Calculating the percentage of dissociation degree

The question is as such:



> 
> Notice the equilibrium expression. $$K\_c=\frac{[NH\_3]^2}{[N\_2][H\_2]^3}.$$ Initially, there were $0.28$ moles for each reactant, after it reaches equilibrium, the number mol of product produced was $0.014$ mol. What is the percentage of dissociation degree?
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> 
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Through an ICE table, I calculated the number of mol of $N\_2$, $H\_2$, and $NH\_3$ at equilibrium. There are $0.273$ mol of $N\_2$ ($0.007$ mol is used), $0.259$ mol of $H\_2$ ($0.021$ mol is used), and $0.014$ mol of $NH\_3$, but I'm not sure how to calculate the percentage of dissociation degree.


My teacher taught that $\alpha=\dfrac{n\_{\text{change}}}{n\_{\text{initial}}}$ (referring to the ICE table), but I'm not sure which $n$ to use. Help would be appreciated, thank you.




For a reaction of the form:


$$\ce{aA +bB<=>cC +dD}$$


You can calculate the dissociation fraction $\alpha$ using its relationship with $K\_C$:


$$K\_C=\frac{\left(\frac{C\_{Co}}{C\_{Ao}}+\frac{c}{a}\alpha\right)^c\;\left(\frac{C\_{Do}}{C\_{Ao}}+\frac{d}{a}\alpha\right)^d}{\left(1-\alpha\right)^a\;\left(\frac{C\_{Bo}}{C\_{Ao}}-\frac{b}{a}\alpha\right)^b}\;\left(C\_{Ao}\right)^{\Delta n}$$


In your case, the relevant reaction is:


$$\ce{3H2(g) + N2(g)<=>2NH3(g)}$$


Where:


$\ce{H2}$ is the limiting reagent A


$\ce{N2}$ is B


$\ce{NH3}$ is C


Using the molar balance equation for $\ce{NH3}$, you can calculate the change in moles $x$ that takes place after equilibrium is reached:


$$N\_C=N\_{Co}+cx\implies x=\frac{N\_C-N\_{Co}}{c}=\frac{0.014-0}{2}=0.007$$


Substituting $x$ into the molar balances of $A$ and $B$:


$$N\_A=N\_{Ao}-ax=0.28-3x=0.259$$


$$N\_B=N\_{Bo}-bx=0.28-x=0.273$$


Since it is mentioned that volume is $\pu{1L}$, all amounts are equal to their respective concentrations, and we can calculate $K\_C$:


$$K\_C=\frac{C\_C^c}{C\_A^a\;C\_B^b}=\frac{N\_C^c}{N\_A^a\;N\_B^b}=\frac{(0.014)^2}{(0.259)^3(0.273)}=0.04132$$


By simplifying and using the equation that relates $K\_C$ and $\alpha$, we can calculate the latter:


$$K\_C=\frac{\left(\frac{2}{3}\alpha\right)^2}{\left(1-\alpha\right)^3\;\left(1-\frac{1}{3}\alpha\right)}\;\left(0.28\right)^{-2}=0.04132$$


Solving for $\alpha$:


$$\alpha=0.075$$


Or equivalently:


$$\alpha=7.5\%$$


Alternatively, you can also use your teacher's equation, although I would argue it's incomplete since it does not account for the molar coefficient $a$ of the limiting reagent A. The complete form would be:


$$\alpha=\frac{ax}{C\_{Ao}}$$


We can verify we get the same result:


$$\alpha=\frac{(3)(0.007)}{0.28}=0.075=7.5\%$$


However, the wording of the problem is in conflict with the definition of $\alpha$, since $\ce{NH3}$ should be dissociating into its constituent gases from a non-zero initial amount with no initial amounts of $\ce{H2}$ or $\ce{N2}$ present, not vice-versa.


Perhaps your teacher overlooked this when designing the problem.


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