3
$\begingroup$

I'm a beginner in chemistry (first year uni course) and I have this exercise that I need some help with. I don't really know where to start so for now. I've tried random approaches to see what works but nothing does.

This is the exercise:

For the reaction $$\ce{N2O4(g) <=> 2NO2(g)}$$

the $K_c$ is $0.060$, at $\pu{80 ^\circ C}$. Evaluate the density of the mixture once the equilibrium has been reached (at equilibrium, $T = \pu{80 ^\circ C}$ and $\pu{494 mmHg}$).

What I've tried to do is:

I have found $K_p$ using the $K_c$ and $K_p$ relation $$K_c = K_p \left(\frac{1}{RT}\right)^N,$$ but now I can't figure out how to proceed.

If receiving the full solution is not an option is it possible to get a step by step guide on what I have to do?

$\endgroup$
7
  • 2
    $\begingroup$ What are the random approaches you tried? Do you know how to calculate the density of pure $\ce{NO2}$ under these conditions? Do you know how to calculate the fraction of reactant and product under these conditions? $\endgroup$
    – Karsten
    Dec 17, 2023 at 2:18
  • $\begingroup$ My best attempt involved using Kp and and I.C.E table to try and find the partial pressures of the substances and then i would use the found partial pressures (The part i couldn't manage to do) to find the mole percentages and then find the molar mass of the substence after it was balanced and then finally tweaking the formula PV=nRT to find the density. $\endgroup$ Dec 17, 2023 at 12:07
  • $\begingroup$ The ICE table is less useful here. You have two unknowns, and you know the total pressure from the ideal gas law. The first equation is that the partial pressures add up to th total pressure, and the second is a quadratic from the equilibrium constant expression. Solve those for the partial pressures (or the mole percentage) first. $\endgroup$
    – Karsten
    Dec 17, 2023 at 13:29
  • $\begingroup$ In principle the constant $K_c$ should have a unit. Is it in mol/L ? $\endgroup$
    – Maurice
    Dec 17, 2023 at 13:55
  • $\begingroup$ Kc should have a unit but in our exercises we are told to use it only as a simple unitless value. $\endgroup$ Dec 17, 2023 at 18:11

3 Answers 3

1
$\begingroup$

This is a more direct calculation (or sketch of a calculation) compared to megamonster68's correct answer:

The partial pressures add up to the given pressure $P$, and are related as part of the equilibrium constant expression:

$$p_{\mathrm{\ce{N2O4}}}= P - p_{\mathrm{\ce{NO2}}}$$

so

$$K_{\mathrm{p}} = \dfrac{{p_{\mathrm{\ce{NO2}}}}^{2}}{P - p_{\mathrm{\ce{NO2}}}}$$

Solving for the partial pressures, you get (as the physically possible solution to the quadratic equation):

$$p_{\mathrm{\ce{NO2}}} =0.504\ \mathrm{atm}$$

$$p_{\mathrm{\ce{N2O4}}} =0.146\ \mathrm{atm}$$

You can calculate the densities separately and add them up:

$$ρ= ρ_\ce{N2O4} + ρ_\ce{NO2}= \frac{1}{R T} (p_\ce{N2O4} M_\ce{N2O4}+ p_\ce{NO2} M_\ce{NO2}) = 1.26\ \frac{\mathrm{g}}{\mathrm{L}} $$

I used three significant figures throughout, even though two of the quantities are given with only two significant figures. In other words, differences in the third digit of the answer are not significant. Surprisingly, the only time the stoichiometric factor 2 appears is in the equilibrium constant expression.

Here is a link to the full calculation.

$\endgroup$
1
$\begingroup$

Deriving $\rho_{\text{mix}}$

Density of mixture $$ \rho_{\text{mix}}= \frac{w_{\ce{N2O4}}+w_{\ce{NO2}}}{V} $$ Let total number of mol at equilibrium be $n$ and $n=w/M$ $$ \rho_{\text{mix}}= \frac{n}{V}(M_{\ce{N2O4}}\frac{n_{\ce{N2O4}}}{n}+M_{\ce{NO2}}\frac{n_{\ce{NO2}}}{n}) $$ From ideal gas equation (the subscript 'eq' is dropped), $$ \frac{n}{V}=\frac{P}{RT} $$ and mole fraction $x_j=n_j/n$, $$ \rho_{\text{mix}}= \frac{P}{RT}(M_{\ce{N2O4}}x_{\ce{N2O4}}+M_{\ce{NO2}}x_{\ce{NO2}}) \tag{1} $$ Assuming initially 1 mol of $\ce{N2O4}$ is present and degree of dissociation is $\alpha$,

Time $\ce{N2O4}$ $\ce{NO2}$
$t=0$ $1$ $0$
At equilibrium $1-\alpha$ $2\alpha$

$$ n=1-\alpha+2\alpha=1+\alpha \\ x_{\ce{N2O4}}=\frac{1-\alpha}{1+\alpha} \\ x_{\ce{NO2}}=\frac{2\alpha}{1+\alpha} \tag{2} $$

We also have $M_{\ce{N2O4}}=2M_{\ce{NO2}}$ then from $(1)$ and $(2)$ $$ \rho_{\text{mix}}=\frac{2PM_{\ce{NO2}}}{RT(1+\alpha)} $$

Calculating $\alpha$

For the equilibrium $$ K_p=\frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}}=\frac{x_{\ce{NO2}}^2}{x_{\ce{N2O4}}}~P $$ where partial pressure $p_j=x_j~P$
In terms of $\alpha$, $$ K_p=\frac{4\alpha^2}{1-\alpha^2}~P $$ Solving for $\alpha$ $$ \alpha=\sqrt{\frac{K_p}{K_p+4P}} $$ Now $K_p=K_c~(RT)^{\Delta n}$ and $\Delta n=2-1=1$. Therefore, $$ K_p=RT~K_c $$

Substituting the above in $\alpha$ $$ \alpha=\sqrt{\frac{RTK_c}{RTK_c+4P}} \\ \begin{align*} & P=\pu{494 mmHg}=\pu{0.66 bar}\\ & T=\pu{353.15 K} \\ & R=\pu{0.08314 bar-\ell~K^{-1} mol^{-1}} \\ & K_c=\pu{0.06 M} \end{align*} $$ From the given values $\alpha=0.63$

Calculating $\rho_{\text{mix}}$

Note: Directly substituting the values will give the density in $\pu{g/\ell}$
$$ \begin{align*} & \alpha=0.63 \\ & M_{\ce{NO2}}=\pu{46.00 g mol^{-1}} \end{align*} $$ Therefore, $$ \rho_{\text{mix}}=\pu{1.27 g/\ell}=\pu{1.27E-3 g/m\ell} $$

$\endgroup$
2
  • $\begingroup$ The assumption of a closed vessel is taken in the solution $\endgroup$ Dec 17, 2023 at 16:19
  • $\begingroup$ Try chemistry markup with mhchem's \ce{…}. If you want to know more, please have a look here and here. $\endgroup$ Dec 19, 2023 at 19:36
0
$\begingroup$

Adding a second alternative method for arriving at the correct answer.

Let: $$A = {\ce{N2O4}}$$ $$B = {\ce{NO2}}$$

We can calculate the total concentration at equilibrium $C$ with the ideal gas law by using the given temperature $T$ and total pressure $P$ at equilibrium:

$$C=\frac{P}{RT}=\frac{\pu{494mmHg}}{\left(\pu{62.36\frac{mmHg\;L}{mol\;K}}\right)\pu{353K}}=\pu{0.02243mol/L}$$

Using the relationship between $K_X$ and $K_C$, and noting that $\Delta n =1$, we can calculate the former:

$$K_C=K_X\;C^{\Delta n}\implies K_X=\frac{K_C}{C}=\frac{\pu{0.06mol/L}}{\pu{0.02243mol/L}}=2.675$$

Next, we can use the expression for $K_X$ and its value to obtain an equation involving the molar fractions of both species. Considering we're dealing with a binary mixture: $X_A+X_B=1$, and:

$$K_X=\frac{X_B^2}{X_A}=2.675\implies \frac{X_B^2}{1-X_B}=2.675\implies X_B=0.775,\;X_A=0.225$$

Next, we can use the molar fractions to calculate the molar mass of the mixture at equilibrium:

$$M=X_AM_A+X_BM_B=(0.225)(\pu{92g/mol})+(0.775)(\pu{46g/mol})=\pu{56.35g/mol}$$

Finally, we can calculate the density of the mixture at equilibrium using the total concentration and molar mass of the mixture:

$$\rho=CM=(\pu{0.02243mol/L})(\pu{56.35g/mol})=\pu{1.263g/L}$$

$\endgroup$

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.