# Finding the kp with density, temperature, and pressure

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.

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

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

• Do you know how to get the partial pressures of the two gases in the mixture? – Chet Miller Jul 13 '17 at 13:22
• If I had the Kp, yes kp=products/reactants and use the ICE chart – S.Polk Jul 13 '17 at 16:24
• 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? – Chet Miller Jul 13 '17 at 22:41
• The molar mass of oxygen is 32 and the molar mass of ozone is 48. – Chet Miller Jul 15 '17 at 20:43
• 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? – Chet Miller Jul 16 '17 at 19:56

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.

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.

• 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 – S.Polk Jul 17 '17 at 17:26
• @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. – bonCodigo Jul 17 '17 at 21:35
• 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 – S.Polk Jul 18 '17 at 0:07
• @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? – bonCodigo Jul 21 '17 at 3:24
• 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. – S.Polk Jul 23 '17 at 13:51

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 $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$