Textbook 4D.23
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Textbook 4D.23
I'm having trouble with textbook problem 4D.23. I put a picture of the problem below that hopefully comes through. Essentially it's two equations which you are supposed to use to derive the enthalpy of formation of N2O5. The solution manual derives an equation for enthalpy of formation that's N2 + (5/2)O2 --> N2O5. It then uses the two given equations to get 2NO + (3/2)O2 --> N2O5, and uses the enthalpy of formation of NO to take it out of the enthalpy for that reaction, saying that that'll give the enthalpy of formation of N2O5. I don't really understand how that would give it as it doesn't match the first equation. Does anyone understand how this works?
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Re: Textbook 4D.23
Recall, the enthalpy of formation of NO(g) is represented below.
However, in order to cancel out the "2 NO(g)" in the equation remaining, the enthalpy of formation of NO(g) must be reversed (i.e. multiply enthalpy by a negative sign) and multiplied by 2 (i.e. multiply enthalpy by 2) to ensure it cancels out into the final equation of interest.
However, in order to cancel out the "2 NO(g)" in the equation remaining, the enthalpy of formation of NO(g) must be reversed (i.e. multiply enthalpy by a negative sign) and multiplied by 2 (i.e. multiply enthalpy by 2) to ensure it cancels out into the final equation of interest.
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Re: Textbook 4D.23
I do not fully understand your question, but I will try to explain why the textbook solves the problem the way it does. Hope it helps!
The standard enthalpy of formation of N2O5 is making 1 mole of N2O5 from its elements in their standard state. So the equation you need to find the enthalpy change of is N2+(5/2)O2-> N2O5 (call it equation A).
To get 1 mole of N2O5 on the right side of the reaction, the textbook divides the second equation by 2 and adds it to the first equation. Things cancel to give 2NO+(3/2)O2->N2O5 (call it equation B).
But NO is not an element. To get the reaction you need, you have to break NO down to N2 and O2. That is where the standard enthalpy of formation of NO comes in.
The standard enthalpy of formation of NO would be the enthalpy change of the reaction (1/2)N2+(1/2)O2->NO (call it equation C). Equation B has 2NO that we want to remove. So we multiply equation C by 2 and add it to equation B. Things cancel and the O2 add up to give equation A, which is what we want.
Keeping track of the enthalpy changes of the whole process will give us our answer.
The standard enthalpy of formation of N2O5 is making 1 mole of N2O5 from its elements in their standard state. So the equation you need to find the enthalpy change of is N2+(5/2)O2-> N2O5 (call it equation A).
To get 1 mole of N2O5 on the right side of the reaction, the textbook divides the second equation by 2 and adds it to the first equation. Things cancel to give 2NO+(3/2)O2->N2O5 (call it equation B).
But NO is not an element. To get the reaction you need, you have to break NO down to N2 and O2. That is where the standard enthalpy of formation of NO comes in.
The standard enthalpy of formation of NO would be the enthalpy change of the reaction (1/2)N2+(1/2)O2->NO (call it equation C). Equation B has 2NO that we want to remove. So we multiply equation C by 2 and add it to equation B. Things cancel and the O2 add up to give equation A, which is what we want.
Keeping track of the enthalpy changes of the whole process will give us our answer.
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