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In the synthesis of ammonia at equilibrium constant is $K_c = 1.2$ according to the reaction: $$\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g)}$$ If an equilibrium is proposed based on initial concentrations of $[\ce{H2}] = \pu{ 0.76 M}$, $[\ce{N2}] = \pu{ 0.60 M}]$, $[\ce{NH3}] = \pu{ 0.48 M}$, with $x = 0.014$ as the change in the concentration of $\ce{N2}$, what are the equilibrium concentrations?

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  • $\begingroup$ We'll guide you to solve such a problem but not just do it for you. // If the initial concentration of $\ce{N2}$ is 0.60 M, and there is a change of 0.014, what is the final concentration of $\ce{N2}$? $\endgroup$ – MaxW Apr 22 at 1:53
  • $\begingroup$ The final concentration of N2 could be 0.60M-0.014M=0.586 $\endgroup$ – Samuel Apr 22 at 2:02
  • $\begingroup$ If that is correct what should I do next? $\endgroup$ – Samuel Apr 22 at 2:04
  • $\begingroup$ What else could he final concentration of $\ce{N2}$ be? $\endgroup$ – MaxW Apr 22 at 2:04
  • $\begingroup$ To me it isn't clear if the change is positive or negative. It seems the problem statement is deliberately ambiguous. $\endgroup$ – MaxW Apr 22 at 2:13
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The given chemical reaction is:

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

So:

  • One molecule of nitrogen reacts with 3 molecules of hydrogen and the reaction yields two molecules of ammonia.

  • A mole is just a big counting unit like a dozen or a million. So one mole of nitrogen reacts with 3 moles of hydrogen and the reaction yields two moles of ammonia.

The problem states "with x = 0.014 as the change in the concentration of $\ce{N2}$". It isn't clear if the change is positive or negative. It seems the problem statement is deliberately ambiguous.

So using the stoichiometry from the chemical reaction, if nitrogen changes by $-x$, then hydrogen changes by $-3x$, and the nitrogen changes by $+2x$. Conversely if nitrogen changes by $+x$, then hydrogen changes by $+3x$, and the nitrogen changes by $-2x$.

Now for the equilibrium:

$$K_\mathrm{c} = 1.2 = \dfrac{\ce{[NH3]^2}}{\ce{[N2][H2]^3}}$$

So we can calculate $K_\mathrm{c}$ for the various compositions of the gases and see if any is equal to 1.2.

\begin{array}{|c|c|c|c|} \hline & \pu{\ce{[N2]}} & \pu{\ce{[H2]}} & \pu{\ce{[NH3]}} & \pu{calc}\ K_\mathrm{c} \\ \hline \pu{initial} & 0.600 & 0.760 & 0.48 & 0.875\\ \hline \pu{\Delta \ce{N2 = +0.014}} & 0.614 & 0.802 & 0.452 & 0.645\\ \hline \pu{\Delta \ce{N2 = -0.014}} & 0.586 & 0.718 & 0.508 & 1.19 \\ \hline \end{array}

So the composition of the gases must be equal to the last line in the table.

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    $\begingroup$ Interestingly the question poses a reaction in the gas phase, but then uses molarity. I don't think it is asked very well. $\endgroup$ – Martin - マーチン Apr 22 at 10:24
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    $\begingroup$ @Martin-マーチン - I agree. I also don't like problem specifying 2 digit concentrations when to follow the change need three digits are needed. So 0.60 molar for $\ce{N2}$ magically must turn into 0.600 molar. $\endgroup$ – MaxW Apr 22 at 10:31

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