Problem 68 - Chemistry: An Atoms-Focused Approach

Question

When \(4.00 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathscr{M}=80.04 \mathrm{g} /\)mol) - the active ingredient in some chemical cold packs is dissolved in \(96.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O},\) the temperature of the resulting solution is \(3.07^{\circ} \mathrm{C}\) colder than the water and ammonium nitrate were before they were mixed together. What is the value of \(\Delta H\) for the following dissolution process? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \quad \Delta H=?$$

Step-by-Step Solution

Answer

Answer: The enthalpy change (ΔH) for the dissolution process of NH4NO3 in water is -24600 J/mol.

Step 1: Calculate moles of NH4NO3

To calculate the moles of NH4NO3, use the given mass (4.00 g) and molar mass (80.04 g/mol): moles_NH4NO3 = (4.00 g) / (80.04 g/mol) = 0.0500 mol

Step 2: Determine the heat released or absorbed

The temperature change (ΔT) is given as -3.07°C. The mass of water is 96.0 g, and the specific heat capacity of water is 4.18 J/g·°C. Calculate the heat released or absorbed (q) using the formula q = mcΔT: q = (96.0 g) * (4.18 J/g·°C) * (-3.07°C) = -1230 J Since the value of q is negative, this indicates that heat is released during the dissolution process.

Step 3: Calculate the enthalpy change per mole of NH4NO3

To determine the enthalpy change (ΔH) per mole of NH4NO3, divide the heat (q) by the moles of NH4NO3: ΔH = (-1230 J) / (0.0500 mol) = -24600 J/mol

Step 4: Report the final value of ΔH

The enthalpy change (ΔH) for the dissolution process of NH4NO3 in water is -24600 J/mol.

Key Concepts

Dissolution Process

In a dissolution process, we refer to the transformation of a solute from its solid form into a solution. This process involves the interaction between the solute particles and the solvent (like water) that leads to the distribution of solute particles around the solvent molecules. For example, when ammonium nitrate (\( \text{NH}_4\text{NO}_3 \)) is dissolved in water, the crystals of \( \text{NH}_4\text{NO}_3 \) separate into ions that then spread throughout the water.Understanding dissolution is important because it can involve energy changes. When a cold pack is activated by dissolving ammonium nitrate, the solution becomes colder as a result of an endothermic reaction, where the system absorbs heat from the surroundings. This absorbed heat is needed to break the solute-solvent interactions, leading to a cooling effect. This understanding of the dissolution process helps explain why the temperature drops when the cold pack is used.

Specific Heat Capacity

Specific heat capacity refers to the amount of heat required to increase the temperature of one gram of a substance by one degree Celsius. Water, for instance, has a specific heat capacity of 4.18 J/g°C. This means 4.18 joules of energy are necessary to raise the temperature of 1 gram of water by 1°C.In the context of the exercise, the specific heat capacity of water is crucial. It allows us to calculate the heat absorbed or released during the dissolution of ammonium nitrate by using the formula: \( q = mc\Delta T \), where \( q \) is the heat change, \( m \) represents the mass of the water, \( c \) is the specific heat capacity of water, and \( \Delta T \) is the change in temperature. By plugging in the numbers, we can determine how much energy the water absorbed or released, leading to the observed temperature change.

Temperature Change

Temperature change in a reaction signifies the gain or loss of heat by a substance. In the dissolution process described in the exercise, the temperature falls by 3.07°C. This indicates that the dissolving ammonium nitrate absorbs heat from the water, causing a decrease in water temperature.This phenomenon is critical for understanding endothermic reactions, where substances absorb energy from their surroundings. The temperature change can thus be directly used to measure the amount of heat energy involved. Using formulas like \( q = mc\Delta T \) allows us to derive the heat \( q \) involved from the observed temperature change, thus helping us understand the enthalpy changes in the system. Observing how temperature changes in various reactions not only helps in chemistry problem-solving — it also provides insight into the fundamental behavior of chemicals in different environments.