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\(M \mathrm{HCl}\) 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 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

Short answer

The final temperature of the mixture when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.100 M Ba(OH)₂ is approximately 27.12°C.

Step-by-step solution

Calculate the limiting reagent

Determine the limiting reagent by calculating the moles of HCl and Ba(OH)₂ in the given volumes and concentrations of each solution. Moles of HCl: moles = Molarity × Volume moles = 0.500 M × 0.100 L = 0.050 moles Moles of Ba(OH)₂: moles = Molarity × Volume moles = 0.100 M × 0.300 L = 0.030 moles Based on the stoichiometry of the reaction (2 moles of HCl react with 1 mole of Ba(OH)₂), twice the moles of HCl are needed in comparison to the moles of Ba(OH)₂. But in this case, there are more moles of HCl present than required, making Ba(OH)₂ the limiting reagent.

Calculate the heat evolved in the reaction

Using the moles of the limiting reactant (Ba(OH)₂), the enthalpy of the reaction (ΔH), and the stoichiometry of the reaction, calculate the heat evolved (q) during the reaction. q (reaction) = Moles of limiting reactant × ΔH q (reaction) = 0.030 moles Ba(OH)₂ × (-118000 J/mol) q (reaction) = -3540 J Notice that the result is negative, which indicates that the system has released heat (exothermic reaction).

Calculate the heat absorbed by the solution

Now, determine the heat absorbed by the solution (q(solution)), which is equal in magnitude but opposite in sign to the heat evolved in the reaction (q(reaction)). Calculate the temperature change in the solution using this heat absorbed and the specific heat capacity provided. q (solution) = -q (reaction) q (solution) = 3540 J ΔT (temperature change) = q(solution) / (mass × specific heat capacity) ΔT = 3540 J / (400.0 g × 4.18 J/g·°C) ΔT = 2.12°C

Calculate the final temperature of the mixture

Using the initial temperature (25.0°C) and the temperature change (ΔT) calculated in the previous step, determine the final temperature of the mixture. Final temperature = Initial temperature + ΔT Final temperature = 25.0°C + 2.12°C Final temperature = 27.12°C Now we have determined that when mixing 100 mL of 0.500 M HCl with 300 mL of 0.100 M Ba(OH)₂, the final temperature of the resulting solution is approximately 27.12°C.

Key concepts

Limiting Reagent

In any chemical reaction, the limiting reagent is the reactant that is completely consumed first. This reactant sets a limit on the amount of products that can be formed. To find the limiting reagent, first calculate the number of moles for each reactant based on the given concentration and volume.
For the reaction between HCl and Ba(OH)₂, we calculate:

• Moles of HCl: 0.500 M × 0.100 L = 0.050 moles
• Moles of Ba(OH)₂: 0.100 M × 0.300 L = 0.030 moles

The balanced equation shows that 2 moles of HCl react with 1 mole of Ba(OH)₂. Therefore, for 0.030 moles of Ba(OH)₂, we need 0.060 moles of HCl, but only 0.050 moles are available. Hence, Ba(OH)₂ is the limiting reagent.
Understanding which reactant is the limiting reagent helps us predict how much product can be formed and how much heat will be released in an exothermic reaction.

Enthalpy Change

Enthalpy change (\( \Delta H \)) indicates whether a reaction releases or absorbs energy when products are formed from reactants. It forms a crucial part of understanding the heat involved in chemical reactions. For our reaction:

• The reaction has an enthalpy change of \( \Delta H = -118 \ \text{kJ/mol} \)

The negative sign shows that the reaction is exothermic, releasing heat to the surroundings.
Calculating the heat released involves using the limiting reagent to determine the amount of energy change:

• Heat released (\( q \)) = Moles of limiting reagent × \( \Delta H \)
• For Ba(OH)₂: \( q \) = 0.030 moles × (-118,000 J/mol) = -3540 J

This calculation reveals the amount of heat released due to the reaction, and since the result is negative, it confirms that the reaction is exothermic. This released heat can also result in a temperature change of the reacting system.

Exothermic Reaction

An exothermic reaction is one that releases heat, resulting in an increase in the temperature of the surroundings. In such reactions, the energy required to break the bonds in reactants is less than energy released when new bonds form in the products. In this particular case, the reaction between HCl and Ba(OH)₂ is exothermic, evident from the negative enthalpy change ( \( \Delta H = -118 \ \text{kJ/mol} \)).
As per the step-by-step method, when \(0.030\) moles of Ba(OH)₂ react, \(3540\) J of heat is released.
The temperature of the solution increases as a result of this heat release.

• Initial temperature: \(25.0^{\circ} \text{C}\)
• Temperature change: \(2.12^{\circ} \text{C}\)
• Final temperature: \(27.12^{\circ} \text{C}\)

The increase in temperature is a direct consequence of the exothermic nature of the reaction.