How to calculate the equilibrium constant for nitrogen, hydrogen, and ammonia?


> 
> $\pu{2 mol}$ of $\ce{N2}$ is mixed with $\pu{6 mol}$ of $\ce{H2}$ in a closed vessel of $\pu{1 L}$ capacity. If $50\%$ of $\ce{N2}$ is converted into $\ce{NH3}$ at equilibrium, what is the value of $K\_c$ (equilibrium constant) for the reaction
>  $$\ce{N2 + 3 H2 <=> 2 NH3}?$$
> 
> 
> 


This is what I tried to solve it:


[![attempt at solution](https://i.stack.imgur.com/BE5dV.jpg)](https://i.stack.imgur.com/BE5dV.jpg)


The answer to this question is $\frac{4}{27}$ as given in the solution key.
It is only possible if the concentration of $\ce{NH3}$ is not raised to its stochiometric coefficient.
Am I doing it wrong or is the answer key wrong?




Few notes about your solution:



> 
> "Both are limiting reactant"
> 
> 
> 


You can neglect this idea in this problem, as the reaction is not a completion reaction, it is an equilibrium reaction.



> 
> Your solution indicates that the amount of $\ce{NH3}~ \text{at equilibrium}=n\_\mathrm{e}(\ce{NH3})=\pu{4mol}$
> 
> 
> 


It is not correct, it is equal $\pu{2mol}$, it can be calculated as the following:


\begin{align}
n\_\mathrm{e}(\ce{NH3}) 
&=n\_\text{ formed}(\ce{NH3})\\
&=2\times{n\_\text{ reacted}(\ce{N2})}\\
&= 2\times \left(2\times\frac{50}{100}\right)\\
&= 2\times{(1)}\\ 
&= \pu{2mol}
\end{align}


Use the molar ratio to calculate the amount of $\ce{H2}$ reacted:
\begin{align}
n\_\text{ reacted} (\ce{H2})
&= 3\times{n\_\text{ reacted}(\ce{I2})}\\
&= 3\times{(1)}\\
&= \pu{3mol}
\end{align}


Calculate The amount of $\ce{H2}$ at equilibrium:
\begin{align}
n\_\mathrm{e}(\ce{H2})
&=n\_\mathrm{I}(\ce{H2}) 
 -n\_\text{reacted}(\ce{H2})\\ 
&= 6-3\\
&=\pu{3mol} 
\end{align}


Use the following table to determine the amount of each species at equilibrium 
in order to calculate $K\_c$ :


\begin{array}{c | c c c c c}
 &\ce{N2} & + &\ce{3H2} &\ce{<=>}& \ce{2NH3} \\\hline
\text{I} &(\pu{2 mol}) & &(\pu{6 mol}) & &(\pu{0 mol}) \\
\text{C} &(\pu{-1 mol}) & &(\pu{-3 mol}) & &(\pu{2 mol}) \\
\text{E} &(\pu{1 mol}) & &(\pu{3 mol}) & &(\pu{2 mol}) 
\end{array}


Use the expression $[i]\_\mathrm{e}= \frac{n\_\mathrm{e}(i)}{V}$
to calculate the concentration of each species at equilibrium:
\begin{align}
 [\ce{N2}]\_\mathrm{e} &= \frac{\pu{1 mol}}{\pu{1 L}}=\pu{1 M}\\
 [\ce{H2}]\_\mathrm{e} &= \frac{\pu{3 mol}}{\pu{1 L}}=\pu{3 M}\\
 [\ce{NH3}]\_\mathrm{e} &= \frac{\pu{2 mol}}{\pu{1 L}}=\pu{2 M}\\
\end{align}


Substitute the concentration of each species at equilibrium in the following formula to calculate $K\_c$:
\begin{align}
K\_c &=\frac{[\ce{NH3}]^2\_\mathrm{e}}{[\ce{N2}]\_\mathrm{e}[\ce{H2}]^3\_\mathrm{e}}\\
&=\frac{(\pu{2 M})^2}{(\pu{ 1M})(\pu{3 M})^3}\\
&=\left(\frac{4}{27}\right)\cdot \pu{L^2//mol^2}
\end{align}


 0