# How do you calculate the delta H for multiple reactions?

I am having difficulty in a question:

Given that

1. $$\ce{N_2(g) \rightarrow 2N(g), \Delta H = 941 kJ/mol}$$

2. $$\ce{N_2(g) \rightarrow N_2^+ + e^-, \Delta H = 1501 kJ/mol}$$

3. $$\ce{N(g) \rightarrow N^+ + e^-, \Delta H = 1402 kJ/mol}$$,

Find the energy interval that produces atomic nitrogen in the flask but not any ions

I think that the energy interval should be $$941 \leq E \leq 941 + 1402$$, but the textbook says that the energy interval should be $$941 \leq E \leq 1402$$.

I think I am correct because $$\ce{1402 kJ/mol}$$ is needed to be supplied after the initial $$\ce{941 kJ/mol}$$ that is used to break the $$\ce{N_2}$$ bond.

Can someone please see the flaw in my reasoning?

• A reaction can proceed with reactants have energy greater than or equal to the activation energy. The enthalpy change of reaction does not need to be taken into consideration. In this case, considering the reaction energy profile of an endothermic reaction, it is likely that the activation energy required is larger than the delta H values. Dec 19, 2018 at 3:34

If you supply $$941~kJ/mol$$, then you have supplied enough energy to move reaction 1 forward. In your flask, you would have $$N(g)$$ inside. Now, if you further supply energy up to $$1402~kJ/mol$$, then you have enough energy to make reaction 3 proceed, and that will convert your atomic nitrogen to nitrogen ions, so that's a no-no.
So you have to maintain $$E$$ that is in between $$941~kJ/mol$$ to $$1402~kJ/mol$$ to prevent reaction 3 from proceeding.
If you supply $$1402~kJ/mol$$ after breaking the bond, then you will definitely produce $$N^+$$ ions as you have met reaction 3's $$E$$ requirement.