enthalpy of the products and reactants both increase, but may do so to different extent.
In each case, the change in enthalpy depends on the heat capacities of the substances.
The change in reaction enthalpy reflects the difference in the changes of the enthalpy.

60. When a substance is heated from T1 to T2, its enthalpy changes from H(T1) to
Eqn 1
Assumed that no phase transition take place in the temperature range of interest) because this equation applies to each substance in the reaction the
Kirchhoff’s Law
standard reaction, the standard reaction enthalpy changes from ∆Hr (T1) to ɵ
Eqn 2
Where the is the difference of the molar heat capacities of product and Reactants under standard conditions weighted by the stoichiometri coefficient that Appear in chemical equation:
Eqn 3

61. Example:
The standard enthalpy of formation of H2O (g) at 298K is kJ mol-1. Estimate its value at 100oC given the following values of the molar heat capacities At constant pressure:
H2O (g) = 33.58JK-1mol-1 H2(g) = JK-1mol-1 O2 = JK-1 mol-1
Assume that the heat capacities are independent of temperature.
Method: When
is independent of temperature in the range T1 to T2, the
Integral in eqn 2 evaluate to (T2-T1)
. Therefore,
To proceed, write the chemical equation, identify the stoichiometri coefficients, and
Calculate from the data.

62. Answer: The reaction is H2 (g) + ½ O2 (g) → H2(g), so
It then follows that
Answer: The reaction is H2 (g) + ½ O2 (g) → H2(g), so

63. State Function
and
Exact Differentials

64. The internal energy and enthalpy are two example of state functions.
Physical quantities that do depend on the path between two state are called
path function.
Example of path functions are the work and the heating that are done when
preparing a state.
We do not speak of a system in a particular state as possessing work or heat.
In each case, the energy transferred as work or heat relates to the path being
taken between states, not the current state itself.
A part of the richness to thermodynamic is that it uses the mathematical properties of state functions to draw far-reaching conclusions about the relations between physical properties and thereby establish connections that may be completely unexpected.
The practical importance of this ability is that we can combine measurements of different properties to obtain the value of a property we required.

65. Exact and Inexact Differentials
Initial state of the system
Work is done by the system as it expands adiabatically to a state f.
In these state the system has an internal energy Uf and the work done on the system along Path 1 from I to f of is w.
Initial and final state are the same as those in Path 1 but in which expansion is not adiabatic.
Initial and final states are the same as before (because U is the state funtion).
U: property of the state, w=property of the path
However, in the second path an energy q’ enter the system as heat and the work w’ is not the same as w. The work and the heat are path functions.

66. If a system is taken along a path (for example, by heating it) U changes from Ui
to Uf, and overall change is the sum (integral) of all the infinestimal changes along the path:
Eqn 4
The value of ∆U depends on initial state and final state of the system but is
independent of the path between them.
An exact differential is an infinestimal quatity that when integrated gives a result
that is independent of the path between the initial and final state.
When the a system is heated the total energy transferred as heat is the sum of all Individual contributions at each point of the path:
Eqn 5
Inexact differential is an infinitesimal quantity that, when integrated gives a result that depends on the path between the initial and final state.

67. The differences between Eqn 4 and Eqn 5:
We do not write ∆q because q is not a state function and energy supplied as heat cannot be expressed as qf-qi.
We must specified the path of integration because q depends on the path selected (example: and adiabatic path has q=0, whereas on the non-adiabatic path between the same two state would have q≠0).
In general, an inexact differential is an infinestimal quantity that, when integrated, gives a result that depends on the path between the initial and final state.
The work done on a system to change it from one state to another depends on the path taken between the two specified states.
example: in general the work is different if the change takes place adiabatically and non-adiabatically. It folows that dw is an exact differential.

68. Changes in Internal Energy
In internal energy U can be regarded as a function of V,T, and p, but because there is an equation of state, stating the values of two of the variables fixes the value of the third.
Therefore, it is possible to write U in terms of just two independent variables:
V and T, p and T, or p and V.
Expressing U as a function of volume and temperature fits the purpose of our discussion.

69. General expression for a change in U with T and V
Eqn 6
Definition of internal pressure
Eqn 7
,eqn 5 can now write as
Eqn 8
In terms of the notification Cv and

70. The Joule Experiment
James Joule thought that he could measure by observing the chenge in Temperature of a gas when it is allowed to expand into vacuum.
When the stopcock opened and
the air expanded into a vacuum.
Filled with air at about 22 atm
No change in temperature
A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expands isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath.

71. The implication as follow:
No work was done in the expansion into a vacuum, so w=0.
No energy entered of left the system (the gas) as heat because the temperature of the bath did not change, so q = 0. Consequently, within the accuracy of the experiment, ∆U=0.
Joule concluded that:
U does not change when a gas expand isothermally and therefore that =0.
His experiment, how ever is crude.
The heat capacity of the apparatus was so large that the temperature change that gases do in fact cause was too small to measure.
Nevertheless, from his experiment Joule had extracted an essential limiting property of a gas, a property of a perfect gas, without a small deviations characteristic of a real gas.

72. Changes in Internal Energy at Constant Pressure
As an example, suppose we want to find out how the internal energy varies with temperature when the pressure rather than volume of the system is kept contant.
If we divided both sides of eqn 8 by dT and impose the condition of constant pressure on the resulting differentials, so that du/dT on the left becomes (∂U/ ∂T)p we obtain:
Eqn 9
It is usually in thermodynamic to inspect the output of a manipulation like this to see if it contains any recognizable physical quantity. The partial derivative on the right in this expression is the slope of the plot of volume against temperature (at constant pressure). This property is normally tabulated as the expansion coefficient, α, of a substance, which is defined as:
Definition of the expansion coefficient
Eqn 10

73. A large value of α means that the volume of the sample respond strongly to
changes in temperature.
List some experimental values of α

74. Isothermal compressibility, KT( kappa) which define as
Eqn 11
Isothermal compressibility is a measure of the fractional change in a volume when the pressure is increase by a small amount; the negative sign in definition ensure that the compressibility is positive quantity; because an increase of pressure, implying a positive dp, bring about reduction of volume, a negative dV.

75. The Joule-Thompson Effect
The closed system of constant composition
Eqn 12
Where the Joule-Thompson coefficient, μ (mu) is define as
Eqn 13
This relation will prove useful for relating the heat capacities at constant volume and for a discussion of liquefaction of gases.

76. Observation of the Joule-Thompson Effect
The analysis of the Joule-Thompson coefficient is central to the technological problems associate with the liquefaction of gases. We need to be able to interpret its physical and measured it.
The apparatus used to for measuring The Joule-Thompson effect. The gas Expands through the porous barrier, Which acts as a throttle, and the whole apparatus is thermally insulated.
This arrangement corresponds to an Isenthalpic expansion
(expansion at constant enthalpy). Whether the expansion results in a heating
or a cooling of the gas depends on the conditions

77. Definiton of the isothermal Joule- Thompson coefficient
The isothermal Joule –Thompson coefficients is the slope of the enthalpy with respect to changing pressure the temperature being held constant.
By compressing Eqn 12 and above equation, we can see that the two coefficients
are related by

78. A schematic diagram of the apparatus used for measuring the isothermal Joule-Thompson coefficient. The electrical heating required to offset the cooling arising from expansion is interpreted as ∆H and used to calculate , which is then converted to μ.

79. Real gases have nonzero Joule-Thompson coefficient
Real gases have nonzero Joule-Thompson coefficient. Depending on the identity of the gas, pressure, the relative magnitudes of the attractive and repulsive intermolecular forces, and the temperature, the sign of the coefficient may be either positive or negative.
+ve sign → dT is –ve when dp is –ve
(in which case the gas cool on expansion)
Heating effect (μ<0) at one temperature show at cooling effect (μ>0) when the temperature is below their upper inversion temperature,T1.
A gas typically has two inversion temperatures, one at high temperature and the other at low.

80. The principle of Linde refrigerator
The gas at high temperature is allowed to expand through a throttle, it cools
and is circulated past the incoming gas.
That gas is cooled, and its subsequent expansion cools it still further.