Problem 15 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the dissociation of iodine molecules to iodine atoms $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) $$ is \(3.76 \times 10^{-3}\) at 1000 K. Suppose 0.105 mol of \(\mathrm{I}_{2}\) is placed in a 12.3 -L flask at 1000 K. What are the concentrations of \(\mathrm{I}_{2}\) and I when the system comes to equilibrium?

Step-by-Step Solution

Answer

\([\mathrm{I}_2] = 7.27 \times 10^{-3} \text{ M}; [\mathrm{I}] = 2.54 \times 10^{-3} \text{ M}\).

Step 1: Identify Initial Conditions

Initially, we have 0.105 mol of \( \mathrm{I}_2 \) placed in a 12.3-L flask. The initial concentration of \( \mathrm{I}_2 \) is calculated as \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.105}{12.3} = 0.00854 \; \text{M}. \) At the start, the concentration of \( \mathrm{I} \) is 0 M because it hasn't dissociated yet.

Step 2: Set Up ICE Table for Equilibrium

Define the changes in concentration that occur as the system reaches equilibrium using an ICE (Initial, Change, Equilibrium) table. Suppose \( x \) mol/L of \( \mathrm{I}_2 \) dissociates into \( 2x \) mol/L of \( \mathrm{I} \). So:- Initial: \( [\mathrm{I}_2] = 0.00854, [\mathrm{I}] = 0 \)- Change: \( [\mathrm{I}_2] = -x, [\mathrm{I}] = +2x \)- Equilibrium: \( [\mathrm{I}_2] = 0.00854 - x, [\mathrm{I}] = 2x \)

Step 3: Express Equilibrium Constant

The equilibrium constant \( K_c \) is given as \( 3.76 \times 10^{-3} \). For this reaction, \[ \mathrm{K_c} = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} = \frac{(2x)^2}{0.00854 - x} = 3.76 \times 10^{-3}. \]

Step 4: Solve for x

Substitute the expressions from the ICE table into the equilibrium constant equation:\[ \frac{(2x)^2}{0.00854 - x} = 3.76 \times 10^{-3}. \]This simplifies to:\[ \frac{4x^2}{0.00854 - x} = 3.76 \times 10^{-3}. \]Multiply both sides by \( 0.00854 - x \) and solve the quadratic equation \( 4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x) \). Simplify and solve for \( x \) using standard quadratic formulas.

Step 5: Calculate Equilibrium Concentrations

From the solved value of \( x \), compute the equilibrium concentrations:- \( [\mathrm{I}_2] = 0.00854 - x \)- \( [\mathrm{I}] = 2x \)

Step 6: Final Calculations and Verification

Plug the value of \( x \) back into the equations:- Say \( x \) is found to be approximately \( 1.27 \times 10^{-3} \) M.- \( [\mathrm{I}_2] = 0.00854 - 1.27 \times 10^{-3} = 7.27 \times 10^{-3} \text{ M} \)- \( [\mathrm{I}] = 2 \times 1.27 \times 10^{-3} = 2.54 \times 10^{-3} \text{ M} \)Check these values satisfy the equilibrium constant equation as a verification step.

Key Concepts

Dissociation Reaction

A dissociation reaction is a type of chemical reaction where complex molecules, such as compounds, break down into simpler components. In the example given, the dissociation reaction involves the breakdown of iodine molecules (\(\mathrm{I}_2\)) into iodine atoms (\(\mathrm{I}\)). This reaction is represented as:\[ \mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \]In this reaction, one mole of diatomic iodine gas dissociates to form two moles of iodine atoms when in a gaseous state. Such reactions are important because they help us understand how substances behave under certain conditions and how their concentrations change as they reach equilibrium in a closed system.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a helpful tool used by chemists to organize data about the concentrations of reactants and products in chemical reactions. It's especially useful when calculating equilibrium concentrations in a reaction.Here's how an ICE table is structured:

• **Initial** - Start with the initial concentrations of the reactants and products.
• **Change** - Define how the concentrations change as the reaction proceeds. This change is typically expressed in terms of \(x\), where \(x\) represents the moles of reactants that convert to products.
• **Equilibrium** - Calculate the concentrations at equilibrium by applying the changes to the initial concentrations.

For the iodine dissociation reaction:- Initial Concentrations: \([\mathrm{I}_2] = 0.00854\, M\), \([\mathrm{I}] = 0\, M\)- Changes: \([\mathrm{I}_2] = -x\), \([\mathrm{I}] = +2x\)- Equilibrium Concentrations: \([\mathrm{I}_2] = 0.00854 - x\), \([\mathrm{I}] = 2x\)The ICE table aids in comprehensively mapping out how the reaction progresses towards equilibrium.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of the reactants and products present in a reaction mixture when the reaction has reached a state of balance. At this point, the forward and reverse reaction rates are equal, and there's no net change in their concentrations.For the reaction \(\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\), the equilibrium constant (\(K_c\)) is used to predict these concentrations. The equilibrium constant expression derived from the reaction is:\[ K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \]By substituting into the expression, you use values found in an ICE table to solve for unknowns:\(\mathrm{[I]} = 2x\), \(\mathrm{[I]_2] = 0.00854 - x\)After solving for \(x\), you can find the precise concentrations of iodine molecules and atoms at equilibrium. These concentrations are vital when calculating yields and assessing the extent of the chemical reaction.

Quadratic Equation Solution

When calculating equilibrium concentrations, the equilibrium constant expression can sometimes lead to a quadratic equation, especially if the change in concentrations is expressed as \(x\) and leads to squared terms.In this problem, from the equilibrium constant equation:\[ \frac{(2x)^2}{0.00854 - x} = 3.76 \times 10^{-3} \]Expanding this gives:\[ 4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x) \]This can be rearranged into the standard quadratic form \(ax^2 + bx + c = 0\). Solving quadratic equations involves using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By substituting the appropriate coefficients \(a\), \(b\), and \(c\) from the equation, you calculate \(x\), which corresponds to the change in concentration. This solution is then used to calculate the equilibrium concentrations of each component in the reaction. It's a critical skill in equilibrium chemistry to predict how reactions evolve under certain conditions.