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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.

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  • $\begingroup$ 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. $\endgroup$ – Shoubhik Raj Maiti Jul 23 at 10:15
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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:

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

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

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

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  • $\begingroup$ 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)? $\endgroup$ – user81232 Jul 22 at 20:59
  • $\begingroup$ @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. $\endgroup$ – andselisk Jul 22 at 21:08
  • $\begingroup$ It seems to me the problem statement implies that the system is isobaric, not isochoric. $\endgroup$ – Chet Miller Jul 26 at 16:31

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