Problem 43 - Chemistry The Molecular Science

Question

\(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask. The flask is sealed and heated to \(700^{\circ} \mathrm{C}\). The initial concentration of each gas is \(0.0088 \mathrm{~mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) has reacted when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K_{c}\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) is approximately 54.3.

Step 1: Understand the Reaction

The balanced chemical equation for the reaction is \[\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\]We need to calculate the equilibrium constant \(K_c\). The equilibrium constant expression \(K_c\) is given by:\[K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\]

Step 2: Calculate Moles of Reacted \(\mathrm{I}_2\)

Since 78.6% of \(\mathrm{I}_2\) has reacted, the amount of \(\mathrm{I}_2\) that reacted is \(0.0088 \times 0.786 = 0.0069208 \ \mathrm{mol/L}\).

Step 3: Determine Change in Concentration

For every \(1\) mole of \(\mathrm{I}_2\) that reacts, \(2\) moles of \(\mathrm{HI}\) are produced. Therefore, the change in concentration for \(\mathrm{HI}\) is \(2 \times 0.0069208 = 0.0138416 \ \mathrm{mol/L}\).

Step 4: Calculate Equilibrium Concentrations

The equilibrium concentration of \(\mathrm{I}_2\) is:\[[\mathrm{I}_2] = 0.0088 - 0.0069208 = 0.0018792 \, \mathrm{mol/L}\] The equilibrium concentration of \(\mathrm{H}_2\) decreases by the same amount as \(\mathrm{I}_2\):\[[\mathrm{H}_2] = 0.0088 - 0.0069208 = 0.0018792 \, \mathrm{mol/L}\]The equilibrium concentration of \(\mathrm{HI}\) is:\[[\mathrm{HI}] = 0 + 0.0138416 = 0.0138416 \, \mathrm{mol/L}\]

Step 5: Plug Equilibrium Concentrations into \(K_c\) Expression

Substitute the equilibrium concentrations into the \(K_c\) expression:\[K_c = \frac{(0.0138416)^2}{(0.0018792)(0.0018792)}\]Calculate \(K_c\):\[K_c = \frac{0.00019153566176}{0.00000353001664} \approx 54.3\]

Step 6: Conclusion

The equilibrium constant \(K_c\) for the reaction \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) at \(700^{\circ} \mathrm{C}\) is approximately 54.3.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential in analyzing how chemical reactions behave. When a reaction reaches equilibrium, the forward and reverse reactions occur at the same rate. This means the concentrations of all reactants and products remain constant over time, even though particles are continuously reacting. In the reaction between \(\mathrm{H}_2(g)\) and \(\mathrm{I}_2(g)\) to form \(2 \mathrm{HI}(g)\), equilibrium is reached when the rate of forming \(\mathrm{HI}\) from \(\mathrm{H}_2\) and \(\mathrm{I}_2\) is equal to the rate at which \(\mathrm{HI}\) decomposes back to \(\mathrm{H}_2\) and \(\mathrm{I}_2\).
The equilibrium constant, \(K_c\), provides a quantifiable measure of this balance. It is calculated using the concentrations of reactants and products at equilibrium. A large \(K_c\) indicates a higher concentration of products compared to reactants, while a small \(K_c\) suggests more reactants than products are present at equilibrium. In this problem, the \(K_c\) value reflects how effectively \(\mathrm{HI}\) is formed at 700 degrees Celsius.

Reaction Stoichiometry

Stoichiometry in chemistry is about understanding the quantitative relationships between the reactants and products in a chemical reaction. It helps us determine how much of each reactant is needed and how much product can be formed, based on the balanced chemical equation.

• In the reaction \(\mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)\), the coefficients tell us that 1 mole of \(\mathrm{H}_2\) reacts with 1 mole of \(\mathrm{I}_2\) to produce 2 moles of \(\mathrm{HI}\).
• When 78.6% of \(\mathrm{I}_2\) has reacted, stoichiometry allows us to calculate how many moles of each substance are present at equilibrium and determine the changes in their concentrations.
• For every mole of \(\mathrm{I}_2\) that reacts, twice as much \(\mathrm{HI}\) is formed, highlighting the stoichiometric relationship, which is crucial for any concentration calculation.

Stoichiometry ensures that the proportions maintained in the reaction are correctly applied, providing accurate data for further calculations.

Concentration Calculation

Concentration calculations are key in finding equilibrium constants and analyzing reactions. These calculations uncover how much of each substance is present in a reaction mixture at equilibrium.
Initially, both \(\mathrm{H}_2\) and \(\mathrm{I}_2\) start with a concentration of 0.0088 mol/L. With 78.6% of \(\mathrm{I}_2\) reacting, we calculate the change in concentration, which is the amount that transformed into \(\mathrm{HI}\). This is done by multiplying the initial concentration by the percentage reacted.
The formation of \(\mathrm{HI}\) involves doubling the moles of \(\mathrm{I}_2\) reacted because of the stoichiometry: one mole of \(\mathrm{I}_2\) produces two moles of \(\mathrm{HI}\). This leads us to the final concentrations:

• \( [\mathrm{H}_2] = [\mathrm{I}_2] = 0.0018792 \, \mathrm{mol/L} \)
• \( [\mathrm{HI}] = 0.0138416 \, \mathrm{mol/L} \)

These concentrations are then plugged into the \(K_c\) formula to find the equilibrium constant, completing our analysis.