1. First Law of Thermodynamics
Doba Jackson, Ph.D.
Dept of Chemistry & Biochemistry
Huntingdon College

2. Outline of Chapter 2 Basic Concepts: Heat, Work, Energy
First Law of Thermodynamics
Expansion work
Measurement of Heat
Enthalpy changes
Adiabatic changes (Cooling)
Thermochemistry
Inexact Differntials

3. System, Surroundings, Universe
System- Part of the world of interest
Surroundings- Region outside the system
Open system- Allows matter and energy to pass
Closed system- Cannot allow matter to pass
Isolated system- Cannot allow matter or energy to pass

5. Work
Work- the amount of energy transferred by motion against an opposing force ▪

7. Heat, and Heat Transfer Heat- is the transfer of energy from one body
to another by thermal contact
Heat can be transferred by
Convection
Conduction
Radiation

8. Internal Energy: Energy that exist internal to the system
First Law: The internal energy of an isolated system is a
constant; Energy cannot be created or destroyed.
ΔU= change in internal energy
q = heat (released or absorbed)
w = work done on or by the system
Internal energy can have different forms:
Translational Kinetic Energy
Rotational Kinetic energy
Vibrational Kinetic energy
*Energy within the atom

9. Internal Energy of a monoatomic gasRT
Monoatomic gas RT
Linear molecule RT
Non-Linear molecule RT RT

10. Expression for Work Change equation to relate pressure and volume
True expression for Work

11. Types of Expansion Work
Free Expansion- work done by expansion against zero opposing force or zero pressure

12. Types of Expansion Work
Expansion against a constant external pressure (irreversible)
If external pressure does, not vary,
this expression can be integrated

13. Types of Expansion Work
Expansion is reversible
In this case, the external pressure Pex
exactly equals the internal pressure P
at each stage of the expansion.
The integral cannot be evaluated because
temperature is not a constant throughout the
integration.

14. Types of Expansion Work
Expansion is Isothermal, reversible
The system is kept at a constant temperature.
The external pressure Pex equals the internal
pressure P at each stage of the expansion.
Moles is always assumed
constant unless otherwise noted.

15. Problems: Calculate the work needed for a 65 kg person to climb up 4
Problems: Calculate the work needed for a 65 kg person to climb up 4.0 meters on (a) the earth, (b) the moon (g=1.60 m/s2)

16. Problem 1: A chemical reaction takes place in a container of cross sectional area of 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an constant external pressure of 121 kPa. Calculate the work (in Joules) done by the system

17. Problem 2: A sample consisting of 2
Problem 2: A sample consisting of 2.00 mol of He is expanded at 22ºC from 22.8 dm3 to 31.7 dm3. Calculate q, w and ΔU (in Joules) when the expansion occurs under the following conditions:   (a) Freely   (b) Isothermal, and constant external pressure equal to the final pressure of the gas   (c) Isothermal, Reversibly  

18. Reversible, Nonspontaneous
Reversible Process: The thermodynamic process in which a system
can be changed from its initial state to its final state then back to its initial
state leaving all thermodynamic variables for the universe (system + surroundings) unchanged.
A truly reversible change will:
- Occur in an infinite amount of time
- All variables must be in equilibrium with each other at every stage of the change
Reversible
Irreversible

19. Irreversible, Spontaneous
Irreversible Process: The thermodynamic process in which a system
that is changed from its initial state to its final state then back to its initial
state will change some thermodynamic variables of the universe.
A truly irreversible change will:
- Occur in an finite amount of time
- All variables will not be in equilibrium with each other at every stage of the change
Reversible
Irreversible

20. Heat exchange can be measured in a Bomb Calorimeter
First Law: The internal energy of an isolated system is a constant.
Constant Volume: no work done

21. Calorimetry is the study of heat transfer during a chemical or physical process
Heat is measured by the
change in temperature of
the water surrounding the
calorimeter
Heat
The calorimeter constant is the
heat capacity of the system

22. Heat Capacity at a constant volume
Heat Capacity (Cv) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant volume.
We need partial derivatives
because the change holds
the volume constant

23. Heat Capacity at a constant volume
We need partial derivatives because
the change holds the volume constant

24. How we can evaluate the Heat capacity equation
If the heat capacity (Cv) does not vary with
temperature during the range of temperatures
used, then the equation can be broken
down to:
Because at constant volume, change in internal
energy is equal to the heat (qv).

25. Problem 2.4: A sample consisting of 1.00 mol of
a perfect gas, for which Cv,m = (3/2)R, initially
at P = 1.00 atm and T = 300 K, is heated reversibly
to 400 K at a constant volume. Calculate the
final pressure, ΔU, q, and w

26. We cannot measure internal energy when the volume is not constant
Typical reactions occur a constant
external pressure

27. Heat exchange can be measured in a Differential Scanning Calorimeter at a constant pressure
First Law: The internal energy of an isolated system is a constant.
Lets define Enthalpy as:

28. Heat Capacity at a constant pressure
Heat Capacity (Cp) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant pressure.
We need partial derivatives because
the change holds the pressure constant

29. How we can evaluate the Heat capacity equation
If the heat capacity (CP) does not vary with
temperature during the range of temperatures
used, then the equation can be broken
down to:
Because at constant volume, change in
internal energy is equal to the heat (qv).

30. Problem 2.20: When 2.25 mg of anthracene, C14H10 (s) was burned in a bomb calorimeter, the temperature rose by 1.35 K. Calculate the calorimeter constant for the system. By how much will the temperature rise if 135 mg of phenol (C6H5OH) is burned in the calorimeter under the same conditions?
(ΔcHΘ(C14H10) = kJ/mol)

31. Adiabatic Expansion: work is done but no heat enters the system
When a gas expands adiabatically, work is done but no heat enters or leaves the system.
The internal energy falls and the temperature also falls.

32. Adiabatic changes to an Ideal Gas can be related by the following equations
C = CV/nR
γ = Cp/CV
The derivation of these equations is
quite long and follows the next slide

33. Adiabatic Expansion of an Ideal Gas (Derivation)
Assume the heat capacity does not
vary during the temperature interval
For an adiabatic change: q = 0

34. Problem 2.1- A 3.75 mole sample of an ideal gas with Cv,m = 3R/2 initially at a temperature Ti=298 K, and Pi= 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 725 kg mass on the piston of diameter 25.4 cm. Calculate the work done in this process and the distance the piston travels. Assume the mass of the piston is negligible.

35. Thermochemistry The study of energy transferred as heat during the
course of chemical reactions.
In thermochemistry, we rely on calorimetry to measure
the internal energy (∆U) enthalpy (∆H).
Changes in matter are:
Physical Changes
Solid  Liquid Melting (Fusion)
Liquid  Gas Boiling (Vaporization)
Solid  Gas Sublimation
Chemical Changes
Formation reactions
Combustion reactions
Other reactions

36. The Thermodynamic Standard State
Chapter 8: Thermochemistry: Chemical Energy
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The Thermodynamic Standard State
The standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. The International Union of Pure and Applied Chemistry (IUPAC) recommends using a standard pressure po = 1 bar (100 kilopascals).
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g)
∆H = kJ
3CO2(g) + 4H2O(l)
C3H8(g) + 5O2(g)
∆H = kJ
The conversion from liquid to gas phases requires energy. Therefore, it’s important that the states are specified.
Putting the “°” superscripted next to the ∆H means standard state.
The standard pressure is actually 1 bar. The enthalpy difference is usually small, however.
Thermodynamic Standard State: Most stable form of a substance at 1 bar and at a specified temperature, usually 25 °C (for non-solutions).
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g)
∆H° = kJ
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37. Physical and Chemical Transitions
Chapter 8: Thermochemistry: Chemical Energy
Physical and Chemical Transitions
4/25/2017
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38. State Functions Internal energy and enthalpy are is a state function.
Work and Heat are path
function. This means we can
never write Δq or Δw

39. Enthalpies of Physical Change
Chapter 8: Thermochemistry: Chemical Energy
Enthalpies of Physical Change
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Hess’s Law
Copyright © 2010 Pearson Prentice Hall, Inc.

40. Chapter 8: Thermochemistry: Chemical Energy
Example of Hess’s Law
Chapter 8: Thermochemistry: Chemical Energy
4/25/2017
Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.
Germain Hess
Example 1:
The two reactions below are known methods to produce ammonia by the
Haber process. Use the information to solve for the enthalpy change in the
reaction below.
Hydrazine, N2H4, is a reactive intermediate (more on that in Ch. 12: Chemical Kinetics). Chemical reactions are far more complicated than discussed in Ch. 6 (stoichiometry).
The overall reaction and step 2 can be measured, but step 1 cannot be measured easily.
A fascinating book about Haber is:
Charles, Daniel; Master Mind: The Rise and Fall of Fritz Haber, The Nobel Laureate Who Launched the Age of Chemical Warfare; HarperCollins Publishers Inc., 2005.
2NH3(g)
3H2(g) + N2(g)
∆H°1 = kJ
2NH3(g)
N2H4(g) + H2(g)
∆H°2 = kJ
Solve for the reaction below
N2H4(g)
2H2(g) + N2(g)
∆H°3 = ?
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41. Example 1: 2NH3(g) 3H2(g) + N2(g) 2NH3(g) N2H4(g) + H2(g) 2NH3(g)
Chapter 8: Thermochemistry: Chemical Energy
Example 1:
The two reactions below are known methods to produce ammonia by
the Haber process. Use the information to solve for the enthalpy change
in the reaction below.
4/25/2017
2NH3(g)
3H2(g) + N2(g)
∆H°1 = kJ
2NH3(g)
N2H4(g) + H2(g)
∆H°2 = kJ
Steps to solve the problem
- Find known reactions that will add up to the desired reaction
- Add up the reactions and add up the ΔH, ΔU, ΔG or other state functions 2
2NH3(g)
3H2(g) + N2(g)
∆H°1 = kJ
∆H°2 = kJ
Solve for the reaction below
∆H°3 =
-92.2 kJ + (187.6 kJ)
= 95.4 kJ
N2H4(g)
2H2(g) + N2(g)
Copyright © 2010 Pearson Prentice Hall, Inc. 41

42. Chapter 8: Thermochemistry: Chemical Energy
Hess’s Law Example
4/25/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

43. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
4/25/2017
Standard Heats of Formation
Standard Heat of Formation (∆fH° ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.
Standard states
CH4(g)
C(s) + 2H2(g)
∆H°f = kJ
1 mol of 1 substance
Students need to remember solid, liquid, gases, and diatomics as discussed previously.
Copyright © 2010 Pearson Prentice Hall, Inc.

44. Chapter 8: Thermochemistry: Chemical Energy
4/25/2017
Standard Heats of Formation are used to determine reaction enthalpies (ΔrHº)
Reaction enthalpies (ΔrHº) can be determined by the difference of the product enthalpies of formation and
the reactant enthalpies of formation.
Using table values.
cC + dD
aA + bB
Example:
∆rH° = {c ∆fH°(C) + d ∆fH°(D)} – {a ∆fHº(A) + b ∆fH°(B)}
Products
Reactants
Copyright © 2010 Pearson Prentice Hall, Inc.

45. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
4/25/2017
Standard Heats of Formation
Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.
C6H12O6(s) + 6*O2(g)
6*CO2(g) + 6*H2O(l)
∆rH° = ?
Standard State
(ΔfHº)= 0
Answer
ΔrH° = [∆H°f (C6H12O6)] - [6*∆H°f (CO2) + 6*∆H°f (H2O)]
Using table values.
The units for standard enthalpies of formation are kJ/mol while the unit for the standard enthalpy change is kJ.
Result is endothermic.
Products
Reactants
∆rH° = [(1 mol)( kJ/mol)] -
[(6 mol)( kJ/mol) + (6 mol)( kJ/mol)]
= kJ
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46. Heats of Combustion (ΔcHº) is a special type of reaction enthalpy
Chapter 8: Thermochemistry: Chemical Energy
Heats of Combustion (ΔcHº) is a special type of reaction enthalpy
4/25/2017
Calculate the standard enthalpy of combustion for methane (CH4(g)).
Answer
CO2(g) + H2O(l)
CH4(g) + O2(g) 2 2
∆cH°= [∆fH°(CO2) + 2 ∆fH°(H2O)] - [∆fH°(CH4)]
= [(1 mol)( kJ/mol) + (2 mol)( kJ/mol)] -
ΔfHº (CH4) = kJ/mol
ΔfHº (CO2) = -393 kJ/mol
[(1 mol)(-74.8 kJ/mol)]
= kJ
ΔfHº (H2O) = -285 kJ/mol
Combustion reactions are always balanced with 1 as the coefficient
of the hydrocarbon
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47. Temperature dependence of reaction enthalpy (ΔrHº)
Often standard reaction enthalpies need to be corrected for temperature differences.
If Cp is independent of temperature during the
temperature range, the integral can be evaluated.
The equation doesn’t work if there is a phase transition
through the temperature range.

48. Problem: Balance the reactions and determine ΔrHΘ and ΔrUΘ
NO: kJ/mol
NH3: kJ/mol
H2O: kJ/mol