Question

For the reaction \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}), \mathrm{K}=3.76 \times 10^{-5}\) at \(1000^{\circ} \mathrm{K}\). Suppose you inject 1 mole of \(I_{2}\) into a \(2.00\) liter-box at \(1000^{\circ}\) \(\mathrm{K}\). What will be the final equilibrium concentrations of \(\mathrm{I}_{2}\) and of I?

Short answer

The final equilibrium concentrations of \(I_{2}\) and I are approximately \(0.50\,\text{M}\) and \(1.12 \times 10^{-2}\,\text{M}\), respectively, at \(1000^{\circ} K\).

Step-by-step solution

Write down the balanced chemical equation and the given information

We are given the following reaction and information: \[I_{2} (g) \rightleftarrows 2I (g)\;\;\; K = 3.76 \times 10^{-5} \text{ at } 1000^{\circ} K\] We are also given that there is 1 mole of \(I_{2}\) initially in a 2.00 L container.

Calculate the initial concentrations of the reactants and products

Using the moles and volume of the container, we can find the initial concentrations: Initial concentration of \(I_{2}\): \[\frac{1\,\text{mol}}{2.00\,\text{L}} = 0.50\,\text{M}\] Since the reaction has not started yet, we assume that there is no I initially (i.e., initial concentration of I = 0).

Set up an ICE table

The ICE table helps us keep track of the changes in concentrations as the reaction progresses. The table looks like this: \[\begin{array}{c|ccc} & I_{2} & \rightleftarrows & 2I \\ \hline \text{Initial (M)} & 0.50 & & 0 \\ \text{Change (M)} & -x & & +2x \\ \text{Equilibrium (M)} & 0.50 - x & & 2x \\ \end{array}\] Here, x represents the change in the concentration.

Write the expression for the equilibrium constant and substitute the equilibrium concentrations

The equilibrium constant K is related to the concentrations of the reactants and products at equilibrium. \[K=\frac{[I]^2}{[I_{2}]}\] Substitute the equilibrium concentrations from the ICE table into the K expression: \[K=\frac{(2x)^2}{(0.50-x)}\]

Solve for x

Substitute the given value for K and solve for x \[3.76\times 10^{-5}=\frac{(2x)^2}{(0.50-x)}\] \[0.50\cdot 3.76\times 10^{-5} = (0.50-x)(2x)^2\] To solve for x, we can first simplify the equation, and then use either quadratic formula or approximation (since K is very small) to find the value of x.

Calculate the final equilibrium concentrations

Once we have the value of x, we can substitute it back into the equilibrium expressions from the ICE table to find the final equilibrium concentrations: \[ [I_{2}]_{eq} = 0.50 - x\] \[ [I]_{eq} = 2x\] Having finished the calculations, we now have the equilibrium concentrations of both the reactant (\(I_{2}\)) and the product (I) at 1000°C.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a number that gives us insight into where the balance of a chemical reaction lies. It is defined for a reversible reaction at a specific temperature. In our given reaction \( I_{2} (g) \rightleftarrows 2I (g) \), \( K = 3.76 \times 10^{-5} \) at 1000°C, which indicates the extent to which the reactants turn into products at equilibrium.
The small value of \( K \) suggests that at equilibrium, the concentration of products is significantly lower than that of the reactants. This knowledge helps us predict how a reaction will behave without needing to do a full analysis immediately.

• If \( K > 1 \), products predominate, meaning the reaction favors product formation.
• If \( K < 1 \), reactants predominate, indicating the reaction does not proceed much towards product formation.

Remember, \( K \) is only affected by changes in temperature and is not altered by concentrations or pressures directly.

ICE Table

The ICE table is a helpful tool in chemistry for keeping track of concentrations as chemical reactions proceed to equilibrium. ICE stands for Initial, Change, and Equilibrium.
In the context of our reaction \( I_{2} (g) \rightleftarrows 2I (g) \), here's how it is employed:

• Initial: Start with the initial concentrations, e.g., 0.50 M for \( I_{2} \), and initially 0 M for I.
• Change: Determine the change in concentrations as the reaction moves towards equilibrium. Typically, this is represented with \( -x \) for reactants and \( +2x \) for products.
• Equilibrium: Write the expressions reflecting concentrations at equilibrium, like \( 0.50 - x \) for \( I_{2} \), and \( 2x \) for I.

This method is crucial as it simplifies the process of calculating unknown equilibrium concentrations from a reversible reaction.

Reaction Quotient

The reaction quotient, \( Q \), is a valuable tool that allows us to predict the direction in which a reaction must shift to reach equilibrium. It is calculated like the equilibrium constant but uses the initial concentrations or pressures.
For our reaction \( I_{2} (g) \rightleftarrows 2I (g) \), the expression of \( Q \) is similar:
\[ Q = \frac{[I]^2}{[I_{2}]}\]
Knowing \( Q \) helps us take the next steps in understanding the equilibrium status:

• If \( Q < K \), the system will move to the right, forming more products.
• If \( Q > K \), the reaction will shift left, forming more reactants.
• If \( Q = K \), the system is already at equilibrium, with no net change in concentrations.

The comparison of \( Q \) and \( K \) thus provides an immediate glimpse into the state of a reaction relative to equilibrium.

Mole Concentration

Mole concentration, often referred to as molarity, involves expressing the amount of a substance in a specific volume of solution. It is measured in moles per liter (M).
In our example, the initial mole concentration of \( I_{2} \) is calculated as follows:

• Given 1 mole of \( I_{2} \) in a 2.00-liter container
• Molarity \( = \frac{1\, \text{mol}}{2.00\, \text{L}} = 0.50\, \text{M} \)

This concentration tells us how much of a given substance is available to participate in a reaction. As concentration changes, they significantly influence the speed (rate) of reaching equilibrium. Monitoring these concentrations with the help of formulas and tools like ICE tables allows us to understand the reaction process deeply.