Question

Hydrogen iodide decomposes according to the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ A sealed 1.50-L container initially holds \(0.00623 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), \(0.00414 \mathrm{~mol}\) of \(\mathrm{I}_{2}\), and \(0.0244 \mathrm{~mol}\) of \(\mathrm{HI}\) at \(703 \mathrm{~K}\). When equilibrium is reached, the concentration of \(\mathrm{H}_{2}(g)\) is \(0.00467 \mathrm{M}\). What are the concentrations of \(\mathrm{HI}(g)\) and \(\mathrm{I}_{2}(g) ?\)

Short answer

Equilibrium concentrations: \( [\mathrm{HI}] = 0.0153 \mathrm{~M} \), \( [\mathrm{I}_2] = 0.00328 \mathrm{~M} \).

Step-by-step solution

Write the balanced chemical equation

The balanced equation for the decomposition of hydrogen iodide is: \[ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \].

Convert initial moles to concentrations

Use the volume of the container (1.50 L) to convert the initial moles of each substance to molar concentrations: Initial concentration of \( \mathrm{H}_{2} \): \[ \frac{0.00623 \mathrm{~mol}}{1.50 \mathrm{~L}} = 0.00415 \mathrm{~M} \] Initial concentration of \( \mathrm{I}_{2} \): \[ \frac{0.00414 \mathrm{~mol}}{1.50 \mathrm{~L}} = 0.00276 \mathrm{~M} \] Initial concentration of \( \mathrm{HI} \): \[ \frac{0.0244 \mathrm{~mol}}{1.50 \mathrm{~L}} = 0.0163 \mathrm{~M} \]

Set up the ICE table

Create an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations: \[ \begin{array}{|c|c|c|c|} \hline & \mathrm{HI}(g) & \mathrm{H}_{2}(g) & \mathrm{I}_{2}(g) \ \hline \text{Initial (M)} & 0.0163 & 0.00415 & 0.00276 \ \hline \text{Change (M)} & -2x & +x & +x \ \hline \text{Equilibrium (M)} & 0.0163-2x & 0.00415+x & 0.00276+x \ \hline \end{array} \]

Determine the change in concentration (x)

At equilibrium, the concentration of \( \mathrm{H}_{2} \) is given as 0.00467 \mathrm{M}. Use this information to find \(x\): \[ 0.00415 + x = 0.00467 \] Solving for \(x\) gives: \[ x = 0.00467 - 0.00415 = 0.00052 \]

Calculate equilibrium concentrations

Using the value of \(x\), calculate the equilibrium concentrations for \( \mathrm{HI} \) and \( \mathrm{I}_{2} \): Equilibrium concentration of \( \mathrm{HI} \): \[ 0.0163 - 2x = 0.0163 - 2(0.00052) = 0.0153 \mathrm{~M} \] Equilibrium concentration of \( \mathrm{I}_{2} \): \[ 0.00276 + x = 0.00276 + 0.00052 = 0.00328 \mathrm{~M} \]

Key concepts

ICE table

When dealing with chemical equilibria, it's essential to measure how concentrations change as the system reaches equilibrium. This can be effectively done using an ICE table. ICE stands for Initial, Change, and Equilibrium.

An ICE table helps organize initial concentrations, changes in concentration, and equilibrium concentrations in a structured manner. Initially, we fill in the table with known concentrations and describe the change in terms of an unknown variable (often x). As the reaction progresses, the concentrations of reactants decrease, while those of the products increase, moving towards a balance.

Visualizing these changes helps in better understanding and calculating equilibrium concentrations accurately. For example, in our exercise, we used the ICE table to track the decomposition of hydrogen iodide (HI) into hydrogen (H₂) and iodine (I₂), allowing us to find the equilibrium concentrations of each species clearly.

Equilibrium concentrations

Equilibrium concentrations refer to the amounts of reactants and products present in a reaction mixture when the forward and reverse reactions occur at the same rate. This state of balance means there's no net change in the concentrations over time.

To determine these concentrations, we start with known initial conditions and use the changes observed as the system moves towards equilibrium. This involves solving algebraic expressions derived from the balanced chemical equation and observed changes.

In our exercise, we initially knew the concentration of H₂ at equilibrium. By plugging this value into our ICE table, we calculated the equilibrium concentrations of both HI and I₂, showing how the mixture balances out.

Molar concentration

Molar concentration (denoted as M or mol/L) represents the number of moles of a solute dissolved in one liter of solution. It's a vital concept in chemistry as it allows us to quantify the amount of reactants and products in a solution and helps in making precise calculations for reactions.

To calculate molar concentration, divide the moles of the substance by the volume of the solution in liters. For instance, in the given problem, the molar concentration of HI, H₂, and I₂ was determined by dividing the mole values by the volume of the container (1.50L). This step is crucial while setting up our ICE table since we needed initial molar concentrations to track and compute equilibrium states accurately.

Reaction stoichiometry

Reaction stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It’s based on the balanced chemical equation and helps us understand how the quantities of different substances are related.

For the decomposition of hydrogen iodide, given by the balanced equation \[ 2 \text{HI}(g) \rightleftharpoons \text{H}_{2}(g) + \text{I}_{2}(g) \], the stoichiometric coefficients tell us that 2 moles of HI decompose to produce 1 mole of H₂ and 1 mole of I₂. Stoichiometry is crucial in setting up the 'Change' row in our ICE table, where we incorporate these coefficients to reflect how much each species increases or decreases as the reaction proceeds towards equilibrium.

By correctly factoring in reaction stoichiometry, we ensure accurate calculations of equilibrium concentrations, critical to understanding the final balanced state of the reaction.