/*! 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 6: Problem 61

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 is approximately 154.4 J/g and 18.375 kJ/mol.

Step by step solution

01

Calculate the heat change of water and KBr

To calculate the heat change, we will use the heat capacity formula: \[q = mc\Delta T\] where q represents heat change, m is mass, c is specific heat capacity, and 螖T is the temperature change. We will first calculate the q for water and then for KBr.
02

Step 1a: Calculate the heat change of water

For water, mass (m) = 125 g, specific heat capacity (c) = 4.18 J/掳C路g, initial temperature (T1) = 24.2掳C, and final temperature (T2) = 21.1掳C. So, the temperature change, 螖T = T2 - T1 螖T = 21.1 - 24.2 = -3.1掳C Now, using the heat capacity formula for water: q_water = (125 g)(4.18 J/掳C路g)(-3.1掳C) = -1621.7 J
03

Step 1b: Calculate the heat change of KBr

Since no heat is transferred to the surroundings or the calorimeter, the heat change of KBr must be equal but opposite in sign to that of water. q_KBr = -q_water = 1621.7 J
04

Calculate the enthalpy change per gram of KBr

Now that we have the heat change of KBr, we can calculate the enthalpy change per gram by dividing the heat change by the mass of KBr. Enthalpy change per gram = q_KBr / mass of KBr Enthalpy change per gram = 1621.7 J / 10.5 g 鈮� 154.4 J/g
05

Calculate the enthalpy change per mole of KBr

Finally, to calculate the enthalpy change per mole, we must first determine the molar mass of KBr. The atomic mass of potassium (K) is 39 g/mol, and the atomic mass of bromine (Br) is 80 g/mol. Therefore, the molar mass of KBr is: Molar mass of KBr = 39 g/mol + 80 g/mol = 119 g/mol Now, to find the enthalpy change per mole, we will multiply the enthalpy change per gram with the molar mass of KBr: Enthalpy change per mole = Enthalpy change per gram 脳 Molar mass of KBr Enthalpy change per mole = 154.4 J/g 脳 119 g/mol 鈮� 18374.6 J/mol We can convert this value to kilojoules per mole by dividing by 1000: Enthalpy change per mole = 18374.6 J/mol 梅 1000 = 18.375 kJ/mol The enthalpy change for dissolving potassium bromide is approximately 154.4 J/g and 18.375 kJ/mol.

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Most popular questions from this chapter

Given the following data $$ \begin{aligned} 2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) & \Delta H &=167.4 \mathrm{~kJ} \\ 2 \mathrm{ClF}_{3}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+3 \mathrm{~F}_{2} \mathrm{O}(g) & \Delta H &=341.4 \mathrm{~kJ} \\\ 2 \mathrm{~F}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~F}_{2} \mathrm{O}(g) & \Delta H &=-43.4 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g) $$

Standard enthalpies of formation are relative values. What are \(\Delta H_{\mathrm{f}}^{\circ}\) values relative to?

Using the following data, calculate the standard heat of formation of \(\operatorname{ICl}(g)\) in \(\mathrm{kJ} / \mathrm{mol}\) : $$ \begin{aligned} \mathrm{Cl}_{2}(g) & \longrightarrow 2 \mathrm{Cl}(g) & \Delta H^{\circ} &=242.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(g) & \longrightarrow 2 \mathrm{I}(g) & \Delta H^{\circ} &=151.0 \mathrm{~kJ} \\ \mathrm{ICl}(g) & \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g) & \Delta H^{\circ} &=211.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(s) & \Delta H^{\circ}=62.8 \mathrm{~kJ} \end{aligned} $$

What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?

A \(5.00-\mathrm{g}\) sample of aluminum pellets (specific heat capacity \(=\) \(0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity \(=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) are heated to \(100.0^{\circ} \mathrm{C}\). The mixture of hot iron and aluminum is then dropped into \(97.3 \mathrm{~g}\) water at \(22.0^{\circ} \mathrm{C}\). Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
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