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
Your line of attack is to construct the required equation one step at a time using the equations that you are given and the four rules of number (x/+-).
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
using data from equation 1 and 2 enthalpy change can be calculated Enthalpy change= -909 + 2(-115) = -1139 KJ/mol
Welcome to TSR.
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
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
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
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
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?
You can look at it from two perspectives, I guess. Like in maths you can cancel things on boths sides or like you say the enthalpy change to 4NO from 4NO is zero - it's like a spectator.