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 when 1.00 L of 2.00 M Na₂SO₄ solution is mixed with 2.00 L of 0.750 M Ba(NO₃)₂ solution, and the temperature increases from 30.0 °C to 42.0 °C is found to be \(304,896 \ \text{J/mol}\).

Step-by-step solution

1. Write the balanced chemical equation

Before starting any calculations, we should write the balanced chemical equation for the reaction. Na₂SO₄ (aq) + Ba(NO₃)₂ (aq) → BaSO₄ (s) + 2 NaNO₃ (aq)

2. Identify the limiting reactant

To find the limiting reactant, we will compare the mole ratios of the two reactants. The volume and concentration of the reactants are given as: Na₂SO₄: 1.00 L × 2.00 M = 2.00 moles Ba(NO₃)₂: 2.00 L × 0.750 M = 1.50 moles The mole ratios from the balanced equation are: Na₂SO₄ : Ba(NO₃)₂ = 1 : 1 Comparing the mole ratios: 2.00 moles Na₂SO₄ / 1 = 2.00 1.50 moles Ba(NO₃)₂ / 1 = 1.50 Since 1.50 (Ba(NO₃)₂) < 2.00 (Na₂SO₄), Ba(NO₃)₂ is the limiting reactant.

3. Calculate moles of BaSO₄ formed

As the reaction has a 1:1 mole ratio between Ba(NO₃)₂ and BaSO₄, the moles of BaSO₄ formed will be equal to the moles of limiting reactant used. Hence: 1.50 moles Ba(NO₃)₂ → 1.50 moles BaSO₄

4. Calculate the heat released

We need to calculate the heat released (q) during the reaction using the given information. The specific heat capacity (c) is given as 6.37 J/°C·g, and the density of the final solution (ρ) is 2.00 g/mL. The temperature change (∆T) experienced by the reaction is: ∆T = 42.0°C - 30.0°C = 12.0°C To find the mass (m) of the final solution, we use the given density and total volume (3.00 L): m = ρ × volume = 2.00 g/mL × 3000 mL = 6000 g Now, we can calculate the heat released using the formula: q = mc∆T q = (6000 g)(6.37 J/°C·g)(12.0°C) q = 457344 J

5. Calculate the enthalpy change per mole of BaSO₄ formed

Now we can calculate the enthalpy change (ΔH) per mole of BaSO₄ by dividing the heat released (q) by the moles of BaSO₄ formed: ΔH = q / moles of BaSO₄ ΔH = 457344 J / 1.50 moles ΔH = 304896 J/mol So, the enthalpy change per mole of BaSO₄ formed is 304,896 J/mol.

Key concepts

Limiting Reactant Determination

Understanding the concept of the limiting reactant is essential for predicting the amount of product that can be formed in a chemical reaction. It refers to the substance that is completely consumed first during the chemical reaction, thus limiting the amount of product that can be formed. This is because no further reaction can occur once the limiting reactant is used up.

To determine the limiting reactant, one must first know the balanced chemical equation, which gives the mole ratios of the reactants and the products. With the mole ratio, you can then compare the number of moles of each reactant available. The reactant that provides the fewest moles of product, according to the stoichiometry of the balanced equation, is the limiting reactant.

In our exercise, the balanced chemical equation is Na₂SO₄(aq) + Ba(NO₃)₂(aq) → BaSO₄(s) + 2 NaNO₃(aq), and by comparing the available moles of Na₂SO₄ and Ba(NO₃)₂, we can see that Ba(NO₃)₂ is the limiting reactant. This is because even though there are more moles of Na₂SO₄, the reaction requires equal amounts of Na₂SO₄ and Ba(NO₃)₂ for stoichiometric conversion to products.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry provides quantitative information about the proportions in which reactants combine to form products during a chemical reaction. It is derived from the balanced chemical equation, which indicates the ratios of moles of substances involved.

Accurate stoichiometric calculations are fundamental to determine theoretical yields, which represent the maximum amount of product that can be obtained if all of the limiting reactant is converted into product. The stoichiometry of a reaction is based on the law of conservation of mass, stating that matter is neither created nor destroyed during a chemical reaction. Therefore, the amounts of various components need to be balanced.

In the given exercise, we conceptualize this through the equation Na₂SO₄(aq) + Ba(NO₃)₂(aq) → BaSO₄(s) + 2 NaNO₃(aq), where the stoichiometry (1:1 ratio) informs us that each mole of Ba(NO₃)₂ will produce one mole of BaSO₄. The stoichiometry plays a crucial role in leading us to calculate the correct amount of product formed, which is vital for later determining the enthalpy change per mole.

Heat Capacity and Density in Thermochemistry

In thermochemistry, the concepts of heat capacity and density are important for calculating the amount of heat absorbed or released during a reaction. Heat capacity, defined as the amount of heat energy required to raise the temperature of a substance by one degree Celsius, is crucial when determining the thermal energy change of the system.

Density, the mass per unit volume, is also significant as it helps in determining the mass of solutions or mixtures when volumes are known, as is often the case in chemical reactions. In thermochemical calculations, it is used along with heat capacity to quantify the heat (q) released or absorbed by the reaction, using the formula: q = mc∆T, where m is the mass, c is the specific heat capacity, and ∆T is the change in temperature.

In our example, the specific heat capacity of the solution is 6.37 J/°C·g, and the density of the final solution is 2.00 g/mL. These values allow us to calculate the total heat change as the product of mass, specific heat capacity, and the temperature difference. This calculation demonstrates how heat capacity and density are applied in practice to determine the enthalpy change for the thermochemical process in question.