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LECTURE 6

This version was saved 14 years, 6 months ago View current version     Page history
Saved by tngn@gmu.edu
on September 28, 2009 at 11:47:37 pm
 

 

 

Thien Nguyen ( Green ) ** Not complete, still missing some figures and graphs.**

 

 

Standard enthalpies of formation

 

-         It is the standard reaction enthalpy written as Deltaf H, 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)      Enthalpy H = 0

 

 

 

 

 

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.

                                    

                                  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.

************ NOT DONE YET***********

 

The Joule Experiment 

- Measures the interaction between molecules

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

 

 

 

Any change of internal energy is associated with heat.

 

Results:

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

 

 

 

 

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

 

               H = constant enthalpy

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n
First balance the two reactions:
 
C3H6 + 9/2O2 à 3CO2 + 3H2O
 
 
nC3H6 + H2 + C3H8 + 5O2 à C3H8 + 3CO2 + 4H2O
nC3H6 + H2 + 5O2 à 3CO2 + 4H2O
 
 
nH2O à H2 + ½ O2 DrHθ= +286 kJ/mol
 
Then calculate the enthalpy for each balanced reaction:
nC3H6 + H2 à C3H8 DrHθ= -124 kJ/mol
nC3H8 + 5O2 à 3CO2 + 4H2O                 DrHθ= -2220 kJ/mol
nH2O à H2 + 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
 
 

 

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.

 

n

Example:

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 à 3CO2 + 4H2O                           DrHθ= -2220 kJ/mol
 

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