Question

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

Short answer

The enthalpy change per mole of BaSO4 formed is \(+305056 \frac{\mathrm{J}}{\mathrm{mol}}\). This indicates an exothermic reaction.

Step-by-step solution

Write the balanced chemical equation for the reaction

: We first need to write the balanced chemical equation for the reaction between Na2SO4 and Ba(NO3)2 which forms BaSO4 precipitate: \(Na_2SO_4 + Ba(NO_3)_2 \rightarrow BaSO_4(s) + 2NaNO_3\)

Calculate the number of moles of reactants

: Next, we calculate the number of moles of Na2SO4 and Ba(NO3)2 in the initial solutions: Moles of Na2SO4 = Volume (in L) × Molarity Moles of Na2SO4 = \(1.00 \mathrm{~L} \times 2.00 \mathrm{M} = 2.00 \mathrm{~mol}\) Moles of Ba(NO3)2 = Volume (in L) × Molarity Moles of Ba(NO3)2 = \(2.00 \mathrm{~L} \times 0.750 \mathrm{M} = 1.50 \mathrm{~mol}\)

Determine the limiting reactant and moles of BaSO4 formed

: From the balanced chemical equation, we see that 1 mole of Na2SO4 reacts with 1 mole of Ba(NO3)2 to form 1 mole of BaSO4. The ratio between reactants and product is 1:1:1. Thus, the limiting reactant is Ba(NO3)2. Moles of BaSO4 formed = Moles of limiting reactant (Ba(NO3)2) Moles of BaSO4 formed = \(1.50 \mathrm{~mol}\)

Calculate the total mass of the final solution

: First, we find the total volume of the final solution: Total volume = Volume of Na2SO4 solution + Volume of Ba(NO3)2 solution Total volume = \(1.00 \mathrm{~L} + 2.00 \mathrm{~L} = 3.00 \mathrm{~L}\) Next, we convert the total volume to mL and multiply it by the given density to find the total mass of the final solution: Total mass = Total volume (in mL) × Density Total mass = \((3.00 \mathrm{~L} \times 1000 \frac{\mathrm{mL}}{\mathrm{L}}) \times 2.00 \frac{\mathrm{g}}{\mathrm{mL}} = 6000 \mathrm{~g}\)

Calculate the heat released during the reaction (q)

: We use the given temperature change, specific heat capacity, and total mass of the solution to calculate the heat released during the reaction (q). q = Total mass × Specific heat capacity × Temperature change q = \(6000 \mathrm{~g} \times 6.37 \frac{\mathrm{J}}{\mathrm{g} \cdot {}^\circ \mathrm{C}} \times \left(42.0^\circ \mathrm{C} - 30.0^\circ \mathrm{C}\right)\) q = \(6000 \mathrm{~g} \times 6.37 \frac{\mathrm{J}}{\mathrm{g} \cdot {}^\circ \mathrm{C}} \times 12^\circ \mathrm{C} = 457584 \mathrm{~J}\)

Calculate the enthalpy change per mole of BaSO4 formed

: Finally, we can calculate the enthalpy change per mole of BaSO4 by dividing the heat released (q) by the number of moles of BaSO4. Enthalpy change per mole of BaSO4 = \(\frac{q}{\text{moles of BaSO}_4}\) Enthalpy change per mole of BaSO4 = \(\frac{457584 \mathrm{~J}}{1.50 \mathrm{~mol}} = 305056 \frac{\mathrm{J}}{\mathrm{mol}}\) Thus, the enthalpy change per mole of BaSO4 formed is \(+305056 \frac{\mathrm{J}}{\mathrm{mol}}\). Note that the positive sign indicates an exothermic reaction.

Key concepts

Calorimetry

Calorimetry is the science of measuring the heat exchanged in chemical reactions or physical changes. It helps us understand the energy transfer between a system and its surroundings. In the context of enthalpy change, calorimetry allows us to calculate the heat absorbed or released by a reaction. When conducting a calorimetry experiment, one considers the specific heat capacity of the substances involved, along with their mass and the temperature change they experience.

• Specific Heat Capacity: This is the amount of energy needed to change the temperature of 1 gram of a substance by 1 degree Celsius.
• Temperature Change: Indicates how much the temperature of the system has increased or decreased during the reaction.
• Total Mass: The entire mass of the participating solutions helps us understand how much material is absorbing or releasing heat.

In our exercise, the temperature rise indicated an exothermic reaction, allowing us to calculate the total heat transfer.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction from continuing further. It determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratio from the balanced chemical equation with the actual moles available. In this exercise, the balanced equation \(Na_2SO_4 + Ba(NO_3)_2 \rightarrow BaSO_4(s) + 2NaNO_3\) indicates that one mole of \(Ba(NO_3)_2\) reacts with one mole of \(Na_2SO_4\).

• Calculation: We have 2 moles of \(Na_2SO_4\) but only 1.5 moles of \(Ba(NO_3)_2\).
• Conclusion: \(Ba(NO_3)_2\) becomes the limiting reactant because it has fewer moles, limiting the amount of \(BaSO_4\) that can be formed to 1.5 moles.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form a solid product, known as the precipitate. This type of reaction is characterized by the formation of an insoluble compound when cations and anions of two different solutions meet. In our exercise, we observe a precipitation reaction where \(Na_2SO_4\) reacts with \(Ba(NO_3)_2\) to form solid \(BaSO_4\).

• Cation-Anion Interaction: \(Ba^{2+}\) ions from the \(Ba(NO_3)_2\) solution react with \(SO_4^{2-}\) ions from the \(Na_2SO_4\) solution.
• Insoluble Product: The reaction of these ions results in the formation of \(BaSO_4\), a white solid that precipitates out of the solution.

This demonstrates the principle of selective solubility, where certain compounds are more prone to form insoluble salts in a solution.

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding how substances absorb and release heat. It is defined as the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius. In the given problem, the specific heat capacity is stated to be \(6.37 \text{ J}/^{\circ}\text{C} \cdot \text{g}\). This value helps us calculate the energy change in the reaction.

• Heat Calculation: The formula \(q = m \cdot C \cdot \Delta T\) allows us to calculate the heat (\(q\)) released, where \(m\) is the mass of the solution, \(C\) is the specific heat capacity, and \(\Delta T\) is the temperature change.
• Practical Application: Knowing the specific heat capacity helps determine how much temperature change a given amount of energy could cause in the substance.

This property is significant in calorimetry because it allows for the precise calculation of heat changes in chemical processes.