Enthalpy change
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10. Look at the enthalpy cycle below...
a) What enthalpy change is represented by delta H1?
b) What are the combined enthalpy changes represented by delta H2?
c. What enthalpy change is represented by delta H3?
d. write and equation that links delta H1, delta H2, and delta H3.
e. use the answers from parts a and d and the following standard enthalpy changes of combustion, calculate the standard enthalpy change of formation of propane.
11. a. write the equation to represent the formation of one mole of butane, C4H10 (g) from its elements in their standard states.
b. draw an enthalpy cycle to show the relationship between the formation of butane from carbon and hydrogen and the combustion of these elements to give carbon dioxide and water.
c. Use you enthalpy cycle to calcuate a value for the standard enthalpy change of formation of butane.
12. Look at the enthalpy cycle below.
a. What enthalpy change is represented by delta H1?
b) What enthalpy change is represented by delta H2?
c. What are the combined enthalpy changes represented by delta H3?
d. write and equation that links delta H1, delta H2, and delta H3.
e. use the answers from parts a and d and the following standard enthalpy changes of combustion, calculate the standard enthalpy change of combustion of methane.
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Solution Summary
It provides examples of calculating the standard Enthalpy change of formation and combustion.
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10a. The enthalpy change of formation of propane. How do you know this? Because enthalpies of formation deal with the formation of a molecule from its elemental constituents, in this case, C and H2.
10b. The enthalpy changes of combustion of carbon and of hydrogen. But it's really "multiplied by three" for carbon and "multiplied by four" for hydrogen. Combusion is the reaction of a compound with molecular oxygen, O2.
10c. The enthalpy change of combustion of propane.
10d. There are two routes to get to the lower box from the upper left box. You can go directly following dH2 or you can take a two step approach first following dH1 and then dH3. Therefore:
dH1 + dH3 = dH2
10e. The standard enthalpy change of formation of propane is dH1.
We're told the following:
C + O2 ----> CO2 -393 kJ/mol
H2 + 1/2 O2 ----> H2O -286 kJ/mol
C3H8 + 5O2 -----> 3CO2 + 4H2O -2220 kJ/mol
So, how do we end up with
3C + 4H2 ------> C3H8 ???
We do it like this. First we multiply the combusion of carbon by three:
3C + 3O2 ----> 3CO2 -1179 kJ/mol
Then we multiply the second ...
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