Problem 67 - General Chemistry

Question

When \(15.3 \mathrm{~g}\) of sodium nitrate, \(\mathrm{NaNO}_{3}\), was dissolved in water in a calorimeter, the temperature fell from \(25.00^{\circ} \mathrm{C}\) to \(21.56^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1071 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), what is the enthalpy change when \(1 \mathrm{~mol}\) of sodium nitrate dissolves in water? The solution process is $$ \mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) ; \Delta H=? $$

Step-by-Step Solution

Answer

The enthalpy change is \(-20.47 \mathrm{~kJ/mol}\).

Step 1: Calculate Temperature Change

First, we need to determine the change in temperature (\(\Delta T\)) of the solution. The initial temperature is \(25.00^{\circ} \mathrm{C}\) and the final temperature is \(21.56^{\circ} \mathrm{C}\). So, the change is:\[\Delta T = T_{\text{final}} - T_{\text{initial}} = 21.56^{\circ} \mathrm{C} - 25.00^{\circ} \mathrm{C} = -3.44^{\circ} \mathrm{C}\]

Step 2: Calculate Heat Absorbed by Calorimeter

The heat absorbed by the solution and calorimeter, \(q\), can be calculated using the formula\[q = C \times \Delta T\]where \(C\) is the heat capacity of the calorimeter, given as \(1071 \mathrm{~J}/{ }^{\circ} \mathrm{C}\). Plugging in the values, we get:\[q = 1071 \mathrm{~J}/{ }^{\circ} \mathrm{C} \times (-3.44^{\circ} \mathrm{C}) = -3685.44 \mathrm{~J}\]

Step 3: Calculate Molar Enthalpy Change

The calculated heat, \(q = -3685.44 \mathrm{~J}\), is for dissolving \(15.3\mathrm{~g}\) of \(\mathrm{NaNO}_{3}\). We first need to find the number of moles of \(\mathrm{NaNO}_{3}\) using its molar mass, which is approximately \(85.0 \mathrm{~g/mol}\).\[\text{Moles of } \mathrm{NaNO}_{3} = \frac{15.3 \mathrm{~g}}{85.0 \mathrm{~g/mol}} = 0.18 \text{ moles}\]To find the enthalpy change per mole, divide the heat absorbed by the number of moles:\[\Delta H = \frac{-3685.44 \mathrm{~J}}{0.18 \text{ moles}} = -20474.67 \mathrm{~J/mol}\]

Step 4: Express Enthalpy Change in kJ/mol

Finally, the enthalpy change is often expressed in kJ/mol. So, we convert \(-20474.67 \mathrm{~J/mol}\) to kJ/mol:\[\Delta H = \frac{-20474.67 \mathrm{~J/mol}}{1000} = -20.47 \mathrm{~kJ/mol}\]

Key Concepts

Calorimetry

Calorimetry is a branch of science that deals with the measurement of heat changes during chemical reactions or physical changes. When sodium nitrate dissolves in water within a calorimeter, the device allows us to observe the heat exchanges between the solution and the surroundings.

The key role of calorimetry is to measure these heat exchanges precisely. This involves using a calorimeter, a device that insulates the system from its environment to prevent any heat loss or gain to the surroundings. The insulated setup ensures that any temperature change measured is solely due to the reaction itself, making it possible to accurately calculate the enthalpy change, which is the change in heat content during a reaction at constant pressure.

To calculate the heat absorbed or released by the calorimeter, we use the formula:

• \( q = C \times \Delta T \)

where \( q \) is the heat absorbed or released, \( C \) is the heat capacity of the calorimeter, and \( \Delta T \) is the temperature change. Calorimetry is fundamental in gaining insights into heat-related properties during chemical reactions.

Sodium Nitrate Dissolution

The dissolution of sodium nitrate (\(\text{NaNO}_3\)) in water is a common example of an endothermic process, where the reaction absorbs heat from the surroundings. When 15.3 grams of sodium nitrate is introduced to water, it splits into sodium ions (\(\text{Na}^+\)) and nitrate ions (\(\text{NO}_3^-\)), which facilitates the dissolution process.

In an endothermic dissolution, the energy required to break the ionic bonds in the solid sodium nitrate and interact with the water molecules is greater than the energy released when the ions are solvated by the water. This results in heat absorption, which manifests as a drop in temperature of the solution.

During the sodium nitrate dissolution, the measurement of heat change helps us determine the enthalpy change, \( \Delta H \), for the process. The heat absorbed by the reaction is calculated, and this value is used to compute \( \Delta H \) per mole of \(\text{NaNO}_3\) using the formula:

• \[ \Delta H = \frac{q}{\text{moles of } \text{NaNO}_3} \]

This gives us a clearer understanding of the energy changes involved during dissolution.

Temperature Change

Temperature change is a crucial indicator in calorimetry experiments as it quantifies the direction and magnitude of heat flow. When sodium nitrate is dissolved in water in this exercise, the temperature decreases from \( 25.00^{\circ} \text{C} \) to \( 21.56^{\circ} \text{C} \).

This temperature change signifies that the dissolution process is endothermic, meaning the system absorbs heat from its surroundings, resulting in a temperature drop. The formula used to determine the temperature change is:

• \[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]

In this case, \( \Delta T \) is calculated as \(-3.44^{\circ} \text{C}\).

The negative value of \( \Delta T \) directly indicates that the surroundings (such as the solution and calorimeter) have lost heat to the dissolving sodium nitrate. The temperature change is directly proportional to the amount of heat absorbed, and it helps us compute the heat absorbed by the calorimeter, further reinforcing the calculations of the reaction's enthalpy change.