Problem 104 - General Chemistry

Question

A 29.1-mL sample of \(1.05 \mathrm{M}\) KOH is mixed with \(20.9 \mathrm{~mL}\) of \(1.07 M \mathrm{HBr}\) 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 \(21.8^{\circ} \mathrm{C}\) 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 \(1.00 \mathrm{~g} / \mathrm{mL}\), and volumes are additive.

Step-by-Step Solution

Answer

The final temperature of the reaction mixture is approximately \( 15.8\degree \text{C} \).

Step 1: Calculate Total Moles of Reactants

To find the moles of each reactant, use the formula \( \text{moles} = \text{volume in liters} \times \text{molarity} \). For KOH: \[ \text{moles of KOH} = 0.0291 \times 1.05 = 0.030555 \text{ moles} \]. For HBr: \[ \text{moles of HBr} = 0.0209 \times 1.07 = 0.022363 \text{ moles} \].

Step 2: Determine Limiting Reactant

The chemical reaction is \( \ce{KOH + HBr -> KBr + H2O} \), which proceeds in a 1:1 molar ratio. Since \(0.022363\) moles of HBr are fewer than \(0.030555\) moles of KOH, HBr is the limiting reactant.

Step 3: Calculate Heat Released by Reaction

The enthalpy change \( \Delta H = -55.8 \mathrm{~kJ/mol} \) refers to the heat released per mole of reaction. Multiply this by the moles of the limiting reactant: \[ q = 0.022363 \times (-55.8) = -1.248524 \mathrm{~kJ} \] or \(-1248.524 \mathrm{~J}\) as heat energy released.

Step 4: Calculate Total Mass of the Solution

Assuming the density is \(1.00 \mathrm{~g/mL}\), the total mass is the sum of the volumes of solutions: \(29.1 + 20.9 = 50.0 \mathrm{~mL}\). Thus, the mass is \(\text{mass} = 50.0 \mathrm{~g}\).

Step 5: Calculate Temperature Change

Use the formula \( q = m \cdot c \cdot \Delta T \), where \( m = 50.0 \mathrm{~g} \), \( c = 4.18 \mathrm{~J/g^{ ext{°}}C} \) (specific heat capacity of water), and \( q = \text{-1248.524 J} \). Solve for \( \Delta T \) as follows: \( \Delta T = \frac{-1248.524}{50.0 \times 4.18} = -5.969 \degree \text{C} \).

Step 6: Calculate Final Temperature

Subtract the temperature change from the initial temperature to find the final temperature: \[ T_{\text{final}} = 21.8 - 5.969 = 15.831 \degree \text{C} \].

Key Concepts

Enthalpy Change

Enthalpy change, represented as \( \Delta H \), is a crucial concept in calorimetry that measures the heat absorbed or released during a chemical reaction. In this particular exercise, the reaction between KOH and HBr releases energy as heat, indicated by a negative \( \Delta H = -55.8 \, \mathrm{kJ/mol} \). Understanding enthalpy change helps us determine the amount of energy associated with the formation or breaking of bonds in a reaction.
The negative sign in the enthalpy value signifies that the reaction is exothermic, meaning it releases heat into the surroundings. For energy calculations, it's important to consider the enthalpy change per mole of a specific reaction as it allows us to scale up or down based on the moles of reactants involved. For this exercise, we multiply the limiting reactant's moles by \( \Delta H \) to find the total heat change for the reaction.

Limiting Reactant

Identifying the limiting reactant is vital in determining the extent of a chemical reaction. It is the reactant that is completely consumed first, limiting the amount of product formed. In our situation, HBr is the limiting reactant because it has fewer moles compared to KOH.
To find the limiting reactant, compare the mole ratios of the reactants to the coefficients in the balanced chemical equation. Since the reaction of \( \mathrm{KOH + HBr \rightarrow KBr + H_2O} \) takes place in a 1:1 ratio, the reactant with the smaller number of moles becomes the limiting reactant. This simplification allows us to focus on only one reactant when calculating the heat released.

Heat Capacity

Heat capacity is an important property that explains how much heat energy a substance can absorb per degree of temperature change. For calorimetry calculations, knowing the specific heat capacity \( c \) allows us to relate heat transfer to temperature changes. In this exercise, the specific heat capacity of the aqueous solution is approximated to that of water, which is \( 4.18\, \mathrm{J/g^{°}C} \).
By incorporating heat capacity into the equation \( q = m \cdot c \cdot \Delta T \), we can solve for the temperature change \( \Delta T \) when the mass \( m \) and heat transfer \( q \) are known. This relationship is key to determining the final temperature of the reaction mixture after the chemical reaction.

Temperature Calculation

Temperature calculation in a calorimetry problem involves using known values and solving step-by-step to find the final temperature. Here, we begin with the heat change (\( q \)) and mass \( m \) of the solution to find \( \Delta T \) using \( q = m \cdot c \cdot \Delta T \). The negative heat change of \(-1248.524 \, \mathrm{J}\) results in a temperature decrease.
Once \( \Delta T \) is determined, it is subtracted from the initial temperature to find the final temperature of the mixture. The calculation ends with \( T_{\text{final}} = 21.8 - 5.969 \approx 15.83^{\circ} \text{C} \). This exercise demonstrates how to translate the enthalpy change of a reaction into observables like temperature change in a controlled setting like a calorimeter.