Finding the kp with density, temperature, and pressure

Question

Consider the following reaction

$$\ce{3O2(g) <=> 2O3(g)}$$

At $\pu{175°C}$ and a pressure of $\pu{128Torr}$, and equilibrium mixture of $\ce{O2}$ and $\ce{O3}$ has a density of $\pu{0.168g/L}$. Calculate $K_p$ for the above reaction at $\pu{175°C}$.

I used the ideal gas law $pV=nRT,$ $K_p=K_c(RT)^n$, and $\rho=M \times c$ but I am not certain what to do after finding the total $c$ of the reaction. I was thinking of using the ICE chart to find the concentration of reactant and products which will lead to the $K_c$ and then $K_p$ value but since I don't know the initial concentration of $\ce{O2}$ I end up with an equation with two variables.

I was thinking of using the density to find the molar mass of the reaction and then somehow the weight of $\ce{O2}$ and $\ce{O3}$ in grams but it seems like a stretch.

$$pV=nRT$$ $$T=175\ \mathrm{^\circ C}=448\ \mathrm K$$ $$p=128\ \mathrm{Torr}=0.1684\ \mathrm{atm}$$ $$0.0821=\frac nV\times\left(0.0821\ \mathrm{\frac{atm\ l}{K\ mol}}\times448\ \mathrm K\right)$$ $$0.0821=\frac nV\times36.78\ \mathrm{\frac{atm\ l}{mol}}$$ $$\frac nV=0.00457\ \mathrm{\frac{mol}l}=c$$

$$ \begin{bmatrix} & \ce{3O2} & \ce{2O3} \\ \mathrm I & y & 0 \\ \mathrm C & -3x & 2x \\ \mathrm E & y-3x & 2x \end{bmatrix}0.00457\ \mathrm{mol/l}=(y-3x)+2x$$

Answer 1

You can use the formula for the ideal gas $$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$

As pressure, temperature and density are given, you get the mean molar mass.

$$M_\mathrm{mean}=\frac{\rho RT}{p}$$

From that, you get molar fractions.

$$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$

$$x_{\ce{O3}}=1-x_{\ce{O2}}$$

From that, you get partial pressures.

$$p_{\ce{O2}}=p\cdot x_{\ce{O2}}$$

$$p_{\ce{O3}}=p-p_{\ce{O2}}$$

From that, you get $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$

Answer 2

Looking at your struggle and @ChesterMiller's support, I am sharing how I would solve it. Anyone else please feel free to point out any mistakes in this answer.

Short Answer

here. (combining Dalton's law, Avagadro's law, ideal gas law, reaction quotient at equilibrium $Q_p = K_p$ in terms of partial pressures of gas mixture) Molecular weight of $\ce{O2} = \pu{32gmol-1},~\ce{O3} = \pu{48gmol-1}$.

$$\text{density}, \: d = \frac{\text{mass}, \: m}{\text{volume}, \: v}$$

$\text{num of moles,} \: n = \frac{\text{mass}, \: m}{\text{Molecular Weight}, \: M}$

Therefore we consider molar density,

$$\frac{d}{M} = \frac{\text{mass}}{\text{volume}}$$

$$\text{total density } T = \frac{\text{mass}( \ce{O2})~+~\text{mass} (\ce{O3})}{V \text{ volume}}$$

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Using ideal gas law, the density of a gas is directly proportional to its molecular mass (M).

$d_\mathrm{i} = \frac{P_\mathrm{i}\times M_\mathrm{i}}{RT}$

The density of a gas mixture is obtained using total mass (m) of the gases in volume V at temperature T.

$P_\mathrm{T} = \frac{d_\mathrm{T}\times RT}{m_\ce{O2} + m_\ce{O3}}$

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Mole fractions, $\chi_\ce{O2} + \chi_\ce{O3} = 1$.

How?

Total number of moles in gas mix = N

If $\ce{O2}$ has n moles, $\ce{O3}$ has = $N-n$ moles

Therefore, $N = n + (N-n)$, get the mole fraction of each gas to total moles,

$\frac{N}{N} = \frac{n}{N} + \frac{N-n}{N}$

$ 1 = \frac{n}{N} + \left(\frac{N}{N}- \frac{n}{N}\right)$

The mole fraction of $\ce{O2}$,

$\frac{n}{N} = \chi_\ce{O2}$

$\frac{N-n}{N} = \chi_\ce{O3}= (1 - \chi_\ce{O2})$

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Now apply this to the ideal gas law equation that we derived earlier,

$\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}} = \chi_\ce{O2} * M_\ce{O2} + \chi_\ce{O2} * M_\ce{O3}$

$\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}} = 32x + 48(1-x)$

$$\chi = \frac{48 - \left(\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}}\right)}{16}$$

Partial pressures of gas mix in the system at equilibrium, (I wouldn't simplify these calculations as total pressure to density cancels out each other at the end)

$p_\ce{O2} = P_\mathrm{T} \times \chi_\ce{O2}$

$p_\ce{O3} = P_\mathrm{T} \times \chi_\ce{O2}$

$K_p = \frac{[p_\ce{O3}]^2}{[p_\ce{O2}]^3}$

$K_p = \frac{[P_\mathrm{T} \times \chi_\ce{O3}]^2}{[P_\mathrm{T} \times \chi_\ce{O2}]^3}$

$$K_p = \frac{\left[P_\mathrm{T} \times \left(1-\left(\frac{48 -\left(\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}}\right)}{16}\right)\right)\right]^2}{\left[P_\mathrm{T} \times \left(\frac{48 - \left(\frac{d_\mathrm{T}\times*RT}{P_\mathrm{T}}\right)}{16}\right)\right]^3}$$

Please plug in the numbers, bear in mind the conversion factors.