Thermodynamics of chemical reactions "at constant pressure"

Question

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?

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