While solving question on calculating degree of dissociation, I came across this question:

At 627 degrees Celsius and 1 atm $\ce{SO_3}$ is partially dissociated into $\ce{SO2}$ and $\ce{O2}$ by the reaction;

$$\ce{SO_3 <=> SO2 + 1/2O2}$$

the density of the equilibrium mixture is 0.925 gram per liter what is the degree of dissociation?"

In answer, they calculate the the molecular mass of mixture at equilibrium(Mmix) by using density of equilibrium which comes 63.348 and we know molecular mass of $\ce{SO3}$ is 80 , vapor density of $\ce{SO3}$ will be 80 by 2 is equal to 40 ,vapour density of mixture is equal to molecular mass of mixture by 2 is equal to 68.348 by 2 is equal to 34.174 and from here we can calculate degree of dissociation by using the formula(x=D-d/(n-1)d ) but my doubt is what is the difference between vapor density of mixture and density of the equilibrium mixture?

  • $\begingroup$ I read your explanation of the answer a half dozen times and I can't follow it. However the gist of the problem is that the various gas and gas mixtures follow ideal gas law PV=nRT. Thus $\ce{SO3}$ is one molecule but $\ce{SO2 + 1/2O2}$ is 1.5 molecules. So let's invoke magic. At 627 C I have pure $\ce{SO3}$ at 1.00 atm. I now cast another spell and convert all of the $\ce{SO3}$ to $\ce{SO2 + 1/2O2}$. The volume, mass and density are all the same. But the number of molecules has gone up by 1.5 times. So the new pressure is 1.5 atm. $\endgroup$
    – MaxW
    Jan 7, 2021 at 6:14
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    $\begingroup$ Please rewrite your calculations ! The value 63.348 suddenly becomes 68.348, as if 3 has been changed into 8 ! $\endgroup$
    – Maurice
    Jan 7, 2021 at 9:32
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    $\begingroup$ Prefer using symbolic algebra to literal numbers. You can easily derive the average molar mass from density via the ideal gas state equation. And then the equilibrium composition. $\endgroup$
    – Poutnik
    Jan 7, 2021 at 9:58
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    $\begingroup$ The two answers below can't both be right, but I can't figure out which is wrong and where the mistake was made. $\endgroup$
    – MaxW
    Jan 8, 2021 at 19:53

2 Answers 2


Obviously we have to assume ideal gas behavior.

The gas constant is $\pu{0.082057 L⋅atm⋅K−1⋅mol−1}$. The molar volume of any gas at 900 K and 1 atm is:

$$V = \dfrac{nRT}{P} = \dfrac{\pu{1 mol}\times \pu{0.082057 L⋅atm⋅K−1⋅mol−1}\times \pu{900 K}}{\pu{1 atm}}= \pu{73.851 L}$$

The molecular mass of $\ce{SO3}$ is 80.066 g/mol, at 900 K and 1 atm so for pure $\ce{SO3}$, which is 0% decomposed, we'd expect a density $\mathrm{D_0}$ of:

$$\mathrm{D_0} = \dfrac{\pu{80.066}}{73.851} = \pu{1.0842 g/L}$$

Now let's consider a mixture of 2/3 $\ce{SO2}$ and 1/3 $\ce{O2}$ where the $\ce{SO3}$ is 100% decomposed. The volume of a mixture of $6.023\times10^{23}$ molecules is still the same, but the density of the mixture, $\mathrm{D_{100}}$ , is different. First we need to calculate the average molecular mass.

$$\mathrm{M_{100}}= \dfrac{2}{3}\times 64.066 + \dfrac{1}{3}\times 15.999 = 48.044$$

$$\pu{D_{100}} = \dfrac{\pu{48.044}}{73.851} = \pu{0.6506 g/L}$$

So at 0% dissociation the density is $\pu{1.0842 g/L}$ and at 100% dissociation the density is $\pu{0.6506 g/L}$. It is linear in between. Let $x$ be the % dissociation. Then:

$$\mathrm{D_x} = 1.0842 - \dfrac{1.0842 - 0.6506}{100}\times x = 1.0842 - 0.004336x$$

For a density of 0.925

$$ 0.925 = 1.0842 - 0.004336x$$

$$\therefore x = 36.7\%$$


Let's start from $m = 0.925 g˙\ce{SO3}$ that will be heated at $900 K$. This is the amount of $\ce{SO3}$ that will occupy a volume of $1.00 L$ after heating it at $900 K$ at $1 atm$. and after partial decomposition. In moles it is also $n_o$ moles $\ce{SO3}$, $$\ce{ $n_o = \frac{m}{80 g/mol} = \frac{0.925 g}{80 g/mol} = 0.01156 mol$}$$ At $900 K$, a fraction $\alpha$ of these moles are decomposed into $\alpha$ moles $\ce{SO2}$ plus $\alpha/2 $ moles $\ce{O2}$. Then the total number of moles at equilibrium will be : $$\ce{$n$ = $n_o$˙($1$ - $\alpha$ + $\alpha$ + \alpha/2}) = n_o(1 + \alpha/2) = 0.01156 mol (1 + \alpha /2)$$ The total volume is : $$\ce{$V = nRT/p= 0.01156 mol$(1 + \alpha/2)$RT/p = 0.001 m^3$}$$ Simple rearrangement of this formula gives : $$\ce{1 + \alpha /2 = \frac{0.001 $m^3·p$}{0.01156 $mol·RT$} = \frac{0.001 $m^3$·101325 $Pa$}{0.01156 $mol$·8.314 $J mol^{-1}K^{-1}·900 K$} = 1.172 }$$ The degree of dissociation $\alpha$ is equal to $\alpha = 2 ·0.172 = 34.4$ %

  • $\begingroup$ 0.925 is density of equilibrium mixture , not mass of SO3.. $\endgroup$
    – Rover
    Jan 7, 2021 at 12:50
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    $\begingroup$ $0.925 g$ is the mass of the $\ce{SO3}$ that will occupy $V = 1 L$ at $900 K$ and $1 atm$. after partial decomposition. $\endgroup$
    – Maurice
    Jan 7, 2021 at 13:37
  • $\begingroup$ Ok , got your logic and answer is also close to given answer 34% ,so density of equilibrium mixture means density of SO3 after partial decomposition?, can we like this.. $\endgroup$
    – Rover
    Jan 7, 2021 at 14:04
  • $\begingroup$ @Rover. OK. So please remove the down point you gave me one hour ago ! $\endgroup$
    – Maurice
    Jan 7, 2021 at 14:06
  • $\begingroup$ that wasn't me.. and can we conclude that eq mix means density of products after partial decomposition ? $\endgroup$
    – Rover
    Jan 7, 2021 at 15:48

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