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The vapour density of $N_2O_4$ at certain temperature is 30.Calculate the percentage of dissociation of $N_2O_4$ at this temperature.$\ce{N2O4_{(g)} <=> 2NO2_{(g)}}$?
I am unable to understand the concept behind vapour density of a the mixture.
Currently I understand that
2 x vapour density=molar mass.
vapour density =
mass of n molecules of gas ÷ mass of n molecules of hydrogen.
vapour density
= molar mass of gas ÷ molar mass of H2.
I am unable to apply the above formula because a mixture does not have molar mass.
And I am also not able to understand that if
2 x vapour density=molar mass
then in the question molar mass of $N_2O_4$ at the certain temperature given would be 60 instead of 92.
The correct answer is 53.33%

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1 Answer 1

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The big mistake you made was assuming a mixture does not have a molar mass.

Molar mass for a mixture is calculated by using the mole fractions and molar masses of each constituent.

Let:

$A$ represent $\ce{N2O4}$

$C$ represent $\ce{NO2}$

Then the reaction becomes:

$$\ce{A<=>2C}$$

First, we calculate the molar mass of the mixture using the given vapor density:

$$M=2v=(30)(2)=60\;g/mol$$

Then, we can set up the following system of equations in terms of molar masses and mole fractions:

$$X_A\;M_A+X_C\;M_C=M$$ $$X_A+X_C=1$$

Substituting all known values, the system looks like this:

$$92\;X_A+46\;X_C=60$$ $$X_A+X_C=1$$

Solving this system, we get the equilibrium molar fractions:

$$X_A=0.3043$$ $$X_C=0.6957$$

Then, we can calculate $K_X$, the equilibrium constant in terms of molar fractions:

$$K_X=\frac{X_C^2}{X_A}=\frac{0.6957^2}{0.3043}=1.5905$$

Finally, we use the relationship between $K_X$ and dissociation fraction $\alpha$ for this reaction to calculate it:

$$K_X=\frac{(2\alpha)^2}{1-\alpha^2}=1.5905$$

Solving for $\alpha$:

$$\alpha=\sqrt{\frac{K_X}{4+K_x}}=\sqrt{\frac{1.5905}{4+1.5905}}$$

$$\alpha=0.5333$$

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  • $\begingroup$ "First, we calculate the molar mass of the mixture using the given vapor density: M=2v=(30)(2)=60g/mol".but the vapour density of value 30 was given for N2O4 and was not given for the mixture of N2O4 and NO2.so doesn't 60g/mol indicate the molar mass of N2O4 at equilibrium instead of the mixture. How did you conclude that 60g/mol indicate the molar mass of the mixture. $\endgroup$
    – hsdfasd
    Commented Oct 26, 2022 at 5:58
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    $\begingroup$ The molar mass of $\ce{N2O4}$ is 92, and 92≠60, so that means it has to be a mixture. Even if we assumed no $\ce{NO2}$ was initially present before decomposition, that would mean $X_A$=1, $X_C$=0 , and we would still get the contradiction: 92≠60. Also, there's no such thing as "molar mass at equilibrium". Molar masses are the same regardless of whether we're at equilibrium or not. $\endgroup$
    – Sam202
    Commented Oct 26, 2022 at 6:05
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    $\begingroup$ Another way to look at it is this: $\ce{N2O4}$ is 92, and $\ce{NO2}$ is 46, so it makes sense that for a mixture of them, the molar mass of the mixture (60) has to be in between 46 and 92. $\endgroup$
    – Sam202
    Commented Oct 26, 2022 at 6:20

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