Hess' Law- Enthalpies of Formation (Help)
Hey guys I'm sort of confused as to how to find the enthalpy of formation here.
Heres the question:
Given the data:
4NH3(g)+3O2(g)= 2N2(g)+6H20, ΔH= -1530kJ/mol
H2(g)+1/2O2(g)= H20(l), ΔH= -288kJ/mol
Calculate the enthalpy of formation of ammonia.
Could someone please tell me how the ΔH values are used in order to find the ΔHf value of ammonia?
Heres the question:
Given the data:
4NH3(g)+3O2(g)= 2N2(g)+6H20, ΔH= -1530kJ/mol
H2(g)+1/2O2(g)= H20(l), ΔH= -288kJ/mol
Calculate the enthalpy of formation of ammonia.
Could someone please tell me how the ΔH values are used in order to find the ΔHf value of ammonia?
Sorry you've not had any responses about this. 
Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. 
I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed.
Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed.
Spoiler
Original post by Ijazx1
Hey guys I'm sort of confused as to how to find the enthalpy of formation here.
Heres the question:
Given the data:
4NH3(g)+3O2(g)= 2N2(g)+6H20, ΔH= -1530kJ/mol
H2(g)+1/2O2(g)= H20(l), ΔH= -288kJ/mol
Calculate the enthalpy of formation of ammonia.
Could someone please tell me how the ΔH values are used in order to find the ΔHf value of ammonia?
Heres the question:
Given the data:
4NH3(g)+3O2(g)= 2N2(g)+6H20, ΔH= -1530kJ/mol
H2(g)+1/2O2(g)= H20(l), ΔH= -288kJ/mol
Calculate the enthalpy of formation of ammonia.
Could someone please tell me how the ΔH values are used in order to find the ΔHf value of ammonia?
You have to used the equations given to construct the equation that represents the formation of ammonia. Remember the definition of formation enthalpy?
You can do anything to the given equations (four rules of number) providing that you do the same to the magnitude and sign of the energy change.
HINT
Spoiler
(edited 8 years ago)
Ok treat it like simultaneous equations, start off by flipping the second equation around (and change the sign)
4NH3(g)+3O2(g)= 2N2(g)+6H20
H2O(l)= H2(g)+1/2O2(g)
multiply second equation by 6
4NH3(g)+3O2(g)= 2N2(g)+6H20
6H2O(l)= 6H2(g)+ 3O2(g)
add equations together to give:
4NH3= 2N2+6H2
flip it around, and change the sign to get enthalpy of formation!!
(leaving the calculations out for you to try)
4NH3(g)+3O2(g)= 2N2(g)+6H20
H2O(l)= H2(g)+1/2O2(g)
multiply second equation by 6
4NH3(g)+3O2(g)= 2N2(g)+6H20
6H2O(l)= 6H2(g)+ 3O2(g)
add equations together to give:
4NH3= 2N2+6H2
flip it around, and change the sign to get enthalpy of formation!!
(leaving the calculations out for you to try)
Original post by Sydian
Ok treat it like simultaneous equations, start off by flipping the second equation around (and change the sign)
4NH3(g)+3O2(g)= 2N2(g)+6H20
H2O(l)= H2(g)+1/2O2(g)
multiply second equation by 6
4NH3(g)+3O2(g)= 2N2(g)+6H20
6H2O(l)= 6H2(g)+ 3O2(g)
add equations together to give:
4NH3= 2N2+6H2
flip it around, and change the sign to get enthalpy of formation!!
(leaving the calculations out for you to try)
4NH3(g)+3O2(g)= 2N2(g)+6H20
H2O(l)= H2(g)+1/2O2(g)
multiply second equation by 6
4NH3(g)+3O2(g)= 2N2(g)+6H20
6H2O(l)= 6H2(g)+ 3O2(g)
add equations together to give:
4NH3= 2N2+6H2
flip it around, and change the sign to get enthalpy of formation!!
(leaving the calculations out for you to try)
you da real mvp man.
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