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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: \(q = mcΔT\) Where: \(q\) is the heat absorbed (J), \(m\) is the mass of the water (g), \(c\) is the specific heat capacity (J °C⁻¹ g⁻¹), and \(ΔT\) is the change in temperature (°C). Given: \(m = 125\) g, \(c = 4.18\) J °C⁻¹ g⁻¹, and \(ΔT = T_{final} - T_{initial} = 21.1 - 24.2 = -3.1\) °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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Most popular questions from this chapter

Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of 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.

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

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=\) the \(0.20 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\)) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C} .\) The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C} .\) What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?

A \(150.0 -\mathrm{g}\) sample of a metal at \(75.0^{\circ} \mathrm{C}\) is added to \(150.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(15.0^{\circ} \mathrm{C}\). The temperature of the water rises to \(18.3^{\circ} \mathrm{C}\). Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

The best solar panels currently available are about \(15 \%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1\) 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 124 for the energy rate supplied by the sun.)
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