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

Question

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:

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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)

Answer 1

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 $q_v$ = $q_p$