Energy changes
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I thought that the enthalpy changes could only be deduced using solely enthalpy changes of combustion, not both combustion and formation values. Please help with the problem given below.
The question is: use the data to calculate the standard enthalpy of formation of gaseous hydrazine
DHc[N2H4(g)] = –568 kJ mol–1
DHf[H2O(g)] = –242 kJ mol–1
Igloo writes
Note that using Hess's law it is possible to work out enthalpy changes of formation from enthalpy changes of combustion. Given the values for the enthalpy changes of formation it is then possible to calculate enthalpy changes for reaction.
It is important for you to realize that the enthalpy change of formation of a substance can often be the enthalpy change of combustion of another allied substance.
For example, H2(g) + xO2(g) -> H2O(l)
The enthalpy change for this reaction is the enthalpy change of formation of liquid water, but it is also the enthalpy change of combustion of gaseous hydrogen.
So, in your question, the second piece of data could equally have been given as DHc[H2(g)]. Do you see now?
Now, to cope with your problem. I should say from the outset that it is impossible to do this question, not because of “mixing” enthalpies of formation and combustion (which, as I have already mentioned is not an issue), but because the enthalpy of combustion – by definition – always quotes chemicals in their normal states at 298 K (standard conditions), so you should have been given the data for liquid water and not gaseous water. I shall alter the state symbols to make this correction, and then I’ll take you through the method, though I’m not going to do the arithmetic for you!
First, construct an energy diagram, with, as usual, the elements underneath the species.
The relevant equation here is for the combustion of hydrazine:
–568 kJ
N2H4(g) + O2(g) –> N2(g) + 2H2O(l) [note that I’ve changed “g” to “l”]
Next, apply Hess’ Law and work out the value for DHf[N2H4(g)]. Don’t forget sign + units!
Risk assessment
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updated: 22 August 2003
