$\ce{N2O4}$ 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: $$K_p = \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{0.1315 \cdot \dfrac{(2\alpha)^2}{1-\alpha} =0.1315 \cdot \dfrac{0.852^2}{0.574}= 0.1663\ bar} \tag{15}$$


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


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 ($\alpha$).


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

  • $\begingroup$ 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 . $\endgroup$ Jan 9 '17 at 1:22
  • $\begingroup$ Good catch both of you. $\endgroup$
    – Ben Norris
    Jan 9 '17 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.