/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Consider the dissolution of \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .\)

Short Answer

Expert verified
The final temperature of the solution after the dissolution of 11.0 g of CaCl2 in 125 g of water at \(25.0^{\circ} \mathrm{C}\) is 40.4 °C.

Step by step solution

01

Calculate the heat released by the dissolution process

To calculate the heat released by the dissolution process, we need to multiply the moles of CaCl2 with the enthalpy of the process. First, we should find the number of moles of CaCl2 used: Molar mass of CaCl2 = 40.08 (Ca) + 2 * 35.45 (Cl) = 110.98 g/mol Number of moles = (mass of CaCl2) / (molar mass of CaCl2) Number of moles = 11.0 g / 110.98 g/mol = 0.0991 mol Now, we can find the heat released by the dissolution process: Heat released (q) = moles * enthalpy change q = 0.0991 mol * -81.5 kJ/mol q = -8.07 kJ Since q is negative, the heat is released during the process.
02

Calculate the heat absorbed by water

Using the equation \(q = mc \Delta T\), we can find the heat absorbed by water: 8.07 kJ = (125 g) * (4.18 J/g·°C) * ΔT 8070 J = (125 g) * (4.18 J/g·°C) * ΔT Now, we need to solve for ΔT (the temperature change of water).
03

Calculate the final temperature

First, we should isolate ΔT: ΔT = 8070 J / (125 g * 4.18 J/g·°C) ΔT = 15.4 °C Since ΔT is positive, the temperature of the water will increase. So, we can compute the final temperature by adding the initial temperature with ΔT: Final Temperature = Initial Temperature + ΔT Final Temperature = 25°C + 15.4 °C Final Temperature = 40.4 °C The final temperature of the solution after the dissolution of CaCl2 is 40.4 °C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
When substances undergo a chemical reaction or state change, they either absorb or release energy. This energy change is known as enthalpy change, denoted by \(\Delta H\). In the dissolution of calcium chloride (\(\mathrm{CaCl}_2\)), the enthalpy change is negative, \(-81.5\) kJ/mol. This indicates that the process is exothermic, meaning it releases heat into the surroundings. Enthalpy change is an important concept because it helps us predict whether a process will absorb or release heat. It is measured in kilojoules per mole (kJ/mol) and is specific to certain conditions, such as temperature and pressure conditions set for the reaction. When \(\mathrm{CaCl}_2\) dissolves, the ionic bonds between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^-\) break, releasing energy. This concept helps us understand how chemical processes can lead to temperature changes in the environment they occur in.
Heat Transfer
In the process of heat transfer, energy moves from one body or system to another. This movement is typically from a region of higher temperature to a region of lower temperature. In the case of the \(\mathrm{CaCl}_2\) dissolution, the heat released by the exothermic reaction moves from the dissolved salts to the water. The heat transfer equation \(q = mc\Delta T\) is a crucial tool here. It allows us to relate the amount of heat transferred (\(q\)) to the mass (\(m\)), specific heat capacity (\(c\)), and change in temperature (\(\Delta T\)). Our calculations show \(8070\) J of heat was transferred. Understanding this concept is vital because it explains how energy exchange can lead to a change in temperature, affecting the physical states of materials.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is needed to change the temperature of a substance by one degree Celsius (°C). In our problem, water has a specific heat capacity of \(4.18\, \mathrm{J} /\mathrm{g} \cdot ^{\circ} \mathrm{C}\). This property indicates that water requires \(4.18\) joules of energy to raise the temperature of \(1\, \text{gram}\) by \(1^{\circ}\text{C}\). This is higher than many other substances, which is why water is often used as a cooling agent. Specific heat capacity helps us determine how substances react under heat exchange, aiding in calculations of heat absorbed or released in processes like the dissolution of \(\mathrm{CaCl}_2\). The specific heat capacity material not only affects calculations but also provides insights into designing efficient thermal systems.
Dissolution Process
The dissolution process involves a solid solute dissolving into a solvent to form a solution. In our example, \(\mathrm{CaCl}_2\) is the solute and water is the solvent. When \(\mathrm{CaCl}_2\) dissolves, it disintegrates into its constituent ions, \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^-\), dispersing uniformly in the water.This process is influenced by factors like temperature, stirring, and the nature of the solute and solvent. The dissolution, being exothermic, releases heat, which is typical when ionic compounds dissolve in water. Understanding dissolutions is key in chemistry, allowing us to predict solubility, mix different substances consciously, and control reactions by managing the energy they release or absorb. This knowledge is especially useful in industries like pharmaceuticals, where precise dissolution rates and behaviors affect product formulation.

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Most popular questions from this chapter

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

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