Problem 51 - Chemistry The Molecular Nature of Matter

Question

The heterogeneous reaction \(2 \mathrm{HCl}(g)+\mathrm{I}_{2}(s) \rightleftharpoons\) \(2 \mathrm{HI}(g)+\mathrm{Cl}_{2}(g)\) has \(K_{\mathrm{c}}=1.6 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). Suppose \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl}\) and solid \(\mathrm{I}_{2}\) are placed in a \(1.00 \mathrm{~L}\) container. What will be the equilibrium concentrations of HI and \(\mathrm{Cl}_{2}\) in the container?

Step-by-Step Solution

Answer

The equilibrium concentrations are \( [\mathrm{HI}] = 3.18 \times 10^{-12} \) M and \( [\mathrm{Cl}_2] = 1.59 \times 10^{-12} \) M.

Step 1: Write the equilibrium expression

For the reaction \( 2 \mathrm{HCl}(g) + \mathrm{I}_{2}(s) \rightleftharpoons 2 \mathrm{HI}(g) + \mathrm{Cl}_{2}(g) \), write the expression for the equilibrium constant \( K_c \):\[ K_c = \frac{[\mathrm{HI}]^2[\mathrm{Cl}_2]}{[\mathrm{HCl}]^2} \]Since \( \mathrm{I}_2 \) is a solid, its concentration is not included in the expression.

Step 2: Set up the ICE table

Set up an ICE (Initial, Change, Equilibrium) table to keep track of the molar concentrations throughout the reaction. Initial concentrations are \( [\mathrm{HCl}] = 0.100 \) M, \( [\mathrm{I}_2] = \) solid (not included), \( [\mathrm{HI}] = 0 \) M, and \( [\mathrm{Cl}_2] = 0 \) M. During the reaction, the concentrations change by \( -2x \) for \( HCl \), \( +2x \) for \( HI \), and \( +x \) for \( Cl_2 \). At equilibrium, the concentrations are \( 0.100 - 2x \), \( 2x \), and \( x \), respectively.

Step 3: Plug the equilibrium concentrations into the expression

Substitute the equilibrium concentrations into the equilibrium expression: \[ K_c = \frac{(2x)^2(x)}{(0.100 - 2x)^2} \]Plug in the value for \( K_c \): \[ 1.6 \times 10^{-34} = \frac{4x^3}{(0.100 - 2x)^2} \].

Step 4: Solve for x

To solve for \( x \), assume that \( 2x \) is much smaller than \( 0.100 \) due to the very small value of \( K_c \). This gives the approximation that \( (0.100 - 2x)^2 \approx (0.100)^2 \), simplifying the equation to: \[ 1.6 \times 10^{-34} = \frac{4x^3}{0.010000} \]Solving for \( x \) yields: \[ x^3 = \frac{1.6 \times 10^{-34} \times 0.010000}{4} \] \[ x^3 = 4.0 \times 10^{-36} \] \[ x = \sqrt[3]{4.0 \times 10^{-36}} \]

Step 5: Calculate the equilibrium concentrations

Calculate the value for \( x \) and then determine the equilibrium concentrations of \( HI \) and \( Cl_2 \), which are \( 2x \) and \( x \), respectively: \[ x = 1.59 \times 10^{-12} \] \[ [\mathrm{HI}] = 2(1.59 \times 10^{-12}) \] \[ [\mathrm{Cl}_2] = 1.59 \times 10^{-12} \]

Key Concepts

ICE Table Method

The ICE table method is an organized way to manage the changes in concentrations of reactants and products as a chemical system reaches equilibrium. ICE stands for Initial, Change, and Equilibrium. To use this method, you begin by noting the initial molar concentrations of the reactants and products before any reaction occurs. Equilibrium concentrations are unknown and so we often represent the changes in concentration that occur as the reaction proceeds with variables such as 'x'.

For instance, if a reactant concentration decreases by a certain amount, we record this as '-x', and if a product concentration increases, this would be '+x'. The purpose of the ICE table is to relate these changes to the initial concentrations and then apply those values to the equilibrium constant expression to solve for 'x' and find the final equilibrium concentrations of all species involved. It is particularly useful for visual learners and helps ensure that no steps are missed during complex calculations.

Equilibrium Constant Expression

The equilibrium constant expression is derived from the law of mass action and plays a pivotal role in understanding the extent of a reaction. It is a mathematical representation that relates the concentrations of the products raised to their coefficients in the balanced equation to the concentrations of the reactants raised to their coefficients. For a given reaction at a specific temperature, the equilibrium constant is a fixed value.

When writing the equilibrium constant expression, it's important to note that concentrations of pure solids and liquids are omitted because their concentrations are effectively constant and do not change during the reaction. In gas and solution phase reactions, the equilibrium constant is denoted as Kc, where 'c' stands for concentration. The expression for the equilibrium constant provides a quantitative measure of the position of equilibrium and allows for calculations required to predict the outcome of reactions upon changes in conditions.

Solving Equilibrium Problems

Solving equilibrium problems often involves a mixture of qualitative understanding and quantitative calculation. If a reversible reaction is at equilibrium and the equilibrium constant is known, you can determine the concentrations of all species at equilibrium by setting up and solving the equilibrium constant expression.

The steps usually include writing the balanced chemical equation, expressing the equilibrium constant, implementing the ICE table to determine changes in concentration, substituting equilibrium concentrations into the equilibrium expression to solve for unknown variables, and finally, evaluating the approximations made. In cases where the equilibrium constant is very small or very large, simplifications can often be made, such as assuming the change in concentration of reactants is negligible. This allows for easier calculation and is justified when the equilibrium lies far to one side (as it does in our provided example).

Heterogeneous Reaction Equilibrium

Heterogeneous reaction equilibrium involves reactions where the reactants and products are in different phases, such as a solid reacting with a gas, as in the provided exercise. In these situations, the equilibrium constant expression includes only the concentrations of the gaseous and aqueous species since the concentration of pure solids and liquids remain constant throughout the reaction. This simplifies the expression and the calculations.

It's important to remember that even though these components are not present in the equilibrium expression, they still participate in the reaction and are essential for the reaction equilibrium. However, they do not affect the numerical value of the equilibrium constant. Understanding heterogeneous equilibria is crucial since many industrial and environmental processes involve reactions between different phases.