Chemical reactions occur at constant temperature and pressure.

Consider a gaseous, equilibrium reaction: $\ce{2NO2(g) <=> N2O4(g)}$. Most questions/textbooks formulate such questions by stating: The reaction happens at ($T$) temperature and ($P$) pressure.

This gives the impression that the surrounding Temperature and pressure are constant.

  • The first confusion that then arises is: For thermodynamic quantities such as enthalpy that involve pressure, which pressure should we use, internal or external?

After reading some chem.SE answers, it seems to me that $P_{sys}=P_{surr}$ is a basic assumption. However this doesn't seem to go well with the example I mentioned: Since the pressure of the system will be given by $(n_1)RT/V + (n_2)RT/v$, where $n_1$ and $n_2$ are the moles of $\ce{NO2}$ and $\ce{N2O4}$. Clearly, as the reaction progresses, ($n_1+n_2$) changes, and thus,the pressure of the system changes. If the surrounding pressure is kept constant, this beaks the "assumption": $P_{sys}=P_{surr}$.

Am I missing something?

  • 2
    $\begingroup$ Constant temperature and pressure with variable summary molar amounts means variable volume. $\endgroup$
    – Poutnik
    Aug 4, 2020 at 11:15
  • $\begingroup$ i dont understand the term "summary" $\endgroup$
    – satan 29
    Aug 4, 2020 at 13:12
  • $\begingroup$ Or rather total molar amount $n_\mathrm{tot} = n_{\ce{NO2(g)}} + n_{\ce{N2O4(g)}}$. By other words, $V = n_\mathrm{tot}RT/p$, implying ideal gas behaviour. $\endgroup$
    – Poutnik
    Aug 4, 2020 at 13:14
  • $\begingroup$ i still dont understand what exactly your statement means. What does "summary mean"? $\endgroup$
    – satan 29
    Aug 4, 2020 at 14:20
  • 1
    $\begingroup$ What I do not get is even if I corrected myself using total instead of summary, and used explicit formula, he remained confused, like if it was his decision. I am Czech. Poutník mean a wanderer or a pilgrim, in outdoor or religious sense. Similarity to the Russian word Sputnik ( cotravelling mate of Earth ) is not accidental, the old Slavic word root "puť" means "a journey". OTOH, there is no relation to the English verb "to pout". $\endgroup$
    – Poutnik
    Aug 5, 2020 at 4:50

1 Answer 1


You have correctly stated the pressure of the system is $$p = \frac {(n_1 + n_2)RT}{V}$$

But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring $p_\mathrm{ext} = p_\mathrm{sys}$. It is similar as a thermostatic water bath keeps the system temperature constant.

So less confusing is to rewrite the ideal gas state equation :

$$V = \frac {(n_1 + n_2)RT}{p}$$

There are 4 variables: $p, V, T, n_\mathrm{tot}=n_1 + n_2$. But there are just 3 degress of freedom = 3 independent variables.

Either $V, n_\mathrm{tot}, T$ are given, and $p=f(V, n_\mathrm{tot},T)$,
either $p, V, T$ are given, and $n_\mathrm{tot}=f(p,V,T)$,
either $p, n_\mathrm{tot}, T$ are given, and $V=f(p,n_\mathrm{tot},T)$,
either $p, n_\mathrm{tot}, V$ are given, and $T=f(p,n_\mathrm{tot},V)$.

We have already given 2 variables: $p, T$, with $n_\mathrm{tot},V$ remaining free.
But there is the equilibrium relation

$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$

$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$

Solving the quadratic equation we would finally get $n_\mathrm{tot}$, so all 3 degrees of freedom are saturated with given values for $p, T, n_\mathrm{tot}$.

So in our case, the following applies:

$p, n_\mathrm{tot}, T$ are given, and $V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$

  • 1
    $\begingroup$ your first paragraph explains it perfectly. I am sorry that i didn't realize this earlier, through your comments. +1. $\endgroup$
    – satan 29
    Aug 5, 2020 at 5:06

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