Problem 37 - Chemistry and Chemical Reactivity

Question

Calorimetry Assume you mix 100.0 \(\mathrm{mL}\) of \(0.200 \mathrm{M}\) CsOH with \(50.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) in a coffee-cup calorimeter. The following reaction occurs: $$ \mathrm{CsOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CsCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The temperature of both solutions before mixing was \(22.50^{\circ} \mathrm{C},\) and it rises to \(24.28^{\circ} \mathrm{C}\) after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all \(1.00 \mathrm{g} / \mathrm{mL}\) and the specific heat capacities of the solutions are \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}.\)

Step-by-Step Solution

Answer

The enthalpy change for the reaction per mole of CsOH is \(-56.13 \mathrm{kJ/mol}\).

Step 1: Calculate the Total Mass of the Solution

Since the density is given as \(1.00 \text{ g/mL}\), the mass of the total solution is simply the volume of the solutions combined. The total volume is \(100.0 \text{ mL} + 50.0 \text{ mL} = 150.0 \text{ mL}\). Thus, the mass of the solution is \(150.0 \text{ g}\).

Step 2: Find the Temperature Change

The temperature change \(\Delta T\) is the final temperature minus the initial temperature. Here, the temperature change is \(24.28^{\circ} \mathrm{C} - 22.50^{\circ} \mathrm{C} = 1.78^{\circ} \mathrm{C}\).

Step 3: Calculate the Heat Absorbed or Released

Use the formula \(q = m \cdot c \cdot \Delta T\) to calculate the heat change, where \(m\) is the mass, \(c\) is the specific heat capacity \((4.2 \mathrm{J/g} \cdot \mathrm{K})\), and \(\Delta T\) is the temperature change. Therefore, \(q = 150.0 \mathrm{g} \cdot 4.2 \mathrm{J/gK} \cdot 1.78\mathrm{K} = 1122.6 \mathrm{J}\).

Step 4: Determine the Limiting Reactant and Moles Reacted

Calculate moles of each reactant. For CsOH: \(M = 0.200 \mathrm{M}\) and \(V = 0.100 \mathrm{L}\), so moles = \(0.200 \times 0.100 = 0.020 \mathrm{mol}\). For HCl: \(M = 0.400 \mathrm{M}\) and \(V = 0.050 \mathrm{L}\), so moles = \(0.400 \times 0.050 = 0.020 \mathrm{mol}\). Both reactants are limiting with \(0.020 \mathrm{mol}\) reacting.

Step 5: Calculate the Enthalpy Change per Mole

Since the reaction involves \(0.020 \mathrm{mol}\) of CsOH, and the heat change \(q\) is \(-1122.6 \mathrm{J}\) (exothermic reaction), the enthalpy change per mole of CsOH is \(\Delta H = \frac{-1122.6 \mathrm{J}}{0.020 \mathrm{mol}} = -56130 \mathrm{J/mol}\) or \(-56.13 \mathrm{kJ/mol}\).

Key Concepts

Enthalpy Change

Enthalpy change is a key concept in calorimetry and thermodynamics. It represents the heat absorbed or released during a chemical reaction at constant pressure. The symbol for enthalpy is \( H \), and the change in enthalpy, denoted as \( \Delta H \), can indicate whether a reaction releases heat or absorbs it. In our exercise, the reaction between cesium hydroxide (CsOH) and hydrochloric acid (HCl) involves an enthalpy change. To calculate the enthalpy change per mole of a reactant, as in this example, we first determine the total heat released, \( q \). Then, we divide \( q \) by the number of moles of the limiting reactant. This gives us \( \Delta H \) per mole, which provides insight into the energy dynamics of the reaction.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance completely consumed first, determining the maximum amount of product that can be formed. In our calorimetry example, both reactants, CsOH and HCl, were present in equivalent molar quantities (0.020 mol each). As they react in a 1:1 ratio, both act as limiting reactants. Finding the limiting reactant is crucial in calorimetry as it lets us determine which reactant dictates the extent of the reaction and thereby helps us compute the enthalpy change accurately. Without this step, calculating the enthalpy change per mole might yield incorrect results relating to the actual energy exchange in the reaction.

Exothermic Reaction

An exothermic reaction is one that releases heat to its surroundings. This is observable when the temperature of a system increases as the reaction proceeds. In the example provided, the reaction between CsOH and HCl increased the temperature from 22.50°C to 24.28°C, suggesting an exothermic nature. The negative sign in the enthalpy change (\( \Delta H \)) indicates the exothermic process. In such reactions, the energy of the products is lower than that of the reactants, and the excess energy is given off as heat. It's important to note that exothermic reactions are common in everyday processes like combustion, and understanding them helps in energy management.

Heat Capacity

Heat capacity is a property that describes the amount of heat required to change the temperature of a substance by a given amount. Specifically, specific heat capacity is the heat needed to raise the temperature of 1 gram of a substance by 1 Kelvin (or 1°C).In our calorimetry case, the specific heat capacity used is 4.2 J/g·K, typical for solutions like water. This value is essential for calculating the total heat change during the reaction using the formula \( q = m \cdot c \cdot \Delta T \), where \( m \) is mass and \( \Delta T \) is the temperature change.

Coffee-cup Calorimeter

The coffee-cup calorimeter is a simple yet effective tool used to measure the enthalpy change of a reaction occurring in solution. It consists of an insulated coffee cup that minimizes heat loss to the environment, ensuring most of the heat exchange occurs within the system. In the context of our exercise, the calorimeter allowed measurement of the temperature rise from 22.50°C to 24.28°C caused by the exothermic reaction of CsOH and HCl. This temperature change, along with the known specific heat capacity and total mass of the solution, facilitated calculation of the enthalpy change. Such setup provides a relatively easeful way for students to understand and explore ubiquitous chemical reactions and their energetic implications.