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:

Question 143 from Chapter 7 of Petrucci et al: General Chemistry, 11th Edition (see full citation below)

  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)
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    $\begingroup$ 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. $\endgroup$ Dec 17, 2019 at 15:07

1 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 $q_v$ = $q_p$

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    $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$ Dec 17, 2019 at 15:08

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