A student carries out the following reaction in the lab by mixing 50.0 mL of a 1.00 M solution of hydrochloric acid with 50.0 mL of a 1.00 M solution of potassium hydroxide:

\[ \text{HCl (aq) + KOH (aq) } \rightarrow \text{ KCl (aq) + H}_2\text{O (l)} \]

The average initial temperature of the two solutions was 25.00 °C. After mixing, the final temperature of the solution was 31.43 °C. The mass of the resulting solution was 100.2 g.

1. Determine the theoretical yield of KCl (in moles) for the student’s reaction.

2. Assuming that the specific heat capacity of the solutions is equal to the specific heat capacity of pure water, calculate the heat (\( q_{\text{solution}} \)) in kJ that was absorbed by the contents of the coffee cup from the reaction.

3. Calculate the value of \( \Delta H_{\text{rxn}} \) in the units kJ/mol KCl.

4. For this reaction, the theoretical value of \( \Delta H_{\text{rxn}} \) is –55.8 kJ/mol KCl. What was the student’s percent error when determining \( \Delta H_{\text{rxn}} \)?

The student's theoretical yield of KCl is 0.0500 mol. The heat absorbed (q_solution) by the solution is 2.687 kJ. The student's calculated enthalpy change (ΔH_rxn) is 53.74 kJ/mol, which results in a percent error of 196.59% compared to the theoretical value of ΔH_rxn, which is -55.8 kJ/mol.

Explanation:

1. The theoretical yield of KCl is determined by the stoichiometry of the reaction. Since both HCl and KOH are present in equal molar amounts, they will react in a 1:1 ratio according to the chemical equation HCl (aq) + KOH (aq) --> KCl (aq) + H2O (l). With 0.0500 mol of each reactant (calculated from the 50.0 mL of 1.00 M solutions), the theoretical yield of KCl is also 0.0500 mol.

2. To calculate the heat absorbed (qsolution) by the solution, use the formula q = m•c•ΔT, where m is the mass of the solution, c is the specific heat capacity (4.184 J/g°C for water), and ΔT is the change in temperature. Here, q = 100.2 g • 4.184 J/g°C • (31.43°C - 25.00°C) = 2686.892 J or 2.687 kJ.

3. To find the enthalpy change (ΔHrxn) per mole of KCl, divide the heat absorbed by the moles of KCl formed. ΔHrxn = 2.687 kJ / 0.0500 mol = 53.74 kJ/mol.

4. To calculate the percent error in the student's determination of ΔHrxn, use the formula: percent error = |(experimental value - theoretical value) / theoretical value| • 100%. Here, the percent error = |(53.74 kJ/mol)- (-55.8 kJ/mol)| / (-55.8 kJ/mol) • 100% = 196.59%.