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. $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})$$ A mixture has \(\left|\mathrm{I}_{2}\right|=0.020 \mathrm{mol} / \mathrm{L}\) and \(|\mathrm{I}|=\) \(2.0 \times 10^{-8} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilibrium (at \(500 \mathrm{K}\) )? If not, which way must the reaction proceed to reach equilibrium?

Step-by-Step Solution

Answer

The reaction is not at equilibrium; it must proceed forward (towards products) to reach equilibrium at 500 K.

Step 1: Write the Expression for Qc

For the reaction \( \mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \), the reaction quotient \( Q_c \) is defined as: \[ Q_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \] where \( [\mathrm{I}] = 2.0 \times 10^{-8} \mathrm{mol/L} \) and \( [\mathrm{I}_2] = 0.020 \mathrm{mol/L}. \) Substituting these values in gives us \( Q_c = \frac{(2.0 \times 10^{-8})^2}{0.020}. \)

Step 2: Calculate Qc

Calculate \( Q_c \) using the expression: \[ Q_c = \frac{(2.0 \times 10^{-8})^2}{0.020} = \frac{4.0 \times 10^{-16}}{0.020} = 2.0 \times 10^{-14}. \]

Step 3: Compare Qc with Kc

The value of \( K_c \) is given as \( 5.6 \times 10^{-12} \). We have calculated \( Q_c = 2.0 \times 10^{-14} \). Since \( Q_c < K_c \), it indicates that the reaction quotient is less than the equilibrium constant.

Step 4: Determine Direction of Reaction

Since \( Q_c < K_c \), the reaction proceeds in the forward direction to reach equilibrium. This means more \( \mathrm{I}_2 \) will dissociate into \( \mathrm{I} \) atoms until equilibrium is reached.

Key Concepts

Reaction Quotient (Qc)

The reaction quotient, symbolized as \( Q_c \), is a measure that helps determine whether a chemical reaction is at equilibrium at a given moment. It is especially useful for reactions that are still ongoing or have not yet reached equilibrium. In a chemical equation, \( Q_c \) compares the concentration of reactants and products at any moment in time. The calculation involves putting the concentrations of the products over the reactants, each raised to the power of their coefficients from the balanced equation.

• For the dissociation of iodine molecules \( \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{I}(g) \), the expression for \( Q_c \) is \( \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \).
• If \( Q_c = K_c \), the reaction is at equilibrium.
• If \( Q_c > K_c \), the reaction goes backward to reach equilibrium.
• If \( Q_c < K_c \), as in this exercise, the reaction continues forward.

Thus, understanding \( Q_c \) helps predict the direction of the reaction needed to reach equilibrium.

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into simpler components. In the case of the iodine dissociation reaction given by \( \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{I}(g) \), it describes the process where iodine molecules split into individual iodine atoms.

Dissociation is important in understanding reaction mechanisms as it often plays a crucial role in both chemical and physical changes. For gases, like iodine in this example, it may involve changes in pressure and concentration. It’s an endothermic process, meaning it requires energy to proceed:

• Energy input is needed to break the \( \mathrm{I}_2 \) bonds.
• As iodine dissociates, more individual iodine atoms are formed.

This type of reaction is reversible, and reaching equilibrium depends on variables such as temperature and pressure.

Equilibrium Constant (Kc)

The equilibrium constant \( K_c \) is an important value that indicates the balance point of a reversible reaction at equilibrium, when concentrations of reactants and products remain constant. It is determined by the specific reaction at a constant temperature. In the example problem, \( K_c \) is given as \( 5.6 \times 10^{-12} \).

This constant tells you several things about the reaction:

• If \( K_c \) is much greater than one, products are favored at equilibrium.
• If \( K_c \) is much less than one, like our example, reactants are favored.
• For equality, if \( K_c \) is around one, then neither side is strongly favored.

Understanding \( K_c \) in context assists prediction of the extent of reactions. Comparing \( Q_c \) and \( K_c \) informs us if the ongoing reaction needs to shift towards reactants or products to reach equilibrium.