Question

At \(1000 \mathrm{~K}\), hydrogen dissociates into gaseous atoms: $$ \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{H}(g) $$ where \(K\) is \(5.0 \times 10^{-18}\). Ten moles of hydrogen gas are pumped into an evacuated \(15.0-\mathrm{L}\) flask and heated to \(1000 \mathrm{~K}\) (a) How many atoms of \(\mathrm{H}\) are in the flask when equilibrium is reached? (b) What percent (in moles) of \(\mathrm{H}_{2}\) dissociated?

Short answer

Answer: The number of hydrogen atoms present in the flask at equilibrium is approximately \(2.7 \times 10^{13}\) atoms. The percentage of H2 dissociated is approximately \(2.25 \times 10^{-7}\%\).

Step-by-step solution

Write the balanced dissociation equation

The balanced equation for the dissociation of hydrogen gas is given by: $$ \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{H}(g) $$

Setup the ICE table

First, we must create an ICE table showing the initial concentration, change in concentration, and equilibrium concentration for each reactant and product. We are given the initial moles of H2 gas (10 moles) and the volume of the flask (15 L). Convert moles to concentration by dividing the moles by the volume. Initial concentration of H2: \(\displaystyle \frac{10\ \mathrm{moles}}{15\ \mathrm{L}} = 0.67\ \mathrm{M}\) Initially, there is no hydrogen gas (H), so its concentration is 0. | | H2 | 2H | |-------------|-------------|-------------| | Initial | 0.67 M | 0 M | | Change | -x M | +2x M | | Equilibrium | 0.67 - x M | 2x M | We're given the equilibrium constant, K = \(5.0 \times 10^{-18}\).

Write the expression of the equilibrium constant

Write the equilibrium constant expression and substitute the equilibrium concentrations from the ICE table: $$ K = \frac{[\mathrm{H}]^2}{[\mathrm{H}_{2}]} $$ Substitute the equilibrium concentrations: $$ 5.0 \times 10^{-18} = \frac{(2x)^2}{0.67 - x} $$

Solve for x

Now we need to solve the equation for x, to find the equilibrium concentrations of H2 and H: $$ 5.0 \times 10^{-18} = \frac{4x^2}{0.67 - x} $$ Since K is very small, we can assume that the value of x is negligible in comparison to 0.67. Therefore, we can simplify the equation: $$ 5.0 \times 10^{-18} \approx \frac{4x^2}{0.67} $$ Rearranging and solving for x: $$ x \approx \sqrt{\frac{5.0 \times 10^{-18} \times 0.67}{4}} \approx 1.5 \times 10^{-9} \mathrm{M} $$

Calculate the number of H atoms

At equilibrium, the concentration of H is \(2x\), which is: $$ 2x \approx 2(1.5 \times 10^{-9}\ \mathrm{M}) \approx 3.0 \times 10^{-9}\ \mathrm{M} $$ Now, calculate the number of H atoms by multiplying the concentration by the volume of the flask and Avogadro's number: $$ \mathrm{H\ atoms} = (3.0 \times 10^{-9}\ \mathrm{M})(15.0\ \mathrm{L})\left(6.022 \times 10^{23}\ \frac{\mathrm{atoms}}{\mathrm{mole}}\right) \approx 2.7 \times 10^{13}\ \mathrm{H\ atoms} $$ So, the number of H atoms in the flask when equilibrium is reached is \(\approx 2.7 \times 10^{13}\) atoms.

Calculate the percent of H2 dissociated

To find the percentage of H2 dissociated, we will use the relationship: $$ \mathrm{Percentage\ dissociated} = \frac{\mathrm{Moles\ of\ H2\ dissociated}}{\mathrm{Initial\ moles\ of\ H2}} \times 100\% $$ The moles of H2 dissociated are equal to x. $$ \mathrm{Moles\ of\ H2\ dissociated} = x \times \mathrm{Volume} \approx (1.5 \times 10^{-9}\ \mathrm{M})(15\ \mathrm{L}) \approx 2.25 \times 10^{-8}\ \mathrm{moles} $$ Substitute the values in the percentage formula: $$ \mathrm{Percentage\ dissociated} = \frac{2.25 \times 10^{-8}\ \mathrm{moles}}{10\ \mathrm{moles}} \times 100\% \approx 2.25 \times 10^{-7}\% $$ Hence, the percentage of H2 dissociated is approximately \(2.25 \times 10^{-7}\%\).