# Explanation of calculating equilibrium constant

Hydrogen iodide is a colourless gas that will decompose into colourless hydrogen gas and purple iodine gas according to the following endothermic reaction.

$$\ce{2 HI (g) <=> H2 (g) + I2 (g)}$$

A $\pu{1.0 L}$ glass container was filled with $\pu{0.60 mol}$ of hydrogen iodide gas. When equilibrium was established, there were $\pu{0.25 mol}$ of iodine gas present in the container. Calculate the equilibrium constant for this reaction. $$\begin{array}{|c|c|c|c|}\hline &\ce{HI}&\ce{H2}&\ce{I2}\\\hline \text{Initial}&0.60&0&0\\\hline \text{Used/made}&0.50&0.25&0.25\\\hline \text{Equilibrium}&0.10&0.25&0.25\\\hline \end{array}$$ $$K=\frac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2}=\frac{0.25\times0.25}{0.10^2}=6.25$$

This is what I understand:

1. Product always goes on the top, reactant on bottom
2. Balancing coefficients eg $\ce{2HI}$ become powers $\ce{-> HI^2}$
3. Initially, 0.6 moles of hydrogen iodide gas was placed, then at equilibrium 0.25 moles of iodine gas.
4. If there was 0.25 moles of iodine gas, there has to be 0.25 moles of hydrogen gas aswell thus the top part of the fraction becomes $0.25 \times 0.25$
5. I have no idea where the 0.1 moles come from …

I don't want to 'understand' these as much as I just want to be able to do them.

Your reasoning is correct. Initially, you saw that there was no iodine in the vessel. After reaching equilibrium, you saw 0.25 mol. Where did this come from?

Answer: Hydrogen iodide broke down to form this iodine and hydrogen. How much hydrogen iodide? Well this is a super simple moles problem.

0.25 moles of iodine was formed, and the mole ratio for $\ce{HI:I_2}$ is $2:1$. For 0.25 moles of iodine to be formed, 0.5 moles of hydrogen iodide must have been used (just look at your balanced equation).

If 0.5 moles of hydrogen iodide was used, how much hydrogen iodide is left? 0.60 moles minus 0.50 moles to leave 0.10 mol.

Then, as you did correctly, you plug it into the $K_c$ formula.

While you have stated that you just want to know how to do it (which you should be able to now), it is useful to know what's going on. Basically, the $K_c$ is a measure of where the equilibrium is. If your $K_c$ value is high, there will be a higher concentration of the products at equilibrium.

You’re pretty good up to step 4. However, your step 5 is somewhat premature:

1. If I have generated $0.25~\mathrm{mol}$ of $\ce{H2}$ and $\ce{I2}$ (each), how much of the original $\ce{HI}$ did I use up?

2. If I used that much $\ce{HI}$, how much of the original portion is remaining at equilibrium.

3. Now that I have amount $x$ of $\ce{HI}$, enter it all into the equation.

4. Calculate, write the answer, score in the exam.

5. Profit.

I’ll leave it to you to decide which of my steps your step 5 corresponds to.