Problem 55 - Chemistry The Central Science

Question

When a 6.50 -g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18\()\) the temperature rises from 21.6 to to \(37.8^{\circ} \mathrm{C}\) . (a) Calculate the quantity of heat (in kJ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} )\) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

Step-by-Step Solution

Answer

The heat released during the reaction is approximately 7.16 kJ, and the enthalpy change for the solution process is approximately 44.06 kJ/mol NaOH.

Step 1: 1. Identify the given information

We are given the following information: - Mass of sodium hydroxide (NaOH) = 6.50 g - Mass of water = 100.0 g - Initial temperature = 21.6 °C - Final temperature = 37.8 °C - Specific heat of water = 4.18 J/g·°C

Step 2: 2. Calculate the mass of the solution

To calculate the heat released, we need the total mass of the solution. Add the masses of NaOH and water to find the mass of the solution. Mass of solution = Mass of NaOH + Mass of water = 6.50 g + 100.0 g = 106.5 g

Step 3: 3. Calculate the change in temperature

Find the difference between the final and initial temperatures to determine the change in temperature. ΔT = T_final - T_initial = 37.8 °C - 21.6 °C = 16.2 °C

Step 4: 4. Calculate the heat released (q)

Use the formula for heat energy, q = mcΔT, with the mass of the solution, specific heat of water, and the change in temperature. q = (mass of solution) × (specific heat of water) × (ΔT) = (106.5 g) × (4.18 J/g·°C) × (16.2 °C) = 7159.71 J Since 1 kJ = 1000 J, convert the heat energy to kJ: q = 7159.71 J × (1 kJ / 1000 J) = 7.16 kJ (approx.)

Step 5: 5. Calculate moles of NaOH

In order to find the enthalpy change per mole of NaOH, we first need to find the number of moles present in the 6.50 g sample. To do this, we'll use the molar mass of NaOH (40.00 g/mol). Moles of NaOH = (Mass of NaOH) / (Molar mass of NaOH) = 6.50 g / 40.00 g/mol = 0.1625 mol

Step 6: 6. Calculate ∆H of the solution process (per mole of NaOH)

Now that we have both the heat released and moles of NaOH, we can find the enthalpy change per mole of NaOH. Divide the heat energy (q) by the moles of NaOH: ΔH = (Heat energy released) / (Moles of NaOH) = 7.16 kJ / 0.1625 mol = 44.06 kJ/mol The enthalpy change for the solution process is ΔH = 44.06 kJ/mol NaOH (approx.).

Key Concepts

Heat of Solution

Understanding the heat of solution is crucial when studying how substances interact with solvents. Simply put, it's the total amount of heat energy that is either absorbed or released when a substance, such as sodium hydroxide (NaOH), dissolves in a solvent like water. This process is highly dependent on the nature of both the solute and the solvent. During dissolution, bonds between solute molecules break and new interactions between solute and solvent molecules form. If the energy required to break the solute bonds is less than the energy released when the new interactions form, the dissolution is exothermic, meaning heat will be released.

In the exercise provided, a 6.50g sample of NaOH dissolves in water, resulting in an increase in the water's temperature—a clear indicator of an exothermic process. To ensure clarity for students, it could be helpful to emphasize the connection between the temperature increase and the exothermic nature of the dissolution process, as this conceptual link can often enhance understanding.

Specific Heat Capacity

At the heart of this problem lies a concept known as specific heat capacity. This fundamental property is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is crucial when calculating the heat energy change in a system. The specific heat capacity of water is famously high—approximately 4.18 J/g·°C—which means water can absorb a lot of heat without a large change in temperature.

Within the context of the exercise, when the solid NaOH dissolves and the solution's temperature rises from 21.6°C to 37.8°C, the specific heat capacity allows us to calculate the actual heat absorbed by the solution. A suggestion to improve the student's understandings would be to explain how the specific heat capacity relates to everyday experiences such as why water in a lake doesn't rapidly fluctuate in temperature.

Calorimetry

Calorimetry is the technique used to measure the heat of chemical reactions or physical changes as well as heat capacity. The coffee-cup calorimeter mentioned in the exercise is a simple, yet effective tool often used in introductory chemistry labs. It assumes that no heat is lost to the surroundings, functioning as an isolated system. The calorimeter enables the determination of the heat involved in the dissolving process of the NaOH. It's important to emphasize to students that while this is an idealized situation, in practice, some heat loss occurs, but the coffee-cup calorimeter is still a valuable learning instrument.

In further detailing this concept, one might explore how the construction of a coffee-cup calorimeter—often with styrofoam cups—helps in minimizing the heat exchange with the environment, thereby making the experiment more accurate.

Stoichiometry

Stoichiometry is a section of chemistry involving the calculation of reactants and products in chemical reactions. It is based on the principle that matter cannot be created or destroyed. The goal of stoichiometry is to predict the amounts of substances consumed and created in a given chemical reaction. In the exercise provided, stoichiometry comes into play as we convert the mass of NaOH into moles using its molar mass. These moles are then used to calculate the enthalpy change per mole of NaOH when it dissolves, providing a molar view of the heat of solution.

For better comprehension, students should be encouraged to understand the significance of stoichiometry in ensuring the balance of equations and predicting the outcomes of reactions, reinforcing the concept that each calculation is a step towards comprehending the broader picture of how reactions proceed quantitatively.