Question

The reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) has \(K_{\mathrm{p}}=45.9\) at \(763 \mathrm{~K}\). A particular equilibrium mixture at \(763 \mathrm{~K}\) contains \(\mathrm{HI}\) at a pressure of \(4.94 \mathrm{~atm}\) and \(\mathrm{H}_{2}\) at a pressure of 0.628 atm. Calculate the equilibrium pressure of \(\mathrm{I}_{2}(g)\) in this mixture.

Short answer

The equilibrium pressure of \(\mathrm{I}_{2}(g)\) in this mixture is approximately \(0.846 \thinspace atm\).

Step-by-step solution

Write the expression for the equilibrium constant, Kp.

Since the reaction is given by \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), we can write the expression for Kp as: \[K_p = \frac{P_{HI}^2}{P_{H_2} \times P_{I_2}}\] Where \(P_{HI}\), \(P_{H_2}\), and \(P_{I_2}\) are the partial pressures of \(\mathrm{HI}(g)\), \(\mathrm{H}_{2}(g)\), and \(\mathrm{I}_{2}(g)\) at equilibrium.

Substitute the given values into the expression.

We are given that \(K_p = 45.9\), \(P_{HI} = 4.94 \thinspace atm\), and \(P_{H_2} = 0.628 \thinspace atm\). Substituting these values into the Kp expression, we get: \[45.9 = \frac{(4.94)^2}{(0.628) \times P_{I_2}}\]

Solve for the equilibrium pressure of I2.

Now, we just need to solve this equation for \(P_{I_2}\): \[\begin{aligned} P_{I_2} &= \frac{(4.94)^2}{(0.628) \times 45.9} \\ P_{I_2} &= \frac{24.4}{28.8552} \\ P_{I_2} &= 0.846 \thinspace atm \end{aligned}\] Therefore, the equilibrium pressure of \(\mathrm{I}_{2}(g)\) in this mixture is approximately \(0.846 \thinspace atm\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\) for reactions involving gases, is a crucial concept in chemical equilibrium. It allows us to predict the extent to which a chemical reaction will occur under a set of conditions. In the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), the equilibrium constant \(K_p = 45.9\) at \(763 \mathrm{~K}\). This specific constant indicates that, at this temperature, the products are favored at equilibrium when compared to the reactants.

• \(K_p\) relates to the pressures of the reactants and products in a balanced chemical equation.
• The larger the \(K_p\), the more the reaction favors the formation of products.

It's important to note that \(K_p\) is only influenced by temperature changes, not by changes in the concentrations or pressures of reactants or products.

Partial Pressures

Partial pressures are essential when dealing with gases at equilibrium. Each gas in a mixture exerts its own pressure, which sums up to give the total pressure of the mixture. This individual pressure contribution is termed as partial pressure.In our reaction example,

• \(P_{HI}\) is given as \(4.94 \thinspace \mathrm{atm}\).
• \(P_{H_2}\) is given as \(0.628 \thinspace \mathrm{atm}\).

These partial pressures are plugged into the equilibrium constant expression to determine the unknown partial pressure of \(\mathrm{I}_{2}(g)\), \(P_{I_2}\).Understanding partial pressures helps us use the formula for \(K_p\) to calculate the pressures of other components in the reaction when some are known.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is similar to the equilibrium constant but applicable at any point during a reaction, not just at equilibrium. It shows the ratio of the concentrations (or pressures for gases) of products to reactants at any given moment.If you were to calculate \(Q\) for the given reaction setup, but without equilibrium, you'd follow the same approach as \(K_p\): \[Q = \frac{P_{HI}^2}{P_{H_2} \times P_{I_2}}\]Comparing \(Q\) to \(K_p\) can tell you which direction the reaction needs to shift to reach equilibrium:

• If \(Q < K_p\), the reaction will proceed forward, forming more products.
• If \(Q > K_p\), the reaction will move in reverse, forming more reactants.
• If \(Q = K_p\), the system is already at equilibrium.

Dynamic Equilibrium

Dynamic equilibrium occurs in a closed system when the forward and reverse reactions happen at the same rate. As a result, the concentrations or pressures of reactants and products remain constant over time.In the example of \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), the system reaches dynamic equilibrium at \(763 \mathrm{~K}\). Here’s what happens at dynamic equilibrium:

• Reactions continue to occur, but there is no net change in the amount of reactants or products.
• The equilibrium constant \(K_p\) is maintained.
• The rates of the forward and backward reactions are equal.

Understanding dynamic equilibrium is crucial because it underlies how systems respond to external changes, maintaining a balance without halting the reaction process.