Problem 48 - Chemistry The Central Science In Si Units

Question

For the reaction, at \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 10.0-L flask containing an equilibrium mixture of the three gases, there are \(1.30 \mathrm{~g} \mathrm{H}_{2}\) and \(21.0 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

Step-by-Step Solution

Answer

The mass of HI in the flask is approximately 219.7 g.

Step 1: Convert Mass to Moles for H₂

First, calculate the number of moles of \ H_{2} using its molar mass. The molar mass of \ H_{2} is approximately 2.02 g/mol. Use the formula:\[ moles = \frac{mass}{molar \ mass} \]For \ H_{2}, this becomes:\[ moles \ of \ H_{2} = \frac{1.30 \, g}{2.02 \, g/mol} \approx 0.644 \ moles \]

Step 2: Convert Mass to Moles for I₂

Now, calculate the number of moles of \ I_{2} using its molar mass. The molar mass of \ I_{2} is approximately 253.8 g/mol. Use the formula:\[ moles = \frac{mass}{molar \ mass} \]For \ I_{2}, this becomes:\[ moles \ of \ I_{2} = \frac{21.0 \, g}{253.8 \, g/mol} \approx 0.0828 \ moles \]

Step 3: Find the Concentration of Reactants

Convert the moles of \ H_{2} and \ I_{2} into concentrations by dividing by the volume of the flask, which is 10.0 L:\[ [H_{2}] = \frac{0.644 \, moles}{10.0 \, L} = 0.0644 \, M \]\[ [I_{2}] = \frac{0.0828 \, moles}{10.0 \, L} = 0.00828 \, M \]

Step 4: Use Equilibrium Expression with Kc

The equilibrium expression for this reaction is given by:\[ K_{c} = \frac{[HI]^2}{[H_{2}][I_{2}]} \]Substitute the given \ K_{c} (55.3) and the concentrations calculated for \ H_{2} and \ I_{2} into this expression:\[ 55.3 = \frac{[HI]^2}{0.0644 \times 0.00828} \]

Step 5: Solve for the HI Concentration

Solve the equilibrium expression for \ [HI]^2:\[ 55.3 = \frac{[HI]^2}{0.0644 \times 0.00828} \]\[ [HI]^2 = 55.3 \times (0.0644 \times 0.00828) \ \approx 0.0295 \]Take the square root to find \ [HI]:\[ [HI] \approx \sqrt{0.0295} \approx 0.1717 \ M \]

Step 6: Convert HI Concentration to Mass

First, find the moles of \ HI by multiplying the concentration by the volume of the flask:\[ moles \ of \ HI = 0.1717 \, M \times 10.0 \, L = 1.717 \, moles \]Convert moles to mass using the molar mass of \ HI (approximately 127.9 g/mol):\[ mass \ of \ HI = 1.717 \, moles \times 127.9 \, g/mol \approx 219.7 \, g \]

Key Concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentration of reactants and products remain constant over time, although they are not necessarily equal. This state of balance is known as chemical equilibrium.

The equilibrium constant, denoted as \(K_c\), is a crucial value that helps us understand the relative concentrations of reactants and products in a chemical equilibrium. In our exercise, the chemical reaction \( ext{H}_2(g) + ext{I}_2(g) \rightleftharpoons 2 ext{HI}(g)\) has an equilibrium constant \(K_c = 55.3\) at \(700 \text{ K}\).

An equilibrium constant greater than one, like in this case, indicates that at equilibrium, the products (HI) are favored over the reactants (H₂ and I₂). However, it's important to note that \(K_c\) depends on temperature; changing the temperature will alter its value. Hence, always ensure conditions like temperature are consistent when working with equilibrium problems.

Concentration Calculation

Calculating the concentration of substances in a reaction is vital for understanding the dynamics of the system at equilibrium. Concentration is typically measured in moles per liter (M).

• First, it's essential to convert the given mass of reactants into moles. This is achieved using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
• After obtaining the moles, divide these by the volume of the container (10.0 L in our problem) to find their concentrations.
  • For \(\text{H}_2\): \( [\text{H}_2] = \frac{0.644 \text{ moles}}{10.0 \text{ L}} = 0.0644 \text{ M} \).
  • For \(\text{I}_2\): \( [\text{I}_2] = \frac{0.0828 \text{ moles}}{10.0 \text{ L}} = 0.00828 \text{ M} \).

These concentrations are then used in the equilibrium expression to find the equilibrium concentration of \([\text{HI}]\). This systematic approach ensures we account for all parts of the reaction.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It's a critical concept in predicting the outcome of reactions and involves using the balanced chemical equation to determine the proportions of substances.

In our problem, the stoichiometry of the reaction \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\) tells us that one mole of \(\text{H}_2\) reacts with one mole of \(\text{I}_2\) to produce two moles of \(\text{HI}\).

Using stoichiometry, we can link the concentrations of reactants to the concentration of products. After finding the concentration of \(\text{HI}\) using the equilibrium expression, we can determine its amount in moles and then convert this to mass as follows:

• Find moles of \(\text{HI}\) by multiplying concentration by volume: \[ \text{moles of HI} = 0.1717 \text{ M} \times 10.0 \text{ L} = 1.717 \text{ moles} \]
• Convert moles to mass using the molar mass: \( \text{mass of HI} = 1.717 \text{ moles} \times 127.9 \text{ g/mol} \approx 219.7 \text{ g} \)

Understanding stoichiometry not only helps us calculate quantities involved but also provides insights into the reaction's efficiency and feasibility.