Problem 26 - General Chemistry: Principles and Modern Applications

Question

The heat of neutralization of \(\mathrm{HCl}(\text { aq) by } \mathrm{NaOH}(\mathrm{aq})\) is \(-55.84 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) produced. If \(50.00 \mathrm{mL}\) of \(1.05 \mathrm{M}\) \(\mathrm{NaOH}\) is added to \(25.00 \mathrm{mL}\) of \(1.86 \mathrm{M} \mathrm{HCl}\), with both solutions originally at \(24.72^{\circ} \mathrm{C},\) what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of \(1.02 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(3.98 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Step-by-Step Solution

Answer

The final temperature of the solution will depend on the given values for volume, molarity, and specific heat, as well as the heat of neutralization. Use the approach outlined in the solution steps to calculate it.

Step 1: Determine the limiting reactant

In this reaction, hydrochloric acid (\(HCl\)) and sodium hydroxide (\(NaOH\)) react to form water and salt. The amount of heat produced depends on the amount of water formed, which in turn depends on the limiting reactant. Draw up a table with the volume and molarity of each reactant to calculate the number of moles of each reactant. The reactant with fewer moles is the limiting reactant.

Step 2: Calculate the heat released by the reaction

Using the heat of neutralization as given (-55.84 kJ/mol of \(H_{2}O\) produced), and the moles of the limiting reactant (the one that generates water), calculate the total heat released by the reaction. Remember the output will need to be converted to joules (1 kJ = 1000 J) to match the required units in subsequent steps.

Step 3: Determine the mass of the solution

Next, calculate the total mass of the final solution using the given densities and the sum of the initial volumes of the reactants. Remember that density = mass/volume and rearrange the formula to find the mass of the solution.

Step 4: Calculate the final temperature

Now you can find the final temperature of the solution. The amount of heat absorbed by the solution will equal the amount of heat released by the reaction. This heat can be calculated using the equation \(q=mc\Delta T\), where m is the mass of the solution, c is its specific heat, and \(\Delta T\) is the temperature difference. Rearrange this formula to find \(\Delta T = q / (mc)\), then add this to the initial temperature to find the final temperature of the solution.

Key Concepts

Heat of Neutralization

The heat of neutralization refers to the energy change when an acid and a base react to form water. In our specific exercise, when hydrochloric acid ( HCl ) and sodium hydroxide ( NaOH ) react, the heat of neutralization is -55.84 ext{kJ/mol} ext{of} H_{2}O ext{produced}. This means for every mole of water created in this reaction, 55.84 kJ of energy is released as heat.
It's important to remember that this energy is released because the reaction is exothermic. Exothermic reactions release energy to the surroundings, which is why the solution’s temperature can increase as the reaction proceeds.
The key takeaway is: the concept is crucial for calculating the total amount of heat exchanged during the reaction, important for subsequently predicting the temperature change.

Limiting Reactant

Identifying the limiting reactant is essential in predicting the amount of product that can be formed in a chemical reaction. In our exercise, we determine the limiting reactant by calculating the number of moles of each reactant from their given concentrations and volumes. Let's break it down:

• For HCl : Convert volume from mL to L, then multiply by molarity to find moles. That gives us: 25.00 ext{mL} imes rac{1}{1000} rac{L}{mL} imes 1.86 ext{M} = 0.0465 ext{moles} .
• Similarly, perform the calculation for NaOH : 50.00 ext{mL} imes rac{1}{1000} rac{L}{mL} imes 1.05 ext{M} = 0.0525 ext{moles} .

In our example, HCl has fewer moles, making it the limiting reactant. This information is crucial as it dictates the maximum amount of product (water in this case) that can be formed and thus the amount of energy that can be released via the heat of neutralization.

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat a substance can absorb before its temperature changes. The specific heat capacity is denoted as "c" in physics, and our exercise uses a specific heat capacity value of 3.98 ext{Jg}^{-1} ^{ ext{°C}}^{-1} for the solution. This means that 3.98 ext{J} of heat is needed to raise the temperature of 1 ext{g} of the solution by 1 ext{°C} .
It's vital to understand how this property is used in predicting temperature changes in reactions. It acts as a buffer, determining how much a temperature rises when a certain amount of heat is absorbed by the system.
This value gets incorporated into the equation for calculating the final temperature of the solution, as it influences how heat energy translates into temperature change, crucial for experiments involving thermal dynamics.

Temperature Change Calculation

For a complete picture, we must understand how to calculate the temperature change as it connects all the previous concepts. Using the equation q=mc riangle T , where:

• q is the heat energy exchanged (in Joules)
• m is the mass of the entire solution
• c is the specific heat capacity
• riangle T is the change in temperature (°C)

To calculate riangle T , rearrange this formula to riangle T = rac{q}{mc} . Here’s a step-by-step guide: 1. First, convert the heat of neutralization from kJ to J (since 1 ext{kJ} = 1000 ext{J} ). 2. Calculate the total mass of the solution by multiplying the sum of its constituent volumes by the given density (in this case, 1.02 ext{g/mL} ). 3. Use the specific heat capacity, 3.98 ext{Jg}^{-1}   ext{°C}^{-1}, in the equation. 4. Add the initial temperature to riangle T , yielding the final temperature of the solution.
This calculation is invaluable for predicting outcomes in lab experiments, ensuring safety and accuracy in observing reactions.