Problem 31 - Chemistry and Chemical Reactivity

Question

Assume you mix \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{CsOH}\) with \(50.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) in a coffee-cup calorimeter. The following reaction occurs: $$\mathrm{CsOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CsCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The temperature of both solutions before mixing was \(22.50^{\circ} \mathrm{C},\) and it rises to \(24.28^{\circ} \mathrm{C}\) after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all \(1.00 \mathrm{g} / \mathrm{mL}\) and the specific heat capacities of the solutions are \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

Step-by-Step Solution

Answer

The enthalpy change is \(-56.01\,\text{kJ/mol}\) for CsOH.

Step 1: Determine Total Mass of the Solution

To determine the total mass of the combined solution, we add the volumes of the solutions, assuming their densities are 1.00 g/mL. Thus, the total volume is \(100.0\,\text{mL} + 50.0\,\text{mL} = 150.0\,\text{mL}\). Since the density is \(1.00\,\text{g/mL}\), the mass is \(150.0\,\text{g}\).

Step 2: Calculate Heat Absorbed by the Solution

Use the formula \( q = mc\Delta T \) to calculate the heat absorbed by the solution, where \( m = 150.0\,\text{g}\), \( c = 4.2\,\text{J/g} \cdot \text{K} \), and \( \Delta T = 24.28^{\circ}\text{C} - 22.50^{\circ}\text{C} = 1.78^{\circ}\text{C} \). Therefore, \( q = 150.0\,\text{g} \times 4.2\,\text{J/g} \cdot \text{K} \times 1.78\,\text{K} = 1120.2\,\text{J} \).

Step 3: Identify Limiting Reactant

Calculate the moles of both reactants to identify the limiting reactant. For CsOH, \( n = 0.200\,\text{M} \times 0.1000\,\text{L} = 0.0200\,\text{mol} \). For HCl, \( n = 0.400\,\text{M} \times 0.0500\,\text{L} = 0.0200\,\text{mol} \). Both have equal moles, hence neither is in excess.

Step 4: Calculate Enthalpy Change per Mole of CsOH

The reaction generates \( 1120.2\,\text{J} \) of heat for \( 0.0200\,\text{mol} \) of CsOH. The enthalpy change per mole would be \( \Delta H = -\frac{1120.2\,\text{J}}{0.0200\,\text{mol}} = -56010\,\text{J/mol} \). Convert this to kilojoules: \( -56.01\,\text{kJ/mol} \). The sign is negative because the reaction is exothermic.

Key Concepts

Calorimetry

Calorimetry involves measuring the heat of chemical reactions or physical changes. In our exercise, calorimetry is used to determine how much heat is absorbed or released during the reaction between \(\text{CsOH}\) and \(\text{HCl}\) within a coffee-cup calorimeter. A coffee-cup calorimeter is a simple device typically used in educational settings. It's made mostly of styrofoam, providing good insulation to keep heat within the system.
To calculate the heat absorbed by the solution, the equation \( q = mc\Delta T \) is used, where:

• \( q \) is the heat absorbed or released.
• \( m \) is the mass of the substance (in grams).
• \( c \) is the specific heat capacity (in \( \text{J/g} \cdot \text{K} \)).
• \( \Delta T \) is the change in temperature (in Kelvin or degree Celsius).

This equation helps us quantify the thermal energy change that occurs during the reaction. In the exercise, we assume the solutions' specific heat capacities are \( 4.2 \text{ J/g} \cdot \text{K} \) and that temperature changes are uniform.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, thus determining the amount of product formed. Identifying the limiting reactant is crucial for understanding how much product a reaction can theoretically produce.
In the problem involving \(\text{CsOH}\) and \(\text{HCl}\), we need to find which reactant runs out first. By calculating the moles of each reactant:

• \(\text{CsOH}: 0.200 \text{ M} \times 0.1000 \text{ L} = 0.0200 \text{ mol}\)
• \(\text{HCl}: 0.400 \text{ M} \times 0.0500 \text{ L} = 0.0200 \text{ mol}\)

Both reactants have the same number of moles, so neither is limiting in this situation. This will yield equal amounts of product in the reaction since the balanced equation indicates a 1:1 mole ratio.

Heat Capacity

Heat capacity is a measure of the amount of thermal energy required to change a substance's temperature by a certain amount. It is an intrinsic property that can depend on the material's density and specific heat. For solutions, we usually deal with specific heat capacity, which refers to the energy needed to change 1 gram of the substance by 1 degree Celsius or Kelvin.
In the given exercise, even though we are dealing with a mixture of solutions, we assume a uniform specific heat capacity of \(4.2 \text{ J/g} \cdot \text{K} \) for simplicity. This means every gram of the solution requires \(4.2\) Joules to increase its temperature by \(1^\circ\text{C}\) or \(1\text{K}\).
Understanding heat capacity helps us calculate how kinematic shifts in molecular structures relate to measurable thermal output in reactions—like the temperature change observed in the calorimeter during the experiment—and it is crucial to accurately calculating the heat absorbed or released in any thermal process.