Problem 3 - Chemistry and Chemical Reactivity

Question

\(K_{c}=5.6 \times 10^{-12}\) at \(500 \mathrm{K}\) for the dissociation of iodine molecules to iodine atoms. $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) $$ A mixture has \(\left[\mathrm{I}_{2}\right]=0.020 \mathrm{mol} / \mathrm{L}\) and \([\mathrm{I}]=2.0 \times\) \(\left.10^{-8} \mathrm{mol} / \mathrm{L} . \text { Is the reaction at equilibrium (at } 500 \mathrm{K}\right) ?\) If not, which way must the reaction proceed to reach equilibrium?

Step-by-Step Solution

Answer

The reaction is not at equilibrium and will shift forward to form more \(\mathrm{I}(\mathrm{g})\).

Step 1: Write the Equilibrium Expression

For the reaction \(\mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g})\), the equilibrium constant expression \(K_c\) is given by: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\]

Step 2: Calculate the Reaction Quotient \(Q_c\)

The reaction quotient \(Q_c\) is calculated in a similar way to \(K_c\) but using the initial concentrations provided: \[Q_c = \frac{([\mathrm{I}])^2}{[\mathrm{I}_2]} = \frac{(2.0 \times 10^{-8})^2}{0.020}\]

Step 3: Solve for \(Q_c\)

Now calculate \(Q_c\): \[Q_c = \frac{(2.0 \times 10^{-8})^2}{0.020} = \frac{4.0 \times 10^{-16}}{0.020} = 2.0 \times 10^{-14}\]

Step 4: Compare \(Q_c\) with \(K_c\)

Compare the values of \(Q_c = 2.0 \times 10^{-14}\) and \(K_c = 5.6 \times 10^{-12}\). Since \(Q_c < K_c\), the reaction is not at equilibrium.

Step 5: Determine the Direction of Reaction Shift

Since \(Q_c < K_c\), the reaction must proceed in the forward direction (toward the formation of \(\mathrm{I}(\mathrm{g})\)) to achieve equilibrium.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \( Q_c \), is a valuable tool used to predict the direction of a chemical reaction. It involves substituting initial concentrations of reactants and products into the equilibrium expression, just like with \( K_c \). However, while \( K_c \) uses concentrations at equilibrium, \( Q_c \) uses the concentrations at any given moment.

• If \( Q_c > K_c \), the reaction must move toward the reactants to achieve equilibrium, shifting left.
• If \( Q_c < K_c \), the reaction needs to produce more products, moving right, toward equilibrium.
• If \( Q_c = K_c \), the system is already at equilibrium. No shift is needed.

In the dissociation of iodine, calculating \( Q_c \) helps us determine if the reaction has reached the equilibrium state. As seen in the exercise, initial concentrations of Iodine and its atoms are plugged into the formula to find \( Q_c \), guiding whether the system needs adjusting.

Equilibrium Constant

The equilibrium constant, \( K_c \), is a fundamental component in studying chemical equilibrium. It is defined as the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients in the chemical equation.
For the dissociation of iodine: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]This constant provides insight into the balance point of the reaction at a given temperature. A larger value of \( K_c \) indicates a reaction favoring product formation, whereas a smaller value suggests favoring reactants. The calculated \( K_c \) of iodine dissociation, \( 5.6 \times 10^{-12} \), implies that under equilibrium, the concentration of iodine atoms in gas is significantly lower than iodine molecules.

To solve the exercise, comparing \( K_c \) with \( Q_c \) allows us to determine if the current system maintains equilibrium or needs adjustments. An essential step in manipulating how \( Q_c \) relates to \( K_c \) is the key to predicting the reaction's shift direction.

Dissociation Reaction

A dissociation reaction involves the separation of a molecule into two or more smaller parts, often referred to as ions or atoms. In this scenario, iodine molecules \( \text{I}_2 \) break down into iodine atoms \( \text{I} \). This reaction can be represented as:
\[\text{I}_2(\text{g}) \rightleftarrows 2\text{I}(\text{g})\] Dissociation is common in gases and involves reversible reactions, meaning the reaction can proceed in both forward and reverse directions. The extent of dissociation depends on the temperature, pressure, and the inherent properties of the system, including the equilibrium constant.

In dissociation reactions like iodine's, the focus is often on the atom or ion concentration resulting from the original molecule's breakdown.

• Understanding equilibrium in these reactions is paramount as it dictates whether the original molecule or its dissociated parts are predominantly present.
• In our task, we observe if the defined conditions allow for an equilibrium or if adjustments are required, guided by measurements of iodine molecules versus atoms.

Such insights help in predicting and controlling reaction behaviors in practical applications.