Problem 27 - Chemistry: Atoms First

Question

The addition of 3.15 g of \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) to a solution of \(1.52 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{SCN}\) in \(100 \mathrm{g}\) of water in a calorimeter caused the temperature to fall by \(3.1^{\circ} \mathrm{C} .\) Assuming the specific heat of the solution and products is \(4.20 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}\) calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: $$\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{NH}_{4} \mathrm{SCN}(a q) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(a q)+2 \mathrm{NH}_{3}(a q)+10 \mathrm{H}_{2} \mathrm{O}(I)$$

Step-by-Step Solution

Answer

The approximate amount of heat absorbed by the reaction is found to be \( q = 104.67 \rm{g} \times 4.20 \frac{J}{g^\circ C} \times 3.1^\circ C = 1361.12 \: J \.}

Step 1: Calculate the heat absorbed

To calculate the heat absorbed by the reaction, use the formula: \( q = m \cdot c \cdot \Delta T \), where \( q \) is the heat absorbed, \( m \) is the mass of the solution and products, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Since the specific heat and change in temperature are provided, we only need to calculate the total mass of the solution and products.

Step 2: Determine the total mass

The total mass of the solution and products is the sum of the mass of the water, the mass of \( \mathrm{Ba}(\mathrm{OH})_2 \cdot 8 \mathrm{H}_2\mathrm{O} \), and the mass of \( \mathrm{NH}_4\mathrm{SCN} \). This is given by: \( 100 \mathrm{g} + 3.15 \mathrm{g} + 1.52 \mathrm{g} = 104.67 \mathrm{g} \).

Step 3: Calculate the heat absorbed by the reaction

Using the masses and specific heat capacity, calculate the heat absorbed: \( q = 104.67 \mathrm{g} \times 4.20 \mathrm{J/g^{\circ}C} \times 3.1^{\circ}\mathrm{C} = \) the heat absorbed by the reaction in Joules.

Step 4: Plug in values and solve

Now plug in the values into the formula \( q = 104.67 \mathrm{g} \times 4.20 \mathrm{J/g^{\circ}C} \times 3.1^{\circ}\mathrm{C} \) and calculate the resulting heat absorption, which will give us the heat absorbed by the reaction in Joules.

Key Concepts

Calorimetry

Calorimetry is an experimental technique that allows us to measure the amount of heat involved in a chemical or physical process. By using a device known as a calorimeter, scientists can track temperature changes in a substance as it undergoes a reaction. This temperature change, when carefully measured, corresponds to the heat absorbed or released during the process.

During a calorimetry experiment, such as the one described in the exercise, the change in temperature of the water solution helps us determine the heat exchange involved in the dissolution and chemical reaction. The experiment took place in a calorimeter, which helped to isolate the system and better measure the temperature changes and therefore the heat involved.

Enthalpy Change

Enthalpy change, often symbolized as \( \Delta H \), embodies the total heat content change within a system at constant pressure. This key term in thermochemistry signifies whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

In this case, since the temperature decreases, we can conclude that the reaction is endothermic—the heat is absorbed by the reaction from the surrounding water. The enthalpy change is a crucial piece of information about a reaction's energy dynamics and can inform us about the stability of the products relative to the reactants.

Specific Heat Capacity

The specific heat capacity (\( c \) is a physical property of materials that describes how much heat energy is required to raise the temperature of one gram of a substance by one degree Celsius. Its units are typically \( \text{J/g}^\circ \text{C} \).

Understanding the specific heat capacity is vital because it tells us how a substance behaves when exchanging heat. In the exercise, the specific heat capacity of the solution and products (\(4.20 \text{J/g}^\circ \text{C}\)) is provided. It helps to calculate the total heat absorbed by the aqueous solution when the temperature of the system decreases, which enables us to solve for the heat absorbed in the reaction.

Chemical Reactions

Chemical reactions, such as the one given in the exercise, involve the breaking and forming of chemical bonds, resulting in new substances. These reactions can be characterized by energy changes, which are often observed as changes in temperature.

Understanding the nature of chemical reactions enables us to predict the energy exchange, reactants' and products' stability, and the spontaneity of the reactions. In the given equation, the combination of \(\mathrm{Ba}(\mathrm{OH})_2 \cdot 8 \mathrm{H}_2\mathrm{O}\) and \(\mathrm{NH}_4\mathrm{SCN}\) results in new products and a net absorption of heat, as demonstrated by the decrease in temperature measured in the calorimetric experiment.