# How would I find the molarity/concentration of a gas using pV = nRT?

The problem I'm working on is:

At $$\pu{1100 K}$$ and a total pressure of $$\pu{1 atm}$$, $$\ce{SO3}$$ is $$75.0\%$$ dissociated.

$$\ce{2SO3(g) <=> 2SO2(g) + O2(g)}$$

Assume $$\pu{1.00 mol}$$ of $$\ce{SO3}$$ was present before dissociation.

Calculate the molarities of $$\ce{SO3},$$ $$\ce{SO2},$$ and $$\ce{O2}$$ at equilibrium.

I've already found the moles at equilibrium to be $$\pu{0.25 mol}$$ for $$\ce{SO3},$$ $$\pu{0.75 mol}$$ for $$\ce{SO2},$$ and $$\pu{0.375 mol}$$ for $$\ce{O2}.$$ I'm just not sure how to go about finding their molarities.

• Molarity is the number of moles of a paticular solute divided by the volume of the solution (in litres). You can find out the volumes of each using the ideal gas law. Then you get the total volume from that and you can calculate the molarities. Jul 23, 2019 at 10:15

I'm only reviving this old question because I got significantly different results compared to the proposed answer.

Let:

A = $$\ce{SO3}$$

C = $$\ce{SO2}$$

D = $$\ce{O2}$$

The change in moles for this reaction is:

$$\Delta n=2+1-2=1$$

The mole fractions at equilibrium are:

$$x_A=\frac{n_A}{n}=\frac{\pu{0.25mol}}{\pu{1.375mol}}=\frac{2}{11}$$

$$x_C=\frac{n_C}{n}=\frac{\pu{0.75mol}}{\pu{1.375mol}}=\frac{6}{11}$$

$$x_D=\frac{n_D}{n}=\frac{\pu{0.375mol}}{\pu{1.375mol}}=\frac{3}{11}$$

The total concentration at equilibrium is:

$$C=\frac{P}{RT}=\frac{\pu{1atm}}{\left(\pu{0.08206\frac{atm\;L}{mol\;K}}\right)\left(\pu{1100K}\right)}=\pu{0.01108mol/L}$$

So the concentrations at equilibrium would be:

$$\boxed{\pmb{C_A=x_A\;C=\left(\frac{2}{11}\right)\pu{0.01108mol/L}=\pu{2.015\cdot10^{-3}mol/L}}}$$

$$\boxed{\pmb{C_C=x_C\;C=\left(\frac{6}{11}\right)\pu{0.01108mol/L}=\pu{6.044\cdot10^{-3}mol/L}}}$$

$$\boxed{\pmb{C_D=x_D\;C=\left(\frac{3}{11}\right)\pu{0.01108mol/L}=\pu{3.022\cdot10^{-3}mol/L}}}$$

As an independent way to verify these answers, we can calculate the volume at equilibrium:

$$V=\frac{nRT}{P}=\frac{\left(\pu{1.375mol}\right)\left(\pu{0.08206\frac{atm\;L}{mol\;K}}\right)\left(\pu{1100K}\right)}{\pu{1atm}}=\pmb{\pu{124.1L}}$$

And we will get the same amounts at equilibrium that were calculated by OP:

$$n_A=V\;C_A=(\pu{124.1L})(\pu{2.015\cdot10^{-3}mol/L})=\pu{0.25mol}$$

$$n_C=V\;C_C=(\pu{124.1L})(\pu{6.044\cdot10^{-3}mol/L})=\pu{0.75mol}$$

$$n_D=V\;C_D=(\pu{124.1L})(\pu{3.022\cdot10^{-3}mol/L})=\pu{0.375mol}$$

We also know that the equilibrium constant in terms of moles is:

$$K_n=\frac{n_C^2\;n_D}{n_A^2}=\frac{(0.75^2)(0.375)}{0.25^2}=\pu{3.375mol}$$

While the equilibrium constant in terms of molar concentration is:

$$K_c=\frac{C_C^2\;C_D}{C_A^2}=\frac{(6.044^2)(3.022)}{2.015^2}\cdot 10^{-3}=\pu{0.02719mol/L}$$

Which means the volume at equilibrium must also be:

$$V=\frac{K_n}{K_c}=\frac{\pu{3.375mol}}{\pu{0.02719mol/L}}=\pmb{\pu{124.1L}}$$

• This is correct. The other answer assumed isochoric, and that the pressure was 1 atm before dissociation. However, the question says 1 atm after dissociation.
– Karsten
Apr 22 at 0:48

Supposing the dissociation process is isochoric, the total volume $$V$$ required for calculation of molarity at equilibrium of $$i$$th component $$c_{\mathrm{eq},i}$$

$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$

can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:

$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\pu{8.21e-2 L atm K-1 mol-1}\cdot\pu{1000 K}}{\pu{1 atm}} \approx \pu{82.1 L}$$

You have found amounts of substances at equilibrium correctly, the rest is just math:

$$c_\mathrm{eq}(\ce{SO3}) = \frac{\pu{0.25 mol}}{\pu{82.1 L}} \approx \pu{3.0e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{SO2}) = \frac{\pu{0.75 mol}}{\pu{82.1 L}} \approx \pu{9.1e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{O2}) = \frac{\pu{0.375 mol}}{\pu{82.1 L}} \approx \pu{4.6e-3 mol L-1}$$

• Thank you! I realize I had been inputting the wrong values for the equation. I had been inputting the specific equilibrium moles instead of the delta n. Just to be clear, when doing this, n is (#moles of gaseous products - #moles of gaseous reactants)? Jul 22, 2019 at 20:59
• @user81232 Yes, this is what you achieve by applying reaction's stoichiometry (e.g. when building up an ICE table). Also, please note that "number of moles" or "# of moles" isn't quite a correct term (see this Meta post) – amount of substance is. Jul 22, 2019 at 21:08
• It seems to me the problem statement implies that the system is isobaric, not isochoric. Jul 26, 2019 at 16:31