Question

When \(25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added to \(25.0 \mathrm{~mL}\) of 1.00 \(M\) KOH in a coffee-cup calorimeter at \(23.50^{\circ} \mathrm{C}\), the temperature rises to \(30.17^{\circ} \mathrm{C}\). Calculate \(\Delta H\) in \(\mathrm{kJ}\) per mole of \(\mathrm{H}_{2} \mathrm{O}\) formed. (Assume that the total volume is the sum of the volumes and that the density and specific heat capacity of the solution are the same as for water.)

Short answer

\Delta H = -55.88 \text{ kJ/mol H2O}

Step-by-step solution

- Write the neutralization reaction

The balanced neutralization reaction between H2SO4 and KOH is:\[H2SO4 + 2 KOH \rightarrow K2SO4 + 2 H2O \]

- Calculate total volume of the solution

The total volume of the solution after mixing is the sum of the volumes of H2SO4 and KOH: \[V_{total} = 25.0 \text{ mL} + 25.0 \text{ mL} = 50.0 \text{ mL} \]

- Determine the limiting reactant

Calculate the moles of H2SO4 and KOH present initially:\[\text{Moles of H2SO4} = 0.500 \text{ M} \times 0.0250 \text{ L} = 0.0125\text{ moles}\]\[\text{Moles of KOH} = 1.00 \text{ M} \times 0.0250 \text{ L} = 0.0250 \text{ moles}\]Given the stoichiometry, 1 mole of H2SO4 reacts with 2 moles of KOH:\[0.0125 \text{ moles H2SO4} \times \frac{2 \text{ moles KOH}}{1 \text{ mole H2SO4}} = 0.025 \text{ moles KOH}\]Since we have exactly enough KOH to react with H2SO4 and vice versa, neither reactant is in excess. Hence, the reaction will consume all H2SO4 and KOH.

- Temperature change (ΔT)

The initial temperature is 23.50°C and the final temperature is 30.17°C. Calculate the temperature change: \[\Delta T = T_{final} - T_{initial} = 30.17\degree \text{C} - 23.50\degree \text{C} = 6.67\degree \text{C} \]

- Calculate heat (q)

Use the formula for heat: \[q = m \cdot c \cdot \Delta T \]Assuming the density of the solution is the same as water (1 g/mL), the mass (m) of the solution is: \[50.0 \text{ mL} \times 1 \text{ g/mL} = 50.0 \text{ g} \]The specific heat capacity (c) of the solution is assumed to be the same as water, 4.18 J/g°C. Now, calculate q: \[ q = 50.0 \text{ g} \times 4.18 \text{ J/g°C} \times 6.67\degree \text{C} = 1396.97 \text{ J} \]

- Calculate ΔH per mole of H2O formed

The reaction forms 2 moles of H2O for every 1 mole of H2SO4 reacted. Since we have 0.0125 moles of H2SO4, it forms: \[ 0.0125 \text{ moles H2SO4} \times 2 \frac{\text{moles H2O}}{1 \text{mole H2SO4}} = 0.025 \text{ moles H2O} \] Converted to kJ, we have: \[1396.97 \text{ J} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 1.39697 \text{ kJ} \] Now, calculate the ΔH per mole of H2O formed: \[ \Delta H = \frac{1.39697 \text{ kJ}}{0.025 \text{ moles H2O}} = 55.88 \text{ kJ/mol} \]

- Adjust the sign of ΔH

Since the reaction is exothermic (releases heat), ΔH should be negative: \[ \Delta H = -55.88 \text{ kJ/mol} \]

Key concepts

Neutralization Reaction

A neutralization reaction occurs when an acid and a base come together to form a salt and water. In this exercise, sulfuric acid (H₂SO₄) reacts with potassium hydroxide (KOH). The balanced equation for this neutralization is:
\[ H_2SO_4 + 2 KOH \rightarrow K_2SO_4 + 2 H_2O \]
This tells us that one molecule of sulfuric acid reacts with two molecules of potassium hydroxide to produce one molecule of potassium sulfate and two molecules of water. Neutralization reactions are usually exothermic, meaning they release heat, which causes the temperature of the solution to increase.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. A calorimeter is the device used for this measurement. In this exercise, a coffee-cup calorimeter is used, which is a simple calorimeter that consists of two Styrofoam cups. When the acid and the base are mixed in this calorimeter, it allows us to measure the temperature change of the solution accurately. The measured temperature rise from 23.50°C to 30.17°C gives a ΔT (temperature change) of 6.67°C. This temperature change lets us calculate the amount of heat exchanged during the reaction.

Specific Heat Capacity

Specific heat capacity (c) is a property of a substance that tells us how much heat energy is needed to raise the temperature of one gram of the substance by one degree Celsius. For water, the specific heat capacity is 4.18 J/g°C. In this calorimetry experiment, we assume that the specific heat capacity of the solution is the same as water. This allows us to use the formula:
\[ q = m \times c \times \triangle T \]
Here,

• q is the heat absorbed or released by the solution
• m is the mass of the solution
• ΔT is the change in temperature of the solution

This formula calculates the total heat change in the reaction based on the known properties of the solution.

Enthalpy Change

The enthalpy change (ΔH) for a reaction is the total heat content change at constant pressure. It is expressed in kilojoules per mole (kJ/mol). In our neutralization reaction, the heat absorbed or released by the solution (q) can be converted to ΔH. Since the reaction forms 0.025 moles of water, and the total heat change q is 1396.97 J (or 1.39697 kJ), the enthalpy change per mole of water is found using:
\[ \triangle H = \frac{q}{\text{moles of H}_2O} = \frac{1.39697 \text{ kJ}}{0.025 \text{ moles H}_2O} = 55.88 \text{ kJ/mol} \text{ H}_2O \]
Since this is an exothermic reaction (it releases heat), the sign of ΔH is negative:
\[ \triangle H = -55.88 \text{ kJ/mol H}_2O \]

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, limiting the amount of product formed. To find the limiting reactant, we compare the number of moles of each reactant based on the stoichiometry of the balanced chemical equation. Here, we calculate the moles of H₂SO₄ (0.0125 moles) and KOH (0.0250 moles). Using the balanced equation:
\[ H_2SO_4 + 2 KOH \rightarrow K_2SO_4 + 2 H_2O \]
For each mole of H₂SO₄, two moles of KOH are required. The amount of KOH needed for 0.0125 moles of H₂SO₄ is:
\[ 0.0125 \text{ moles H}_2SO_4 \times 2 \frac{\text{moles KOH}}{\text{mole H}_2SO_4} = 0.025 \text{ moles KOH} \]
Here, the moles of KOH match exactly with the moles required. Therefore, both reactants are completely used up with neither in excess, meaning each reactant limits the other's amount.