Another Equilibrium Calculation

Hydrogen and iodine gas react according to the following expression:

\(\text{H}_{2}(\text{g}) + \text{I}_{2}(\text{g})\) \(\leftrightharpoons\) \(2\text{HI}(\text{g})\)

When \(\text{0.496}\) \(\text{mol}\) \(\text{H}_{2}\) and \(\text{0.181}\) \(\text{mol}\) \(\text{I}_{2}\) are heated at \(\text{450}\)\(\text{°C}\) in a \(\text{1}\) \(\text{dm$^{3}$}\) container, the equilibrium mixture is found to contain \(\text{0.00749}\) \(\text{mol}\) \(\text{I}_{2}\). Calculate the equilibrium constant for the reaction at \(\text{450}\)\(\text{°C}\).

Step 1: Draw a RICE table

Step 2: Fill in the balanced chemical equation on the table:

Step 3: Fill in the number of moles of \(\text{H}_{2}\), \(\text{I}_{2}\) and \(\text{HI}\) present at the beginning of the reaction

There are \(\text{0.496}\) \(\text{moes}\) of \(\text{H}_{2}\), \(\text{0.181}\) \(\text{moles}\) of \(\text{I}_{2}\) and no moles \(\text{HI}\) at the beginning of the reaction.

Step 4: Fill in the change in the number of moles of \(\text{H}_{2}\), \(\text{I}_{2}\) and \(\text{HI}\) in terms of x

The mole ratio of \(\text{H}_{2}\) : \(\text{I}_{2}\) : \(\text{HI}\) is \(1:1:2\).

Therefore, for every \(\text{1}\) mole of \(\text{H}_{2}\) used, \(\text{1}\) mole of \(\text{I}_{2}\) is used and \(\text{2}\) moles of \(\text{HI}\) will be formed. If the number of moles of \(\text{H}_{2}\) decreases by x (\(-\text{1}\)x), then the number of moles of \(\text{I}_{2}\) also decreases by x (\(-\text{1}\)x) and the number of moles of \(\text{HI}\) increases by 2x (\(\text{+2}\)x).

Step 5: Fill in the number of moles of \(\text{H}_{2}\), \(\text{I}_{2}\) and \(\text{HI}\) at equilibrium in terms of x

To get the number of moles of a substance at equilibrium you take the number of moles initially, and add the change in number of moles in terms of x.

Step 6: Calculate the value of x

You know that number of moles of \(\text{I}_{2}\) at equilibrium is \(\text{0.00749}\) \(\text{mol}\). From the RICE table we can see that there are \(\text{0.181}\) - x moles of \(\text{I}_{2}\) at equilibrium. Therefore:

\(\text{0.00749}\) \(\text{mol}\) = \(\text{0.181}\) \(\text{mol}\) - x

x = \(\text{0.181}\) \(\text{mol}\) - \(\text{0.00749}\) \(\text{mol}\) = \(\text{0.1735}\) \(\text{mol}\)

Step 7: Calculate the concentration of \(\text{H}_{2}\) at equilibrium.

The volume of the container is \(\text{1}\) \(\text{dm$^{3}$}\).

From the RICE table we can see that the number of moles of \(\text{H}_{2}\) at equilibrium = \(\text{0.496}\) \(\text{mol}\) - x. Therefore:

n(\(\text{H}_{2}\)) = \(\text{0.496}\) \(\text{mol}\) - \(\text{0.1735}\) \(\text{mol}\) = \(\text{0.3225}\) \(\text{mol}\)

C(\(\text{H}_{2}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.3225}\text{ mol}}{\text{1}\text{ dm$^{3}$}} =\) \(\text{0.3225}\) \(\text{mol.dm$^{-3}$}\)

Step 8: Calculate the concentration of \(\text{I}_{2}\) at equilibrium.

C(\(\text{I}_{2}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.00749}\text{ mol}}{\text{1}\text{ dm$^{3}$}} =\) \(\text{0.00749}\) \(\text{mol.dm$^{-3}$}\)

Step 9: Calculate the concentration of \(\text{HI}\) at equilibrium.

From the RICE table we can see that the number of moles of \(\text{HI}\) at equilibrium = 2x. Therefore:

n(\(\text{HI}\)) = \(\text{2}\) x \(\text{0.1735}\) \(\text{mol}\) = \(\text{0.347}\) \(\text{mol}\)

Step 10: Complete the RICE table

Step 11: Write the expression for \(\text{K}_{\text{c}}\) for this reaction

\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[HI]}^{2}}{\text{[H}_{2}{\text{][I}}_{2}{\text{]}}}\)

Step 12: Calculate \(\text{K}_{\text{c}}\)

\(\text{K}_{\text{c}}\) \(= \dfrac{(\text{0.347})^{2}}{(\text{0.3225})(\text{0.00749})}\) = \(\text{49.85}\)