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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.

Short Answer

Expert verified
The enthalpy change for dissolving potassium bromide in water is approximately \(-34.43 \, \mathrm{J/g}\) and \(-120.6\, \mathrm{kJ/mol}\).

Step by step solution

01

Calculate the heat absorbed by the potassium bromide

We'll start by determining the heat absorbed by the potassium bromide during the dissolving process. To do this, we'll use the formula: heat absorbed (q) = mass (m) × specific heat capacity (C) × temperature change (ΔT). The temperature change can be found by subtracting the initial temperature from the final temperature.
02

Calculate the heat absorbed by water found in the calorimeter

Now, let's calculate the heat absorbed by water in the calorimeter using the same formula as in Step 1. It's important to note that the heat should be negative since the water cools down.
03

Determine the total heat absorbed

Next, let's find the total heat absorbed by adding the heat absorbed by potassium bromide and water.
04

Calculate enthalpy change in J/g

To find the enthalpy change in Joules per gram (J/g), we need to divide the total heat absorbed by the mass of potassium bromide.
05

Calculate the molar mass of potassium bromide

To convert our enthalpy change from J/g to kJ/mol, first, we need to find the molar mass of potassium bromide. The molar mass can be calculated by adding the molar masses of potassium (K) and bromine (Br), which are 39.10 g/mol and 79.90 g/mol, respectively.
06

Calculate enthalpy change in kJ/mol

Now that we have the molar mass of potassium bromide, we can calculate the enthalpy change in kilojoules per mole (kJ/mol) by multiplying the enthalpy change in J/g by the molar mass of potassium bromide and converting the units to kJ/mol. Following these steps, you can calculate the enthalpy change for dissolving the salt in both J/g and kJ/mol.

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Key Concepts

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

Understanding the Coffee-Cup Calorimeter
The coffee-cup calorimeter is a simple yet effective tool used to measure the heat changes during chemical reactions. It consists of two nested Styrofoam cups, a lid, a thermometer, and sometimes a stirrer. This setup works well because Styrofoam acts as an insulator, minimizing heat exchange with the environment.

When you perform an experiment with a coffee-cup calorimeter, like dissolving a salt in water, any temperature change can be attributed to the reaction itself, assuming no energy is lost to surroundings.

In our specific exercise, we used a coffee-cup calorimeter to observe how the dissolution of potassium bromide affects the temperature of the surrounding water. This helps us compute how much heat was either absorbed or released by the system.
Specific Heat Capacity and Its Role
Specific heat capacity is a crucial property that tells us how much heat energy is needed to change the temperature of one gram of a substance by one degree Celsius. It's denoted as C and its unit is J/g°C.

In the exercise, we assume the specific heat capacity of the solution (water with dissolved potassium bromide) is the same as pure water, which is 4.18 J/g°C. This assumption helps simplify calculations because water makes up most of the solution's mass.

Knowing the specific heat capacity lets us calculate the heat exchanged during the dissolution process just by observing temperature changes.
The Dissolution Process
The dissolution process involves breaking down the crystal structure of the solute—in this case, potassium bromide—into individual ions that are uniformly distributed in the solvent (water here). This process can either absorb or release heat, affecting the surroundings.

If the solution's temperature decreases, the dissolution is endothermic, meaning it absorbs heat from the surroundings. Conversely, if the temperature increases, the dissolution is exothermic, releasing heat into the surroundings.

In our experiment, the final temperature decreased, indicating that the dissolution of potassium bromide is an endothermic process.
Exploring Temperature Change
Temperature change is pivotal in understanding energy exchanges in a reaction. You calculate it by subtracting the initial temperature from the final temperature of the reaction system.

For our experiment with potassium bromide and water, the initial temperature was 24.2°C, and the final temperature was 21.1°C. The temperature change (ΔT) is -3.1°C.

This negative value supports our finding that the dissolution absorbed heat, as indicated by the solution's cooling.
Calculating Molar Mass
Molar mass is the mass of one mole of a substance, giving a bridge between grams and moles in chemical calculations. We find it by summing the atomic masses of each element in a molecule from the periodic table.

In potassium bromide (KBr), the molar mass is calculated by adding the molar mass of potassium (39.10 g/mol) and bromine (79.90 g/mol). The result is a molar mass of 119.00 g/mol.

Knowing the molar mass lets us convert our findings from J/g to kJ/mol, which is a common format in chemical thermodynamics.

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Most popular questions from this chapter

Are the following processes exothermic or endothermic? a. When solid \(\mathrm{KBr}\) is dissolved in water, the solution gets colder. b. Natural gas \(\left(\mathrm{CH}_{4}\right)\) is burned in a furnace. c. When concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is added to water, the solution gets very hot. d. Water is boiled in a teakettle.

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{~kJ}\) : a. What quantity of heat is required to produce 1 mole of mercury by this reaction? b. What quantity of heat is required to produce 1 mole of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$ 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s) $$

Consider the following cyclic process carried out in two steps on a gas: Step 1: \(45 \mathrm{~J}\) of heat is added to the gas, and \(10 . \mathrm{J}\) of expansion work is performed. Step \(2: 60 . \mathrm{J}\) of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in Step \(2 .\)

The best solar panels currently available are about \(19 \%\) efficient in converting sunlight to electricity. A typical home will use about \(40 . \mathrm{kWh}\) of electricity per day \((1 \mathrm{kWh}=1 \mathrm{kilowatt}\) hour; \(1 \mathrm{~kW}=1000 \mathrm{~J} / \mathrm{s}\) ). Assuming \(8.0\) hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 126 for the energy rate supplied by the sun.)

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

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