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Chapter 6: Problem 64

In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with \(75.0 \mathrm{~g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\). Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-25.92 \mathrm{~kJ/mol}\).

Step by step solution

01

Calculate the change in temperature.

To calculate the heat absorbed or released, we first need to find the change in temperature: ΔT = T_final - T_initial ΔT = 23.34°C - 25.00°C ΔT = -1.66°C
02

Calculate the heat absorbed or released by the solution.

Next, use the heat capacity (C) of the solution, the mass (m) of the water, and the change in temperature (ΔT) to find the heat absorbed or released (q): q = m × C × ΔT q = 75.0 g × 4.18 J/g °C × (-1.66°C) q = -518.37 J
03

Calculate the moles of the dissolved ammonium nitrate salt, NH4NO3.

To find the moles (n) of the dissolved ammonium nitrate salt, divide the mass (m=1.60 g) by the molar mass: NH4NO3 has molar mass of 14.01 (N) + 4(1.01) (4H) + 14.01 (N) + 3(16.00) (3O) = 80.05 g/mol n = m / molar_mass n = 1.60 g / 80.05 g/mol n = 0.0200 mol
04

Calculate the enthalpy change for the dissolution of NH4NO3.

To find the enthalpy change (ΔH) for the dissolution of NH4NO3, divide the heat absorbed or released (q) by the moles (n) of the salt: ΔH = q / n ΔH = -518.37 J / 0.0200 mol ΔH = -25918.5 J/mol
05

Convert the enthalpy change to kJ/mol.

Finally, convert the enthalpy change from J/mol to kJ/mol by dividing by 1000: ΔH = -25918.5 J/mol / 1000 ΔH = -25.92 kJ/mol So, the enthalpy change for the dissolution of NH4NO3 is -25.92 kJ/mol.

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Most popular questions from this chapter

The sun supplies energy at a rate of about \(1.0\) kilowatt per square meter of surface area \((1\) watt \(=1 \mathrm{~J} / \mathrm{s})\). The plants in an agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\right)\). Assum- ing that sucrose is produced by the reaction \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g)\) \(\Delta H=5640 \mathrm{~kJ}\) calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis.

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{\mathrm{f}}^{\circ}\) for the hydrocarbon:. $$ \begin{aligned} \Delta H_{\text {reaction }}^{\circ} &=-2044.5 \mathrm{~kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at \(1.00 \mathrm{~atm}\), \(200 .{ }^{\circ} \mathrm{C}=0.751 \mathrm{~g} / \mathrm{L}\) The density of the hydrocarbon is less than the density of \(\mathrm{Kr}\) at the same conditions.

The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{~kJ} $$ a. How much heat is released when \(4.00\) moles of iron are reacted with excess \(\mathrm{O}_{2}\) ? b. How much heat is released when \(1.00\) mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when \(1.00 \mathrm{~g}\) iron is reacted with excess \(\mathrm{O}_{2}\) ? d. How much heat is released when \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

Hess's law is really just another statement of the first law of thermodynamics. Explain.

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.
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