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$\ce{N2O4}$ is dissociated $20\ \%$ at 27 degrees Celsius and 1 bar. Find the equilibrium constant of that reaction. What percentage of $\ce{N2O4}$ dissociates at $13.15\ \mathrm{kPa}$?

An explanation of the solution is below, but there are two things I do not understand.

Why do we have $1 + \alpha$ in the denominator in equations ${\text(6)}$–${\text(9)}$?

How do we get the fraction $\dfrac{(2\alpha)^2}{1-\alpha}$ in equation $\text{(11)}$?

If the degree of dissociation is $\alpha$, then the amounts of substances present $n_0$ dissociated is $n_0 \alpha$, and the remaining amount of undissociated substance $n_0 (1-\alpha)$. Since each molecule $\ce{AB}$ dissociates into molecules of $\ce{A}$ and $\ce{B}$, the amounts of each substance is $$\begin{align} n_{\ce{A}} &= n_{\ce{B}} = n_0 \cdot \alpha \tag{1}\\ n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{2} \end{align}$$ In the case of the dissociation of dinitrogen tetroxide, $\ce{N2O4}$, it will be $$\begin{align} n_{\ce{A}} &= 2n_0 \alpha \tag{3}\\ n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{4}\\ \sum{n} &=n_0 (1+\alpha) \tag{5}\\ x_{\ce{A}} &= \dfrac{2\alpha}{(1+\alpha)} \tag{6}\\ x_{\ce{AB}} &= \dfrac{1-\alpha}{1+\alpha} \tag{7}\\ x_{\ce{NO2}} &= \dfrac{2\alpha}{(1+\alpha)} =\dfrac{0.40}{1.20}=0.3333 \tag{8}\\ x_{\ce{N2O4}} &= \dfrac{1-\alpha}{1+\alpha}=\frac{0.80}{1.20}=0.6666 \tag{9}\\ K_p &= \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{ 1\ bar \cdot \dfrac{(0.3333)^2}{0.6666} = 0.1667\ bar} \tag{10} \end{align}$$ We obtain the degree of dissociation of $\ce{NO2}$ at a pressure of $13.15\ \mathrm{kPa}$: $$\dfrac{K_p}{P} = \dfrac{(2\alpha)^2}{(1-\alpha)}=\mathrm{\dfrac{0.1667\ bar}{0.1315\ bar}=1.268} \tag{11}$$ From here, we get: $$ 4\alpha^2 = 1.268 (1-\alpha) = 1.268 -1.268\alpha \tag{12}$$ Which becomes $$ 4\alpha^2 =1.268\alpha – 1.268 = 0 \tag{13}$$ For the degree of dissociation, we get: $$\alpha =\dfrac{-b + \sqrt{b^2} -4ac}{2a} = \dfrac{-1.268 +\sqrt{1.268^2}+20.288}{8} \\= \dfrac{-1.268 +4679}{8} = 0.426 \tag{14}$$ Check your results: $$K_p = \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{0.1315 \cdot \dfrac{(2\alpha)^2}{1-\alpha} =0.1315 \cdot \dfrac{0.852^2}{0.574}= 0.1663\ bar} \tag{15}$$

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It looks like alpha $\alpha$ is the degree of dissociation. The way it is explained in your image is a little odd, so let me see if I can do a better job. What seems like it might be problematic for you is that the explanation you have been given skips some algebra along the way. Below, I am showing all of the algebra up until your points of confusion.

For your reaction: $\ce{N2O4 <=> 2NO2}$

In the first part of your question, we need to determine the equilibrium constant

$$K_p=\frac{P^2_{\ce{NO2}}}{P_{\ce{N2O4}}}$$

We know the total pressure, but we do not know the partial pressures of each species. Fortunately, thanks to Raoult's Law, we can substitute in mole fractions since $P_i =P \chi_i $.

$$K_p=\frac{P^2 \chi^2_{\ce{NO2}}}{P\chi_{\ce{N2O4}}}=P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

We aren't given any number of moles, but let's say that initially there was some number of moles of $\ce{N2O4}$ before any dissociation occurred. We'll call that number $n_0$. We can write expressions for the number of moles of $\ce{N2O4}$ and $\ce{NO2}$ after dissociation using $n_0$ and $\alpha$. Since $\alpha\%$ of $\ce{N2O4}$ is dissociated, $(1-\alpha\%)$ remains undissociated:

$$n_{\ce{N2O4}}=n_0 (1-\alpha)$$

The moles of $\ce{NO2}$ are $n_0$ times $2\alpha$, since $\alpha\%$ of $\ce{N2O4}$ was converted into $\ce{NO2}$ and there are two $\ce{NO2}$ molecules for every $\ce{N2O4}$ molecule:

$$n_{\ce{NO2}} = 2n_0 \alpha$$

The total moles (need this for mole fraction) is

$$\sum{n}=n_{\ce{N2O4}} + n_{\ce{NO2}} = n_0 (1-\alpha) + 2n_0\alpha = n_0 (1-\alpha +2 \alpha) = n_0 (1+\alpha)$$

When we calculate the mole fractions of each species, the $n_0$ term cancels out, which is good since we do not know what its value is:

$$\chi_{\ce{N2O4}}=\frac{n_{\ce{N2O4}}}{\sum{n}}=\frac{n_0 (1-\alpha)}{n_0 (1+\alpha)}=\frac{1-\alpha}{1+\alpha}$$ $$\chi_{\ce{NO2}}=\frac{n_{\ce{NO2}}}{\sum{n}}=\frac{2n_0\alpha}{n_0 (1+\alpha)}=\frac{2\alpha}{1+\alpha}$$

Now you know where $1+\alpha$ came from. These values are plugged into the law of mass action for $K_p$ to determine the equilibrium constant.

In the second part of the question, you use the value of the equilibrium constant and the new pressure to determine a new value of $\alpha$. It all starts with the law of mass action and we manipulate some algebra from there. First the law of mass action using mole fractions:

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

Next, substitute in the expressions for mole fraction in terms of $\alpha$.

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}=P\frac{\frac{(2\alpha)^2}{(1+\alpha)^2}}{\frac{1-\alpha}{1+\alpha}}$$

The term $1+\alpha$ appears in both the numerator and the denominator and can be canceled.

$$\require{cancel} K_p = P\frac{\frac{(2\alpha)^2}{\cancel{(1+\alpha)}(1+\alpha)}}{\frac{1-\alpha}{\cancel{(1+\alpha)}}}=P\frac{(2\alpha)^2}{(1-\alpha)1+\alpha}=P\frac{(2\alpha)^2}{1-\alpha^2}$$

Now we move $P$ over to the left side of the equation to separate the values we know ($K_p$ and $P$) from those we do not know ($\alpha$).

$$\frac{K_p}{P}=\frac{(2\alpha^2)}{1-\alpha^2}$$

Now you have something like your second equation that you are confused about. Perhaps there is a mistake in the original source?

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  • $\begingroup$ I still do not get it. Why haven't we squared 1 + alpha when you have written that xNO2 is squared? Shouldn't then the fraction go (2alpha/1+alpha) ^2 ? Why do we square only the numerator? $\endgroup$ – user33683 Aug 11 '16 at 13:52
  • $\begingroup$ The equation is wrong, both terms should be squared as you noted. The bottom term should be 1-alpha^2 . $\endgroup$ – Cliff Stamp Jan 9 '17 at 1:22
  • $\begingroup$ Good catch both of you. $\endgroup$ – Ben Norris Jan 9 '17 at 1:30

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