/*! 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 61 A coffee-cup calorimeter initial... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 61

A coffee-cup calorimeter initially contains 125 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 J/g and kJ/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 the potassium bromide in the water is \(154.33\) J/g or \(18.36\) kJ/mol.

Step by step solution

01

Calculate heat absorbed by the water

To calculate the heat absorbed by the water, we use the formula: \(q = mc螖T\) Where: \(q\) is the heat absorbed (J), \(m\) is the mass of the water (g), \(c\) is the specific heat capacity (J 掳C鈦宦� g鈦宦�), and \(螖T\) is the change in temperature (掳C). Given: \(m = 125\) g, \(c = 4.18\) J 掳C鈦宦� g鈦宦�, and \(螖T = T_{final} - T_{initial} = 21.1 - 24.2 = -3.1\) 掳C. Now, calculate the heat absorbed by the water: \(q_{water} = (125\,\text{g}) (4.18\,\text{J}\,^{\circ}\mathrm{C}^{-1}\,\text{g}^{-1})(-3.1\,^{\circ}\mathrm{C}) = -1620.45\) J
02

Calculate heat absorbed by the potassium bromide

Since there is no heat transfer to the surroundings or the calorimeter, the heat absorbed by the potassium bromide is equal and opposite of the heat absorbed by the water: \(q_{KBr} = -q_{water} = 1620.45\) J
03

Calculate enthalpy change in J/g

To calculate the enthalpy change in J/g, we need to divide the heat absorbed by the potassium bromide by its mass: \(m_{KBr} = 10.5\) g Now, calculate the enthalpy change in J/g: \(\Delta H_{J/g} = \frac{q_{KBr}}{m_{KBr}} = \frac{1620.45\,\text{J}}{10.5\,\text{g}} = 154.33\) J/g
04

Calculate molar mass of potassium bromide

To calculate the molar mass of potassium bromide (KBr), we will consider the molecular weights of potassium (K) and bromine (Br). Molecular weight of potassium (K) = 39.1 g/mol Molecular weight of bromine (Br) = 79.9 g/mol Molar mass of potassium bromide (KBr) = Molecular weight of potassium (K) + Molecular weight of bromine (Br) Molar mass of (KBr) = 39.1 g/mol + 79.9 g/mol = 119 g/mol
05

Calculate enthalpy change in kJ/mol

To determine the enthalpy change in kJ/mol, we must multiply the enthalpy change in J/g by the molar mass of KBr and then convert J to kJ: \(\Delta H_{kJ/mol} = \Delta H_{J/g} \times \text{Molar Mass}_{KBr} \times \frac{1 \,\text{kJ}}{1000\,\text{J}}\) \(\Delta H_{kJ/mol} = 154.33\,\text{J/g} \times 119\,\text{g/mol} \times \frac{1 \,\text{kJ}}{1000\,\text{J}} = 18.36\) kJ/mol The enthalpy change for dissolving the potassium bromide in the water is 154.33 J/g or 18.36 kJ/mol.

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Key Concepts

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

Calorimetry
Calorimetry is a technique used to measure the amount of heat transferred during a chemical or physical process. In a coffee-cup calorimeter, which is a simple and accessible version, a reaction taking place at constant pressure allows us to measure heat changes by observing temperature variations in the liquid (usually water). This setup assumes no heat loss to the surroundings, focusing solely on the heat exchange between the substances in the reaction.
  • The key to calorimetry is the relationship between heat (q), mass (m), specific heat capacity (c), and temperature change (螖T). The formula is expressed as: \[ q = mc螖T \] where \( q \) represents the heat absorbed or released by a system.
  • In this exercise, when potassium bromide dissolves in water, the process either absorbs or releases heat, causing a measurable temperature change in the water.
By calculating how much the temperature of the water changes, and knowing the specific heat capacity of water, we can determine the heat involved in the dissolution of the salt. This proficiency in calorimetry allows us to quantify the natures of reactions, such as endothermic or exothermic reactions.
Specific Heat Capacity
Specific heat capacity is a property that defines the amount of heat required to change the temperature of 1 gram of a substance by 1掳C. It plays an essential role in calorimetry calculations, indicating how heat affects a substance's temperature.
  • The unit for specific heat capacity is typically J/g掳C. Water, commonly used in calorimetry, has a specific heat capacity of 4.18 J/g掳C.
  • This means for each gram of water, 4.18 joules are required to change its temperature by 1掳C, making water a practical heat reservoir for energy transfer measurements.
In our context, the specific heat capacity of the solution (water mixed with KBr) is assumed to be that of pure water. The use of specific heat capacity in calculations helps determine the total heat change when the dissolution process causes a temperature drop, as seen in this exercise.
Dissolution Process
The dissolution process is a chemical reaction where a solute (in this case, potassium bromide) is dissolved in a solvent (water), breaking down into its constituent ions. This process often involves changes in energy, resulting in endothermic or exothermic reactions.
  • An endothermic process absorbs heat from the surroundings, leading to a temperature drop, whereas an exothermic process releases heat, causing a temperature increase.
  • The observed temperature decrease from 24.2掳C to 21.1掳C indicates that dissolving potassium bromide is an endothermic process.
  • Calculating the enthalpy change gives insight into how much energy per gram and per mole is absorbed during dissolution.
Understanding the dissolution process and its associated enthalpy changes helps chemists predict reaction behavior and conditions necessary for a reaction to occur or be controlled effectively.

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Most popular questions from this chapter

A \(30.0 -\mathrm{g}\) sample of water at \(280 .\) K is mixed with \(50.0 \mathrm{g}\) water at \(330 . K\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: $$\begin{array}{lr}\text { Equation } & \Delta H(\mathrm{kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & -286 \end{array}$$

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.

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.

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.
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