Question

Consider the reaction $$2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \Delta H=-118 \mathrm{kJ}$$ Calculate the heat when \(100.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) HCl is mixed with \(300.0 \mathrm{mL}\) of \(0.100 M \mathrm{Ba}(\mathrm{OH})_{2} .\) Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{g}\) and a specific heat capacity of \(4.18 \mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g},\) calculate the final temperature of the mixture.

Short answer

The final temperature of the mixture is approximately \(22.88\,^{\circ}\mathrm{C}\).

Step-by-step solution

Determine the limiting reactant

First, we need to determine the limiting reactant in this chemical reaction. Moles of reaction = moles (Volume × Molarity) HCl: $$100.0\,\mathrm{mL} * 0.5\,\mathrm{M} = 50.0\,\mathrm{mmol}$$ Ba(OH)2: $$300.0\,\mathrm{mL} * 0.1\,\mathrm{M} = 30.0\,\mathrm{mmol}$$ Calculate the required ratio of reactants to complete the reaction: $$2\,\mathrm{HCl} \longleftrightarrow 1\,\mathrm{Ba(OH)_2}$$ Required ratio: $$50\,\mathrm{mmol HCl} / 30\,\mathrm{mmol Ba(OH)_2} = 1.67$$ Since the required ratio (1.67) is greater than the actual ratio (2), \(\mathrm{Ba(OH)_2}\) is the limiting reactant.

Calculate the heat produced

Now, we'll calculate the heat produced by the reaction: Given: $$\Delta H = -118\,\mathrm{kJ/mol}$$ Calculate the heat produced by the reaction: $$q = moles \times \Delta H$$ $$q = 30\,\mathrm{mmol} \times -118\,\mathrm{kJ/mol} = -3.54 \times 10^3 \,\mathrm{J}$$

Calculate the temperature change

Next, we have to calculate the temperature change using the heat and specific heat capacity (Cp). Given: - Mass (m) of the mixture: \(400.0\,\mathrm{g}\) - Specific heat capacity (Cp): \(4.18\,\mathrm{J/g°C}\) Calculate the temperature change: $$q = m \times Cp \times \Delta T$$ $$\Delta T = \dfrac{q}{m \times Cp}$$ $$\Delta T = \dfrac{-3.54 \times 10^3 \,\mathrm{J}}{400.0\,\mathrm{g}\times 4.18\,\mathrm{J/g°C}} = -2.12\,^{\circ}\mathrm{C}$$

Calculate the final temperature

Finally, let's calculate the final temperature of the mixture. Given the initial temperature: $$T_{initial} = 25.0\,^{\circ}\mathrm{C}$$ Calculate the final temperature: $$T_{final} = T_{initial} + \Delta T $$ $$T_{final} = 25.0\,^{\circ}\mathrm{C} - 2.12\,^{\circ}\mathrm{C} = 22.88\,^{\circ}\mathrm{C}$$ The final temperature of the mixture is approximately \(22.88\,^{\circ}\mathrm{C}\).

Key concepts

Limiting Reactant

In a chemical reaction involving multiple reactants, the limiting reactant is the substance that gets entirely consumed first. It essentially determines the maximum amount of product that can be formed. The concept is crucial because it helps predict the outcome of reactions and avoids wastage of other reactants.
To find the limiting reactant, we compare the mole ratio of the reactants in the reaction to their actual proportions. In our exercise, we calculate the number of moles for each reactant:

• For HCl: 100 mL × 0.5 M = 50 mmol
• For Ba(OH) ₂: 300 mL × 0.1 M = 30 mmol

The balanced equation says that 2 moles of HCl react with 1 mole of Ba(OH) ₂. This means we need 50 mmol of HCl per 25 mmol of Ba(OH) ₂, requiring a 1:1.67 ratio. Since Ba(OH) ₂ has fewer moles than necessary, it becomes the limiting reactant.

Enthalpy Change

Enthalpy change is the heat absorbed or released during a chemical reaction at constant pressure. In exothermic reactions, like the one in this exercise, it is released, and \(\Delta H \) is negative.
Here, the enthalpy change \(\Delta H = -118 \text{ kJ/mol} \) implies that for every complete reaction described by the balanced equation, 118 kJ of heat is released. Once the limiting reactant is identified, we can compute the total heat produced based on the amount of the limiting reactant that actually reacted:

• Heat produced (\(q\)): 30 mmol × -118 kJ/mol = -3.54 kJ

Thus, the complete reaction of 30 mmol of Ba(OH)₂ leads to the release of 3.54 kJ of heat.

Specific Heat Capacity

Specific heat capacity (\(C_p\)) is an intrinsic property that measures how much heat energy is needed to raise the temperature of a substance by one degree Celsius. The higher the specific heat capacity, the more heat is required to change the temperature.
In this exercise, our mixture has a specific heat capacity of 4.18 J/g°C, which is typical for water. This means each gram of the mixture requires 4.18 joules to increase its temperature by 1°C.
Understanding \(C_p\) helps to determine the degree of temperature change a substance undergoes when a certain amount of heat is absorbed or released, which plays a pivotal role in calorimetry applications.

Temperature Change

Temperature change (\(\Delta T\)) during a chemical reaction can be calculated once we know the heat exchanged (\(q\)), the mass, and the specific heat capacity of the substance involved. This reaction is exothermic, so heat is lost to the surroundings, and we expect a temperature drop.
The equation used is:\[\Delta T = \frac{q}{m \times C_p}\]

• Where \(m = 400.0 \text{g}\) and \(C_p = 4.18 \text{ J/g°C}\).
• Substituting the values gives us:\[\Delta T = \frac{-3.54 \times 10^3 \text{ J}}{400 \times 4.18} = -2.12 \text{°C}\]

Therefore, the mixture's temperature decreases by approximately 2.12°C.

Heat Calculation

Calculating heat involves determining the energy exchange between the system and its surroundings during a reaction. For our exercise, this is imperative for finding the final temperature of the resultant mixture.
We first calculate the heat involved using the reactant's enthalpy change information:

• Identify the limiting reactant and determine its quantity in moles.
• Use the enthalpy change (\(\Delta H\)) to compute the total heat (\(q\)).
• Calculate \(q\) using: \(q = moles \times \Delta H\).

In the given problem, using 30 mmol of the limiting reactant, we find \(q = -3.54 \text{ kJ} = -3.54 \times 10^3 \text{ J}\). This negative sign indicates that the reaction releases heat, leading to a temperature drop in the solution.