# How to calculate the concentration of reagents and products from Kc? [closed]

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?

• 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 $\ce{N2}$? – MaxW Apr 22 at 1:53
• The final concentration of N2 could be 0.60M-0.014M=0.586 – Samuel Apr 22 at 2:02
• If that is correct what should I do next? – Samuel Apr 22 at 2:04
• What else could he final concentration of $\ce{N2}$ be? – MaxW Apr 22 at 2:04
• To me it isn't clear if the change is positive or negative. It seems the problem statement is deliberately ambiguous. – MaxW Apr 22 at 2:13

## 1 Answer

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.

• Interestingly the question poses a reaction in the gas phase, but then uses molarity. I don't think it is asked very well. – Martin - マーチン Apr 22 at 10:24
• @Martin-γγΌγγ³ - 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. – MaxW Apr 22 at 10:31