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Page history last edited by Xuan-Dao Nguyen 11 years, 7 months ago

Dr Coopers comments: Make sure to label and title all figures


Thien Nguyen ( Green )



The First Law:  Hess’s Law


-         It is the combined enthalpy of all the standard enthalpies of each reaction.

-         The most important thing to keep in mind before doing the calculation for Hess’s Law is that all the chemical equations must be balance.

             -         The reason why Hess’s Law is importance is that it can foretell what the enthalpy of the product is going to be from other reactions.




                                                                                                                        Illustration of Hess's Law

                                                                                                                                                                Figure from power point slide in lecture 




What is the standar enthalpy of the combustion of propene, given the enthalpy of hydrogenation of propene and the enthalpy of combustion of propane?

C3H6 + H2 à C3H8 DrHθ= -124 kJ/mol
nC3H8 + 5O2 à3CO2 + 4H2O                          DrHθ= -2220 kJ/mol
First balance the two reactions:
C3H6 + 9/2O2 à 3CO2 + 3H2O
nC3H6 + H2 +C3H8 + 5O2 àC3H8 +3CO2 + 4H2O
nC3H6 + H2 + 5O2 à3CO2 + 4H2O
delta sign required - not "D" 
nH2OàH2 + ½ O2 DrHθ= +286 kJ/mol
Then calculate the enthalpy for each balanced reaction:
nC3H6 + H2 àC3H8 DrHθ= -124 kJ/mol
nC3H8 + 5O2 à3CO2 + 4H2O                DrHθ= -2220 kJ/mol
nH2OàH2 + ½ O2                                 DrHθ= +286 kJ/mol
Finally combine the three reactions into one and add the all the enthalpies together to find the total enthalpy of the reaction.
C3H6 + 9/2O2 à 3CO2 + 3H2O              DrHθ= -2058 kJ/mol


Standard enthalpies of formation


-         It is the standard reaction enthalpy written as Delta H, (DH), and is determine by the formation of the compound from its elements in their reference states.

-         The most stable state at a specified temperature and 1 bar is called the reference state.


o       When calculating the standard enthalpies of formation keep in mind that the standard enthalpies of formation of elements in their reference states are zero because they remain at a constant phase. For example, Nitrogen gas will stay as Nitrogen gas will have an enthalpy of zero at all temperatures in their reference state.


                                   N2 (g) ---> N2 (g) subscripts?     Enthalpy H = 0





                                                                                                                                  The Reaction of enthalpy in term of enthalpies of formation 

                                                                                          Atkins Physical Chemistry, Eight Edition

                                                       Copyright 2006 Peter Atkins and Julio de Paula




The Temperature-dependence of reaction enthalpies


-         Temperatures are needed to help measure the standard enthalpies of different reactions but when there no temperatures present, heat capacities can fill in the role to help determine the value of the standard reaction enthalpies at different temperatures. This can be proven with Kirchhoff’s law in Fig. 2.19 below.


                                                                                                              Kirchhoff's Law                                


                                       Fig. 2.19

Atkins Physical Chemistry, Eight Edition

Copyright 2006 Peter Atkins and Julio de Paula




                                  Fig. 2.19. This graph shows the relation ship between the Enthalpy, H, and the Temperature, T. The products and reactants enthalpy increase as the temperature is increased, and the change in enthalpy is based on the heat capacities of the substances presence in the reaction.



-         Kirchhoff’s Law can be explain through mathematic equations.

-         The relationship between initial enthalpy, H(T1) and H(T2) can be rewritten when a substance is heated from T1 to T2 as below.






Phuong Nguyen


Path Functions and Inexact Differentials:

  • State functions is the property of a system that is not based on how it accquired in each state but only depended on its physical or chemical state. In another, path functions are the processes of the preparation of the state and dependent upon the path leading from one state to another.
  • For example 2 different people want to go up to the top of the mountain. They start from the same spot and will meet up at same place on the top of the mountain. The altitude is the state function because doesn't matter how they took to get up there, their changing in atitude will be the same. But, the distance will be the path function because the distance depends on how each person took to get up to the mountain.


  • In the figure 2-20, the internal energy, U, is the state function, there is 2 different paths that were done the different amount of work. Both paths were started at state 'i' and finished at state 'f', the only different is the path of the expansion. 

Exact and Inexact Differentials:



  • Exact differential: the change in internal energy, the integral of all the possible changes along the path from the starting state, i, to the finished state, f, will be the same.
  • Inexact differential: the change in internal energy will be different depends on the path which is used between the initial and the final points.
  • đq and đw is symbolize as path dependent.



Changes in Internal Energy:


  • The figure 2-21 describe the change in internal energy, U, at the constant temperature, T, then the final internal energy will be equal to inital internal energy plus the slope of the plot of internal energy and volume which is the partial derivative of U with respect to V.


  • The figure 2-22 describe the change in internal energy, U, at the constant volume, V, then the final internal energy will be equal to inital internal energy plus the slope of the plot of internal energy and temperature which is the partial derivative of U with respect to T.


  • The figure 2-23 describe the change in internal energy, U, at the temperature, T, and the volume, V, is freely to change. Then the final internal energy will be equal to inital internal energy plus the partial derivative of U with respect to V and the partial derivative of U with respect to T.





Xuan-Dao Nguyen (someone has overwritten your images - can you re-upload them?)


The Joule Experiment 

- Measures the interaction between molecules

- Measures the change in temperature with volume at constant internal energy

- Purpose was to measure internal pressure



This is not the correct figure 


A high pressure gas being released into a vacuum would cause an increase in temperature due to the gas being expanded.  If heat was released then there should be an increase in temperature.


Any change of internal energy is associated with heat.



Constant internal energy therefore NO heat (q= 0) and NO work (w = 0)




No recognizable temperature change, due to the lack of a precise thermometer

The heat capacity of the whole system was too large to see a change in heat.




The Joule -Thomson Effect

-The change of a gas or liquid when forced through a porous barrier while insulated to prevent the exchange of heat to the environment (adiabatic expansion).

- Important in the liquefication of gases




 Joule - Thomson coefficient

               - the rate change of temperature with respect to pressure


               H = constant enthalpy









Comments (1)

Xuan-Dao Nguyen said

at 2:50 pm on Dec 18, 2009

Hey Professor, I'll upload the pictures again tonight. Thanks

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