1
$\begingroup$

Given

$$ \begin{align} \Delta_\mathrm f H^\circ(\ce{N2O4}) &= \pu{9.16 kJ mol^-1} &\quad \Delta_\mathrm f H^\circ(\ce{NO2}) &= \pu{33.18 kJ mol^-1} \\ S^\circ(\ce{N2O4}) &= \pu{304.3 J K^{-1} mol^{-1}} &\quad S^\circ(\ce{NO2}) &= \pu{204.1 J K^-1 mol^-1} \end{align} $$

for the reaction $\ce{N2O4 <=> 2 NO2},$ find the percentage of dissociation of $\ce{N2O4}$ at $\pu{1 bar}$ and $T = \pu{333 K}.$

I formed the equilibrium:

$$ \begin{array}{ccc} \ce{&N2O4 &<=> &2 NO2,} \tag{1} \\ &n_0\left(1 - \frac{\alpha}{2}\right) & & n_0\alpha \end{array} $$

where $n_0 $ are the initial amount of $\ce{N2O4}$ and $\alpha$ the dissociation value:

$$\alpha = \frac{2x}{n_0}. \tag{2}$$

After doing

$$\Delta G^\circ = -RT\ln K_p \quad\implies\quad K_p = \exp{\frac{-\Delta G^\circ}{RT}} = \pu{7.32E21}, \tag{3}$$

and having to find $\alpha,$ I get stuck with two unknown variables $n_0$ and $\alpha$:

$$K_p = p_\mathrm{tot}\frac{x(\ce{NO2})^2}{x(\ce{N2O4})} = \ldots = p_\mathrm{tot}\frac{n_0\alpha^2}{1 - \alpha/2}. \tag{4}$$

How to proceed?

$\endgroup$
3
  • 1
    $\begingroup$ Ideal gas law, $pV=nRT$, to find $n_0$? $\endgroup$
    – Domen
    Commented Oct 5, 2022 at 22:29
  • $\begingroup$ But we don't know the volume? $\endgroup$
    – Acedium 20
    Commented Oct 6, 2022 at 6:31
  • 1
    $\begingroup$ Your $K_p$ is vast but $K_p \approx 0.14$ so it looks like you forget that some values are in kJ and others in J . It is easier if you use $1-\alpha$ and $2\alpha$ and get partial pressure as $P_{NO2}=2\alpha P_{tot}/(1+\alpha) $ and $P_{N2O4}=(1-\alpha)P_{tot}/(1+\alpha)$ and $K_p=P_{NO2}^2/P_{N2O4}$. $\endgroup$
    – porphyrin
    Commented Oct 6, 2022 at 7:28

2 Answers 2

2
$\begingroup$

Let:

$A$ represent $\ce{N_2O_4}$

$C$ represent $\ce{NO_2}$

Then the reaction becomes:

$$\ce{A <=> 2C}$$

Since no initial amounts are given, I'd suggest using mole fractions for the ICE table rather than moles. Assuming we start with pure $A$:

$X_{A^o}=1$

$X_{C^o}=0$

Writing equations for the equilibrium mole fractions of $A$ and $C$:

$X_A=X_{A^o}-y=1-y$

$X_C=X_{C^o}+2y=2y$

Then, we calculate $K_x$ by using its relationship with $K_p$:

$K_x=\frac{K_p}{P^{\Delta n}}$

($\Delta n=1$ for this reaction)

Then we can use the equilibrium expression for $K_x$ and substitute the expressions of the equilibrium molar fractions:

$K_x=\frac{X_C^2}{X_A}=\frac{(2y)^2}{1-y}$

We can then solve for $y$, and finally use its value to find $\alpha$:

$\alpha=\frac{y}{X_{A^o}}$

But since $X_{A^o}=1$, $\alpha$ is simply:

$\alpha = y$

Note 1: The answer will depend on whatever units your calculated $K_p$ value has, so make sure it matches the units of total pressure $P$ when calculating $K_x$.

Note 2: The assumption that we initially start with pure $A$ can be found to be correct or incorrect by evaluating: $X_A+X_C≈1$

$\endgroup$
0
$\begingroup$

The total number of moles of gas is $n_0(1+\frac{\alpha}{2})$. So the mole fraction of $\ce{NO_2}$ is $\frac{\alpha}{(1+\frac{\alpha}{2})}$ and the mole fraction of $\ce{N_2O_4}$ is $\frac{(1-\frac{\alpha}{2})}{(1+\frac{\alpha}{2})}$. The corresponding partial pressures are equal to these mole fractions times the total pressure, and do not contain $n_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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