Problem 12 - Chemistry: Principles and Reactions

Question

When one mole of KOH is neutralized by sulfuric acid, \(q=-56 \mathrm{~kJ} .\) At \(22.8^{\circ} \mathrm{C}, 25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is neutralized by \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) \(\mathrm{KOH}\) in a coffee-cup calorimeter. What is the final temperature of the solution? (Use the assumptions in Question 11.)

Step-by-Step Solution

Answer

Answer: The final temperature of the solution is 27.27°C.

Step 1: Determine the moles of limiting reactant

To find out which reactant is limiting, examine the moles of each reactant: Moles of H₂SO₄ = Volume × Concentration = (25 mL) × (0.5 mol/L) = 0.0125 mol Moles of KOH = Volume × Concentration = (50 mL) × (0.5 mol/L) = 0.025 mol Since KOH reacts with H₂SO₄ in a 2:1 ratio, the amount of KOH required for 0.0125 mol of H₂SO₄ is 0.025 mol which is exactly the available amount of KOH. Therefore, KOH is the limiting reactant.

Step 2: Calculate the heat change during the reaction

The heat change (q) per mole of KOH during the neutralization is given as -56 kJ. Since all 0.025 mol of KOH will be reacting, the total heat change produced by the reaction is: Total heat change, q = moles × heat change per mole = 0.025 mol × (-56 kJ/mol) = -1.4 kJ

Step 3: Calculate the mass of the solution

The total volume of the solution is given by adding the volume of H₂SO₄ and KOH: Total volume = 25 mL + 50 mL = 75 mL Assuming 1 g/mL as the density of the solution (which is very close to the density of water), we can find the mass of the solution: Mass = Volume × Density = 75 mL × 1 g/mL = 75 g

Step 4: Calculate temperature change

We are given that the calorimeter is well-insulated, so we can assume that all the heat released by the reaction is absorbed by the solution. To calculate the temperature change of the solution, we can use the formula: ΔT = q / (mass × C), where ΔT is the temperature change, q is the heat change, mass is the mass of the solution, and C is the specific heat of water (4.18 J/gC). Since the heat change is negative (exothermic reaction), the temperature will increase. Convert q to J by multiplying with 1000: -q = 1.4 kJ × 1000 J/kJ = 1400 J ΔT = 1400 J / (75 g × 4.18 J/gC) = 4.47 °C

Step 5: Calculate final temperature

The initial temperature of the solution is given as 22.8°C. The final temperature of the solution can be calculated by adding the temperature change to the initial temperature: Final temperature = Initial temperature + ΔT = 22.8°C + 4.47°C = 27.27°C Therefore, the final temperature of the solution is 27.27°C.

Key Concepts

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and salt. In this exercise, sulfuric acid (\(\text{H}_2\text{SO}_4\)) is neutralized by potassium hydroxide (\(\text{KOH}\)). These reactions are of great interest in chemistry because they typically release or absorb heat.
This exercise is a classic example of an exothermic reaction, where heat is released. This release of energy is reflected in the negative sign of the heat change (\(q = -56 ext{ kJ}\)). This negative sign indicates that the system loses heat during the reaction, warming the surrounding solution.

• Neutralization can drive changes in temperature, making it essential in calorimetry studies.
• It is also used to determine heat capacities and understand material properties.

Limiting Reactant

The limiting reactant is the substance that is entirely consumed first during a chemical reaction. It limits the amount of product that can form. In our mini-experiment, \(\text{KOH}\) is the limiting reactant because it completely reacts with the \(\text{H}_2\text{SO}_4\) present.
To find the limiting reactant, we calculate the moles of each reactant, considering their concentrations and volumes:

• Moles of \(\text{H}_2\text{SO}_4\) = \(0.0125 ext{ mol}\)
• Moles of \(\text{KOH}\) = \(0.025 ext{ mol}\)

Since \(\text{KOH}\) perfectly neutralizes the available \(\text{H}_2\text{SO}_4\), it is the limiting reactant. Identifying the limiting reactant helps predict the reaction's completion and the extent to which reactants are consumed.

Specific Heat Capacity

The specific heat capacity (\(C\)) is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.
In chemistry, water is frequently used in reactions, and its specific heat capacity is \(4.18 ext{ J/g°C}\).

• This property is vital because it allows us to calculate how much energy a substance can absorb before changing temperature.
• It helps in predicting temperature changes during reactions in calorimeters.

To measure temperature change in this reaction, we use the formula: \[\Delta T = \frac{q}{\text{mass} \times C}\]where \(q\) is the heat exchanged.

Temperature Change

Temperature change (\(\Delta T\)) in a reaction shows how much the surroundings are heated or cooled due to the reaction. It's a crucial aspect of calorimetry studies. The formula to determine this change is \(\Delta T = \frac{q}{\text{mass} \times C }\).
In our example, the heat released by the neutralization reaction increases the temperature of the solution from 22.8°C to 27.27°C.

• Such changes help confirm the exothermic nature of reactions.
• Understanding temperature change guides safe lab practices.

The calculated \(\Delta T\) of 4.47°C provides insight into the energy dynamics of the reacting mixture, reaffirming how calorimetry can predict outcomes based on reactant properties and quantities.