Problem 103 - General Chemistry

Question

A 14.1-mL sample of \(0.996 \mathrm{M} \mathrm{NaOH}\) is mixed with \(32.3 \mathrm{~mL}\) of \(0.905 M \mathrm{HCl}\) in a coffee-cup calorimeter (see Section \(6.6\) of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest wholenumber coefficients, is \(-55.8 \mathrm{~kJ} .\) Both solutions are at \(21.6^{\circ} \mathrm{C}\) prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), the specific heat of all solutions is the same as that of water, and volumes are additive.

Step-by-Step Solution

Answer

The final temperature of the reaction mixture is approximately \(17.6^{\circ}C\).

Step 1: Calculate Moles of Reactants

First, calculate the moles of NaOH and HCl. For NaOH, use the formula: \( \text{moles} = \text{Molarity} \times \text{Volume (L)} \). For NaOH, \( \text{moles of NaOH} = 0.996 \mathrm{M} \times 0.0141 \mathrm{L} = 0.014046 \text{ moles} \). For HCl, \( \text{moles of HCl} = 0.905 \mathrm{M} \times 0.0323 \mathrm{L} = 0.0292315 \text{ moles} \).

Step 2: Identify the Limiting Reactant

Since \( 0.014046 \text{ mol of NaOH} < 0.0292315 \text{ mol of HCl} \), NaOH is the limiting reactant. The reaction that occurs is \( \text{Neutralization: NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \).

Step 3: Determine Heat Released by the Reaction

Use the enthalpy change provided: \(-55.8 \text{ kJ/mol} \). The total heat released by the reaction is calculated using the limiting reactant.\[ q = \Delta H \times \text{moles of limiting reactant} = -55.8 \text{ kJ/mol} \times 0.014046 \text{ mol} = -0.783768 \text{ kJ} \].

Step 4: Convert Heat Quantity to Joules

Since the specific heat capacity and final temperature is usually discussed in joules, convert the heat released to joules: \(-0.783768 \text{ kJ} = -783.768 \text{ J} \).

Step 5: Calculate Total Volume and Mass of Solution

Add the volumes of the solutions: \(14.1 \text{ mL} + 32.3 \text{ mL} = 46.4 \text{ mL} \). Assuming the density is \(1 \text{ g/mL} \), the mass is approximately \(46.4 \text{ g} \).

Step 6: Calculate the Final Temperature

To find the final temperature, use the formula for heat transfer involving specific heat capacity: \[ q = m \cdot c \cdot \Delta T \] where \( q = -783.768 \text{ J} \), \( m = 46.4 \text{ g} \), and \( c = 4.18 \text{ J/g°C} \).Solve for \( \Delta T \):\[ \Delta T = \frac{q}{m \cdot c} = \frac{-783.768 \text{ J}}{46.4 \text{ g} \times 4.18 \text{ J/g°C}} \approx -4.029°C \] The initial temperature was \(21.6°C \) so the final temperature is \( 21.6°C + (-4.029°C) = 17.571°C \).

Step 7: Final Temperature Adjustment

Since the reaction is exothermic, and heat is released to the surroundings, the total temperature change will be lower. Calculate the estimated temperature of the surroundings equilibrium: Since the computation already assumed no energy loss, we maintain the effective value (in adjustments).Thus, the final temperature of the reaction mixture is approximately: \(17.6°C \).

Key Concepts

Coffee-cup calorimeter

A coffee-cup calorimeter is a simple device used to measure the heat change of a reaction at constant pressure. It's typically made from a styrofoam cup, which acts as an insulating container to minimize heat loss to the surroundings. This setup is ideal for reactions in solution because it keeps the system isolated enough to provide fairly accurate measurements of temperature changes.

When you perform a reaction in a coffee-cup calorimeter, the heat of the reaction causes a change in temperature of the solution. By measuring this change, we can calculate the heat absorbed or released, which represents the enthalpy change for the reaction. Remember that in a coffee-cup calorimeter, the amount of heat absorbed by the solution is usually assumed to be negligible, focusing on measuring the heat generated by the reaction itself.

Enthalpy change

Enthalpy change, often symbolized as \(\Delta H\), is the heat change at constant pressure. It is a measure of the total energy change during a reaction, including both the internal energy and the work done by the system. In exothermic reactions, \(\Delta H\) is negative because energy is released as heat, making the surrounding temperature rise.

Given the reaction enthalpy of \(-55.8 \, \text{kJ/mol}\), this means that for every mole of the limiting reactant consumed, 55.8 kJ of heat is released. This released heat is the driving factor that can increase the temperature of the contents within the calorimeter. However, in many cases, instead of trying to measure such a large energy change directly, we use the equation from calorimetry: \ q = m \cdot c \cdot \Delta T\, and we work backwards to find the specific change in temperature that correlates with the measured heat.

Neutralization reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. In the context of our exercise, the reaction between \(\text{NaOH}\) and \(\text{HCl}\) is a classic example of a neutralization reaction: \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\).

This type of reaction is typically exothermic, meaning it releases heat. As shown in the step-by-step solution, we calculated the heat released by finding the enthalpy change associated with the formation of water and sodium chloride. This is then used to determine how much the temperature of the solution would increase, given no heat is lost to the surroundings.

Specific heat capacity

Specific heat capacity, denoted as \(c\), is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is a property that varies from substance to substance. The specific heat capacity of water, which is often used as a standard, is \(4.18 \, \text{J/g°C}\).

In calorimetric calculations, the specific heat capacity is essential because it allows you to relate the absorbed or released heat to a change in temperature, through the formula \ q = m \cdot c \cdot \Delta T\. In the example from the exercise, the specific heat capacity of the solution was assumed to be the same as water, which simplifies calculations and allows us to determine the final temperature of the reaction mixture.