## Homework problem 8.65

Miya Eberlein 1J
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

### Homework problem 8.65

Can someone help clarify #65 for me? We're trying to get to the equation N2 + 5/2 O2 --> N205
but at the end of the problem the solution is to subtract twice the enthalpy of NO from the formation of N205 because we don't need it... but we don't have N2 in the equation. Do we not need it because we wouldn't factor it in anyways, as that's its natural state? and does it not matter that taking away the NO would mean we only have 3 O2 molecules on the left side, even though we need 5 on the right?
thank you!

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### Re: Homework problem 8.65

Always post the entire question so that Chemistry Community posts are self-sufficient.

In this problem, we are asked to find the enthalpy of formation of dinitrogen pentoxide from two given reactions ($2NO_{(g)}+O_{2(g)}\rightarrow 2NO_{2(g)} \Delta H^{\circ}=-114.1 kJ$ and $4NO_{2(g)}+O_{2(g)}\rightarrow 2N_{2}O_{5(g)} \Delta H^{\circ}=-110.2 kJ$) and the entropy of formation of NO.

By adding the first reaction to 1/2 the second reaction, we get the reaction $2NO_{(g)}+\frac{3}{2}O_{2(g)}\rightarrow N_{2}O_{5(g)}$. The enthalpy of this reaction is equal to $\Delta H^{\circ}_{f}(N_{2}O_{5})-2\Delta H^{\circ}_{f}(NO)$.

Therefore, to find $\Delta H^{\circ}_{f}(N_{2}O_{5})$, we simply need to subtract $2\Delta H^{\circ}_{f}(NO)$.

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