Back with another question similar to the one I asked before, but this one's a bit more confusing
4NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H= –909 kJ/mol
2NO(g) + O2(g) → 2NO2(g) ∆H= –115 kJ/mol
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
Thank you so much for dealing with this question :D
eqn 1 + 2*eqn 2 = ? See if doing this doesn't give you the correct equation you want.
Ignore eqn 3. You don't need it.
A favorite trick when profs want you to think REALLY hard. Eqn 3 is a red herring.
To solve this problem, we need to use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the reactions that make up the overall reaction.
Here's how we can approach the problem step by step:
1. Start with the given reactions:
a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = -909 kJ/mol
b) 2NO(g) + O2(g) → 2NO2(g) ∆H = -115 kJ/mol
c) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) ∆H = -117 kJ/mol
2. We need to manipulate these reactions to get the desired overall reaction:
a) Multiply reaction b) by 2 to have the same number of moles of NO2:
2 (2NO(g) + O2(g) → 2NO2(g)) ∆H = -230 kJ/mol
3. Combine reaction a) and the modified reaction b):
4NH3(g) + 5O2(g) + 2(2NO(g) + O2(g)) → 4NO(g) + 6H2O(g) + 2(2NO2(g)) ∆H = -909 kJ/mol + (-230 kJ/mol)
4. Simplify the equation by canceling out common terms:
4NH3(g) + 7O2(g) → 4NO(g) + 6H2O(g) + 4NO2(g) ∆H = -909 kJ/mol - 230 kJ/mol
5. Add the enthalpy changes of the reactions to get the enthalpy change for the overall reaction:
∆H = -909 kJ/mol - 230 kJ/mol
∆H = -1139 kJ/mol
Therefore, the enthalpy change for the reaction 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) is -1139 kJ/mol.
So, the answer is D) -1139.