# Is the heat for constant pressure the same as for constant volume in the formation of nitric oxide?

In the reaction $$\ce{N2 (g) + O2(g) -> 2NO (g)}$$ my textbook[1] says $$q_v$$ and $$q_p$$ are not the same, without any explanation.

I know that \begin{align} \Delta U &= q_p + w\\ \Delta U &= q_p - P\Delta V\\ \Delta U &= q_p - Δn_\mathrm{gases}RT \end{align}

I also know that $$\Delta U = q_v.$$

because under constant volume, work done ($$-P\Delta V$$) is zero, since there's no change in volume.

In the reaction mentioned, $$n_\mathrm{gases} = 0,$$ so shouldn't
$$\Delta U = q_p$$ and therefore $$q_p = q_v?$$

Here's a picture of question 143 in Chapter 7 of [1]. The answers say a) and d) are false:

1. Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette: General Chemistry: Principles and Modern Applications, 11th edition. Published by Pearson (February 23rd 2016)
• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup (more also here). I have also included a real reference to the book you are using, I hope it is correct. States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. Dec 17, 2019 at 15:07

Kevin is right. In that reaction, the number of moles does not change. So carrying it at constant pressure or at constant volume does not make any difference. And $$q_v$$ = $$q_p$$