$\pu{2 mol}$ 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

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?

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    $\begingroup$ $[\ce{I2}]\mathrm{e}=(2-1)=\pu{1M}$,$[\ce{H2}]\mathrm{e}=(6-3)=\pu{3M}$,$[\ce{NH3}]\mathrm{e}=(2\times{1})=\pu{2M}$ $$k_\mathrm{C}=\frac{(2)^2}{(1)(3)^3 }=\frac{4}{27}$$ $\endgroup$ – Adnan AL-Amleh Mar 29 '19 at 16:41
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    $\begingroup$ A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$ – Martin - マーチン Apr 1 '19 at 12:41
  • $\begingroup$ @Martin-マーチン Can you Please suggest a title for this question $\endgroup$ – pra9 Apr 1 '19 at 17:14
  • $\begingroup$ I have already changed the title of the question; I am more concerned with the image, currently I am somewhat guessing what you are trying to convey. $\endgroup$ – Martin - マーチン Apr 1 '19 at 17:18
  • $\begingroup$ @Martin-マーチン Should I add that "if only 50% of N2 is used" $\endgroup$ – pra9 Apr 1 '19 at 17:26

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}

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    $\begingroup$ The answer would be better, if it were defining the quantities and then using actual equations instead of 'word equations', like amount of $\ce{NH3}$ at equilibrium would be $n_\mathrm{e}(\ce{NH3})$. That would also get rid of MathJax not breaking properly/ flowing into the margins. ($\ce{I2}$ is probably a typo?) $\endgroup$ – Martin - マーチン Apr 1 '19 at 13:43

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