# Thermodynamics of chemical reactions "at constant pressure"

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?

• Constant temperature and pressure with variable summary molar amounts means variable volume. Aug 4 '20 at 11:15
• i dont understand the term "summary" Aug 4 '20 at 13:12
• 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. Aug 4 '20 at 13:14
• i still dont understand what exactly your statement means. What does "summary mean"? Aug 4 '20 at 14:20
• 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". Aug 5 '20 at 4:50

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

• your first paragraph explains it perfectly. I am sorry that i didn't realize this earlier, through your comments. +1. Aug 5 '20 at 5:06