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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.

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

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

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    $\begingroup$ Do you know how to get the partial pressures of the two gases in the mixture? $\endgroup$ – Chet Miller Jul 13 '17 at 13:22
  • $\begingroup$ If I had the Kp, yes kp=products/reactants and use the ICE chart $\endgroup$ – S.Polk Jul 13 '17 at 16:24
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    $\begingroup$ If x is the mole fraction of O2 and (1-x) is the mole fraction of O3, what is the mole fraction of O2 in the mixture if the molar density of the mixture is 0.00457 moles/liter and the mass density is 0.168 g/liter? Based on this, what are the partial pressures? $\endgroup$ – Chet Miller Jul 13 '17 at 22:41
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    $\begingroup$ The molar mass of oxygen is 32 and the molar mass of ozone is 48. $\endgroup$ – Chet Miller Jul 15 '17 at 20:43
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    $\begingroup$ If you are trying to calculate Kp, all you need to know are the partial pressures. Do you know how to calculate the partial pressure of a species once you know the total pressure and the mole fraction of the species? $\endgroup$ – Chet Miller Jul 16 '17 at 19:56
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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 $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$

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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,

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

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


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.

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


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 $\ce{O2}$,

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

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


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.

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  • $\begingroup$ I honest most of your steps but I am still confuse on how the molecular mass of rxn equal to 32x+48(1-x). As you written. XO2+xO3=1 which is just x% + X%=100% of the rxn mass. Let's say x is 60%. The equation then tells me that O2 is 60% of the rxn total mass and 03 is 40%. Which makes sense. This is where 32x+48(1-x) confuses me because it's asking for 60% of O2 mass and 30% of O3 mass instead of 60% and 30% of rxn mass. If the total rxn mass was 70, 70(x)+70(1-x)=70 makes 32(x)+48(1-x)=70 doesn't. Also, I got x=[(dt x RT/PT)/-16 ]-48. My 16 and 48 are negative $\endgroup$ – S.Polk Jul 17 '17 at 17:26
  • $\begingroup$ @S.Polk Let's consider this: What is the difference between mass of a molecule/compound, molar mass and molecular mass? What does this mean to you? Have you learnt about mole fraction? Isn't it in your gen-chem text book? Because if you can grab this concept, you are through. $\endgroup$ – bonCodigo Jul 17 '17 at 21:35
  • $\begingroup$ Yes I do and that still help me understand that specific part of the equation or why you don't have a negative 16 or 48 when 32x+48(1-x) will be 32x+48-48x-->48-16x $\endgroup$ – S.Polk Jul 18 '17 at 0:07
  • $\begingroup$ @S.Polk have you resolved your doubts? Do you know how to measure the mass of a molecule in a molecule mixture? There's a mole ratio and there's partial pressure ratio, and you also know the molar mass of that molecule. Does this ring any bell to you? $\endgroup$ – bonCodigo Jul 21 '17 at 3:24
  • $\begingroup$ A little. You can use the M and density to get the mass of a molecule. I'm still confuse on how they are getting their mole ratio formula. $\endgroup$ – S.Polk Jul 23 '17 at 13:51

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