Is the heat for constant pressure the same as for constant volume in the formation of nitric oxide? In the reaction $\ce{N2 (g) + O2(g) -&gt; 2NO (g)}$
my textbook[1]  says $q_v$ and $q_p$ are not the same, without any explanation.

I know that 
\begin{align}
\Delta U &amp;= q_p + w\\
\Delta U &amp;= q_p - P\Delta V\\
\Delta U &amp;= 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:






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)
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 $q_v$ = $q_p$