Question

P4.18 \(\quad\) A sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) is dissolved in 225 g of water at \(298 \mathrm{K}\) such that the solution is 0.325 molar in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) A temperature rise of \(0.146^{\circ} \mathrm{C}\) is observed. The calorimeter constant is \(330 .\) J \(\mathrm{K}^{-1}\). Calculate the enthalpy of solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in water at this concentration. Compare your result with that calculated using the data in Table 4.1 (Appendix B, Data Tables).

Short answer

The enthalpy of the solution can be calculated as follows: Enthalpy of the solution = \( \frac{330 \times 0.146}{(0.325 \times 0.225)} \) Joules per mole After calculating the enthalpy of the solution, compare it to the given values in Table 4.1.

Step-by-step solution

Calculate the heat absorbed by the solution

To find the heat absorbed by the solution, we can use the following formula: \( q = C \Delta T \) Where: - \(q\) is the heat absorbed by the solution (in Joules), - \(C\) is the calorimeter constant (in J/ºC), - \(\Delta T\) is the temperature change (in ºC). We are given the calorimeter constant as 330 J/K, and the temperature change is 0.146 ºC. We can plug in these values to calculate the heat absorbed: \( q = 330 \times 0.146 \)

Calculate the number of moles of Na₂SO₄ dissolved

We are given the molarity of the solution, which is 0.325 mol/L. Molarity is defined as the number of moles of solute per liter of solution. We can use this information to find the number of moles of Na₂SO₄ dissolved: Molarity = (Number of moles of solute) / (Volume of solution in liters) We are given the mass of water used to create the solution as 225 g. We will assume the solution's volume is approximately that of the water used, and we can calculate the volume in liters as follows: Volume of solution in liters = \( \frac{Mass \, of \, water}{Density \, of \, water} \) The density of water at room temperature is approximately 1 g/mL or 1000 g/L. Therefore: Volume of solution in liters = \( \frac{225}{1000} = 0.225\, L \) Now we can plug in the molarity and volume values to find the number of moles of Na₂SO₄ dissolved: 0.325 mol/L = (Number of moles of Na₂SO₄) / (0.225 L) (Number of moles of Na₂SO₄) = (0.325 mol/L) × (0.225 L)

Calculate the enthalpy of the solution

The enthalpy of the solution is defined as the heat absorbed per mole of solute. We already found the heat absorbed and the number of moles of Na₂SO₄ in the previous two steps, so now we can find the enthalpy of solution: Enthalpy of the solution = \( \frac{Heat \, absorbed}{Moles \, of \, solute} \) Plugging in the values calculated in previous steps: Enthalpy of the solution = \( \frac{330 \times 0.146}{(0.325 \times 0.225)} \) Joules per mole Enthalpy of the solution can be compared to the given values in Table 4.1.

Key concepts

Calorimetry

Calorimetry is a branch of thermodynamics devoted to the measurement of the amount of heat released or absorbed during a chemical reaction, physical change, or heat capacity. In our exercise, a calorimeter is used to determine the heat associated with the dissolution of sodium sulfate in water. The principle behind calorimetry is energy conservation; the heat lost by the reaction (or dissolution, in this case) is gained by the calorimeter and its contents.

If we imagine the calorimeter as an isolated system, any heat produced by the dissolving solute is absorbed by the water and the calorimeter itself. By multiplying the calorimeter constant, which accounts for its heat-absorbing capacity, by the observed change in temperature, we attain the total heat absorbed by the solution. This method of thermal analysis is a cornerstone for understanding enthalpy changes in various chemical processes.

Molarity

Molarity, often symbolized as M, is a concentration unit used in chemistry to describe the amount of a substance per volume of solution. It's defined as moles of a solute per liter of solution. In physical chemistry calculations, understanding molarity is essential when dealing with reactions and solutions like in our exercise where we have a 0.325 molar solution of sodium sulfate.

The relationship between moles, molarity, and the volume of the solution simplifies the determination of how many moles of a solute are present in a given volume. By knowing the molarity, we directly relate the volume of solution to the amount of solute, which is crucial for calculating the enthalpy of solution per mole of the solute.

Thermal Analysis

Thermal analysis encompasses a range of techniques that measure the physical properties of a substance as it changes with temperature. In this context, we are particularly interested in how the temperature of a solution changes when a solute dissolves, reflecting the enthalpy change of the solution process.

The small temperature rise observed when sodium sulfate dissolves indicates that heat is absorbed from the surroundings (an endothermic process), leading to a change in enthalpy. Such thermal analysis enables us to quantify this change and understand the energetics of the dissolution process. Using the data obtained from calorimetry, we perform thermal analysis to deduce the enthalpy change for the sodium sulfate solution.

Physical Chemistry Calculations

Physical chemistry calculations allow the quantification of chemical concepts to understand and predict the behavior of systems. In this exercise, we combine various physical chemistry concepts, starting from the basic heat exchange formula in calorimetry, understanding molarity for concentration calculations, and analyzing thermal changes to quantify the enthalpy of the solution.

The enthalpy of solution is a critical value that provides insight into the energetics of a substance dissolving in a given solvent. Upon calculating the enthalpy of the solution, we achieve a deepened understanding of the interactions between solute and solvent molecules. These calculated values can be compared with standard data, enhancing our understanding of the solution's properties and guiding predictions in diverse chemical applications.