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Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
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#3
Look at the equation you're given, and try to see it as a combination of the given equations above it. What do you notice?
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#4
(Original post by oxe)
Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
I would work from the left to the right in a sequential manner, but leaving oxygen to last. (doing the same to the energy value at each step)
eg
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
You need four ammonia molecules so you start with equation 1.
Then you need 4NO molecules on the RHS so you double equation 2 and reverse it before adding to equation 1.
etc
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#6
using data from equation 1 and 2 enthalpy change can be calculated Enthalpy change= -909 2(-115) = -1139 KJ/mol
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#7
(Original post by Nadia2020)
using data from equation 1 and 2 enthalpy change can be calculated Enthalpy change= -909 + 2(-115) = -1139 KJ/mol
using data from equation 1 and 2 enthalpy change can be calculated Enthalpy change= -909 + 2(-115) = -1139 KJ/mol
A word of advice before replying to threads is check the date. This one is 4 years old. I doubt oxe needs help with this one, since s/he's probably finished university by now.
You also missed out a +, but I've added it for you
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#9
A reply, however old could be useful, as I was looking for the answer to the same problem, and I am grateful for the clear answer from Nadia2020
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#11
(Original post by oxe)
Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
Any ideas how to attack this monster..?
5. Nitric acid is produced industrially from ammonia, air and water using the following sequence of reactions:
(1) 4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –909 kJ mol−1
(2) 2NO(g) + O2(g) → 2NO2(g) ∆H = –115 kJ mol−1
(3) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = –117 kJ mol−1
Which is the enthalpy change (in kJ mol−1) for the following reaction?
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
A −679
B −794
C −1024
D −1139
thanks
Multiply equation 2 by 2 to get 4NO +2O2 --> 4NO2 delta H for this is 2 x -115Kj = -230KJ
Then combine this with equation 1, the 4NO's appear on both sides and cancel out to give:
4NH3 + 7O2 --> 4NO2 + 6H20 delta H is -230 + (-909) = -1139
See attachment for details
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#12
(Original post by Davies Chemistry)
Firstly, you don't need equation 3- that is a red herring
Multiply equation 2 by 2 to get 4NO +2O2 --> 4NO2 delta H for this is 2 x -115Kj = -230KJ
Then combine this with equation 1, the 4NO's appear on both sides and cancel out to give:
4NH3 + 7O2 --> 4NO2 + 6H20 delta H is -230 + (-909) = -1139
See attachment for details
Firstly, you don't need equation 3- that is a red herring
Multiply equation 2 by 2 to get 4NO +2O2 --> 4NO2 delta H for this is 2 x -115Kj = -230KJ
Then combine this with equation 1, the 4NO's appear on both sides and cancel out to give:
4NH3 + 7O2 --> 4NO2 + 6H20 delta H is -230 + (-909) = -1139
See attachment for details
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#13
(Original post by interlanken-fall)
love your method, but how is it possible that we can cancel both sides, wouldn't the NO2 of both sides still change the entahlpy? is it because you don't need no enthaply to change no2 to no2?
love your method, but how is it possible that we can cancel both sides, wouldn't the NO2 of both sides still change the entahlpy? is it because you don't need no enthaply to change no2 to no2?
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