0
$\begingroup$

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?

$\endgroup$
9
  • 2
    $\begingroup$ Constant temperature and pressure with variable summary molar amounts means variable volume. $\endgroup$ – Poutnik Aug 4 '20 at 11:15
  • $\begingroup$ i dont understand the term "summary" $\endgroup$ – satan 29 Aug 4 '20 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 '20 at 13:14
  • $\begingroup$ i still dont understand what exactly your statement means. What does "summary mean"? $\endgroup$ – satan 29 Aug 4 '20 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 '20 at 4:50
3
$\begingroup$

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

$\endgroup$
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 '20 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.