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 \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in \(125 \mathrm{~g}\) 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 \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\).

Short Answer

Expert verified
The final temperature of the solution when an 11.0 g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water is \(9.6^{\circ}\mathrm{C}\).

Step by step solution

01

1. Calculate moles of the substance

First, we need to find the number of moles of \(\mathrm{CaCl}_{2}\). In order to do this, we will use the formula: $$ n = \frac{m}{M} $$ where n is the moles, m is the mass, and M is the molar mass. The molar mass of \(\mathrm{CaCl}_{2}\) is: $$ M = 40.08 (\mathrm{Ca}) + 2 \times 35.45 (\mathrm{Cl}) = 110.98 \frac{\mathrm{g}}{\mathrm{mol}} $$ Thus, the number of moles is: $$ n = \frac{11.0 \mathrm{~g}}{110.98 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0991 \mathrm{~mol} $$
02

2. Calculate heat released when dissolving

Now, we need to find the heat released when dissolving the substance. We will use the formula: $$ q = n \cdot \Delta H $$ where q is the heat released, n is the number of moles, and \(\Delta H\) is the dissolving heat. Given \(\Delta H = -81.5~\mathrm{kJ}\), we have: $$ q = 0.0991 \mathrm{~mol} \cdot -81.5 \frac{\mathrm{kJ}}{\mathrm{mol}} = -8.07 \mathrm{~kJ} $$
03

3. Convert heat to Joules

The heat released should be in Joules, so let's convert it: $$ q = -8.07 \mathrm{~kJ} \cdot 1000 \frac{\mathrm{J}}{\mathrm{kJ}} = -8070 \mathrm{~J} $$
04

4. Calculate heat absorbed by the water

Now, we need to find the heat absorbed by the water using the heat capacity equation: $$ q = mc \Delta T $$ where m is the mass of water, c is the specific heat capacity, and \(\Delta T\) is the temperature difference. Given the mass of water \(m = 125 \mathrm{~g}\) and specific heat capacity of water \(c = 4.18 \mathrm{~J}/({}^{\circ}\mathrm{C} \cdot \mathrm{g})\), we have: $$ -8070 \mathrm{~J} = 125 \mathrm{~g} \cdot 4.18 \frac{\mathrm{J}}{{}^{\circ}\mathrm{C} \cdot \mathrm{g}} \cdot \Delta T $$
05

5. Calculate the final temperature

We will now solve for \(\Delta T\) and find the final temperature: $$ \Delta T = \frac{-8070 \mathrm{~J}}{125 \mathrm{~g} \cdot 4.18 \frac{\mathrm{J}}{{}^{\circ}\mathrm{C} \cdot \mathrm{g}}} = -15.4{}^{\circ}\mathrm{C} $$ Calculate the final temperature by adding the initial temperature to the calculated temperature difference: $$ T_{f} = T_{i} + \Delta T = 25.0^{\circ}\mathrm{C} - 15.4^{\circ}\mathrm{C} = 9.6^{\circ}\mathrm{C} $$ The final temperature of the solution is \(9.6^{\circ}\mathrm{C}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is essential for various chemistry problems, including calculating the amounts of materials required for a reaction. Molar mass is the mass of one mole of a substance, measured in grams per mole (\text{g/mol}). A mole is a fundamental unit in chemistry that stands for Avogadro's number of entities (atoms, molecules, etc.), which is approximately \(6.022 \times 10^{23}\).

To find the molar mass of a compound such as calcium chloride (\text{CaCl}_2), we add the atomic masses of calcium (\text{Ca}) and chlorine (\text{Cl}) found on the periodic table of elements. Calcium has an atomic mass of 40.08 g/mol, and chlorine has an atomic mass of about 35.45 g/mol. Since there are two chlorine atoms in calcium chloride, we multiply the mass of chlorine by two. The molar mass calculation becomes:\begin{align*}M &= 40.08 (\text{Ca}) + 2 \times 35.45 (\text{Cl}) \&= 110.98 \frac{\text{g}}{\text{mol}}\end{align*}
Thus, the molar mass of \text{CaCl}_2 is \(110.98 \text{g/mol}\). This molar mass is integral to converting grams of \text{CaCl}_2 to moles, a step needed to understand the subsequent behavior of the substance when it undergoes a chemical reaction or process such as dissolution.
Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), indicates the amount of heat released or absorbed during a chemical process, measured in joules per mole (\text{J/mol}) or kilojoules per mole (\text{kJ/mol}). A negative \(\Delta H\) signifies an exothermic process where heat is released to the surroundings while a positive \(\Delta H\) indicates an endothermic process where heat is absorbed from the surroundings.

In the dissolution of calcium chloride (\text{CaCl}_2), the enthalpy change is given as \(-81.5 \text{kJ/mol}\). This tells us that for every mole of \text{CaCl}_2 that dissolves, \(81.5 \text{kJ}\) of heat is released into the surroundings. To calculate the total heat change (\text{q}) when a known amount of \text{CaCl}_2 dissolves, we multiply the number of moles (\text{n}) by the enthalpy change:\(\text{q} = \text{n} \cdot \Delta \text{H}\).
The step-by-step calculation provided exemplifies how to use the molar mass and moles of \text{CaCl}_2 to find the total heat released upon dissolution, an essential part of determining the solution's final temperature.
Heat Capacity
Heat capacity is a property of a substance that indicates how much heat energy is required to raise the temperature of that substance by one degree Celsius. The specific heat capacity (\text{c}) is defined as the amount of heat per unit mass needed to raise the temperature by one degree Celsius, expressed in joules per gram degree Celsius (\text{J/g}^\circ\text{C}).

Water has a relatively high specific heat capacity, which is typically \(4.18 \text{J/g}^\circ\text{C}\). This means that it takes \(4.18 \text{J}\) of energy to raise one gram of water by one degree Celsius. This property is crucial in the dissolution exercise because the heat released by the dissolving \text{CaCl}_2 is assumed to be entirely absorbed by the water, leading to a change in the water's temperature.

By using the specific heat capacity of water, we can calculate the heat absorbed and thus the change in temperature when a substance like \text{CaCl}_2 dissolves in water. The relationship between heat (\text{q}), mass (\text{m}), specific heat capacity (\text{c}), and temperature change (\(\Delta T\)) is given by the formula: \(\text{q} = \text{m} \text{c} \Delta T\).
Temperature Change Calculation
Calculating temperature change is a pivotal task that correlates the heat absorbed or released by a substance to the resultant temperature difference. By rearranging the heat capacity equation (\text{q} = \text{mc}\Delta T), you can solve for the temperature change:\(\Delta T = \frac{\text{q}}{\text{mc}}\).

In practical terms, if we know the specific heat capacity of the solvent (water in this case) and the amount of heat released during the dissolution of a compound like \text{CaCl}_2, we can determine how much the temperature of the solvent will change. Remember, the sign of the heat (\text{q}) should reflect exothermic (\text{q} is negative) or endothermic (\text{q} is positive) processes.

Using the heat absorbed by the water and its specific heat capacity, you can find the temperature change of the water. This change in temperature is critical to determine the final temperature of the solution. For example, if the heat released is sufficient to reduce the water temperature by \(15.4^\circ\text{C}\), and the initial temperature was \(25.0^\circ\text{C}\), the final temperature will be the initial temperature minus the temperature change:\(\text{T}_{\text{f}} = \text{T}_{\text{i}} + \Delta T\).
This equation yields the final outcome for the temperature of the solution after the dissolution process, which is central to understanding the thermal effects in chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A coffee-cup calorimeter initially contains \(125 \mathrm{~g}\) water at \(24.2^{\circ} \mathrm{C}\). Potassium bromide \((10.5 \mathrm{~g})\), also at \(24.2^{\circ} \mathrm{C}\), is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\). Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and \(\mathrm{kJ} / \mathrm{mol}\). Assume that the specific heat capacity of the solution is \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

At \(298 \mathrm{~K}\), the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{~kJ} / \mathrm{mol}\) and \(49 \mathrm{~kJ} / \mathrm{mol}\), respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of \(\mathrm{NaCl}\), \(\mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), and \(\mathrm{PbSO}_{4}\) that have \(\Delta H^{\circ}\) values equal to \(\Delta H_{\mathrm{f}}^{\circ}\) for each compound.

A sample consisting of \(22.7 \mathrm{~g}\) of a nongaseous, unstable compound \(\mathrm{X}\) is placed inside a metal cylinder with a radius of \(8.00 \mathrm{~cm}\), and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance between the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in \(10.00 \mathrm{~kg}\) water at \(25.00^{\circ} \mathrm{C}\). The barometric pressure is 778 torr. When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of \(29.52^{\circ} \mathrm{C}\), and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is \(59.8 \mathrm{~cm}\). Chemical analysis shows that the cylinder contains \(0.300 \mathrm{~mol}\) carbon dioxide, \(0.250\) mol liquid water, \(0.025\) mol oxygen gas, and an undetermined amount of a gaseous element \(\mathrm{A}\). It is known that the enthalpy change for the decomposition of \(X\), according to the reaction described above, is \(-1893\) \(\mathrm{kJ} / \mathrm{mol} \mathrm{X}\). The standard enthalpies of formation for gaseous carbon dioxide and liquid water are \(-393.5 \mathrm{~kJ} / \mathrm{mol}\) and \(-286 \mathrm{~kJ} / \mathrm{mol}\), respectively. The heat capacity for water is \(4.184 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\). The conversion factor between \(\mathrm{L} \cdot \mathrm{atm}\) and \(\mathrm{J}\) can be determined from the two values for the gas constant \(R\), namely, \(0.08206 \mathrm{~L}\). \(\mathrm{atm} / \mathrm{K} \cdot \mathrm{mol}\) and \(8.3145 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). The vapor pressure of water at \(29.5^{\circ} \mathrm{C}\) is 31 torr. Assume that the heat capacity of the pistonand-cylinder apparatus is negligible and that the piston has negligible mass. Given the preceding information, determine a. The formula for \(\mathrm{X}\). b. The pressure-volume work (in \(\mathrm{kJ}\) ) for the decomposition of the \(22.7-\mathrm{g}\) sample of \(\mathrm{X}\). c. The molar change in internal energy for the decomposition of \(X\) and the approximate standard enthalpy of formation for \(X\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free