(a) When a 4.25 -g sample of solid ammonium nitrate dissolves in \(60.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from \(22.0^{\circ} \mathrm{C}\) to \(16.9^{\circ} \mathrm{C}\). Calculate \(\Delta H\left(\right.\) in \(\left.\mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) for the solution process $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

Short Answer

Expert verified
The enthalpy change for the solution process of solid ammonium nitrate in water is \(-25.7\,\text{kJ/mol}\) and the process is exothermic.

Step by step solution

01

Calculate the Heat Gained or Lost in the process

First, let's use the heat exchange formula to calculate the heat gained or lost in the process: $$ q = mc\Delta T $$ where: - \(q\) is the heat gained or lost (in J) - \(m\) is the mass of the solution (in g) - \(c\) is the specific heat of the solution (in J/g°C) - \(\Delta T\) is the change in temperature (in °C) Since the specific heat of the solution is the same as that of pure water, we can use the specific heat of water which is \(4.18 \mathrm{J/g\cdot\degree C}\). Also, the mass of the solution can be obtained by adding the mass of ammonium nitrate and the mass of water: $$ m = m_{NH4NO3} + m_{H2O} = 4.25\,\text{g} + 60.0\,\text{g} = 64.25\,\text{g} $$ The change in temperature is given by: $$ \Delta T = T_{final} - T_{initial} = 16.9^{\circ}\text{C} - 22.0^{\circ}\text{C} = -5.1^{\circ}\text{C} $$ Now, let's calculate the heat gained or lost: $$ q = (64.25\,\text{g})\,(4.18\,\text{J/g}\cdot\degree \text{C})\,(-5.1\,\degree\text{C}) = -1364.345\,\text{J} $$ The negative sign indicates that the heat was lost by the solution during the process.
02

Calculate the Amount of Moles of Ammonium Nitrate

Next, we need to find the amount of moles of ammonium nitrate. The molar mass of ammonium nitrate \((NH_{4}NO_{3})\) is: $$ M_{NH_{4}NO_{3}} = (1 \times 14) + (4 \times 1) + (1 \times 14) + (1 \times 16) + (3 \times 16) = 80\,\text{g/mol} $$ Now, let's find the number of moles in a 4.25-g sample: $$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.25\,\text{g}}{80\,\text{g/mol}} = 0.053125\,\text{mol} $$
03

Calculate the Enthalpy Change ΔH

Now, we can find the enthalpy change ΔH by dividing the heat gained or lost by the amount of moles: $$ \Delta H = \frac{-1364.345\,\text{J}}{0.053125\,\text{mol}} = -25,703\,\text{J/mol} $$ As the question asks for the answer in kJ/mol, we will convert J/mol to kJ/mol: $$ \Delta H = \frac{-25,703\,\text{J/mol}}{1000} = -25.7\,\text{kJ/mol} $$
04

Endothermic or Exothermic Process?

Since the enthalpy change ΔH is negative, the process is exothermic, meaning that heat is released when ammonium nitrate dissolves in water. So, the enthalpy change for the solution process is \(-25.7\,\text{kJ/mol}\) and the process is exothermic.

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Key Concepts

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

Thermochemistry
Thermochemistry is the branch of chemistry that deals with the study of heat and energy changes associated with chemical reactions and physical transformations. Understanding thermochemistry is crucial when exploring the intricate dance of molecules during reactions. For instance, when solid ammonium nitrate dissolves in water, we look at the heat exchange to determine whether energy is absorbed from or released to the surroundings. By defining the enthalpy change (\(\textstyle Delta H\)), which is the total heat content of a system, we can quantify this heat exchange in units of joules per mole (J/mol) or kilojoules per mole (kJ/mol).

To calculate the enthalpy change, we apply the formula \(q = mc\textstyle Delta T\), where 'q' represents the heat change, 'm' stands for the mass of the substance involved, 'c' is the specific heat capacity, and \(/\textstyle Delta T\) is the temperature change. This formula is a pillar in thermochemistry and allows us to track energy flow during a reaction, elucidating whether it is endothermic (absorbs heat) or exothermic (releases heat).
Calorimetry
Calorimetry is an experimental technique used to measure the amount of heat transferred in a chemical or physical process. In our example of ammonium nitrate dissolving, a coffee-cup calorimeter serves as the stage where this heat transfer is theatrically revealed. This simple, yet effective device measures temperature changes of the solution, which, when paired with the mass and specific heat capacity, allows us to calculate the lost or gained heat.

The specific heat capacity, a substance's heat absorption characteristic, is pivotal in this calculation. By using water's specific heat capacity, we can infer the heat change since the solution's heat capacity is assumed to mimic that of pure water. Through calorimetry, we thereby ascertain the thermal narrative of the dissolving process, marking the magnitude of \(\textstyle Delta H\), which in this case is negative due to the heat loss.
Solution Chemistry
Solution chemistry focuses on the interactions and behaviors of substances dissolved in solvents to form solutions. The dissolving of ammonium nitrate in water is a captivating episode in the saga of solution chemistry. When a solid ionic compound like ammonium nitrate dissolves, it dissociates into its constituent ions, ammonium (\(NH_4^+\)) and nitrate (\(NO_3^-\)), releasing or absorbing heat in the process.

Engaging with solution chemistry entails predicting the outcomes of solute-solvent interactions. The property of being an endothermic or exothermic process has profound implications for the enthalpy change. Endothermic reactions result in a temperature drop of the solution because they require heat from the surroundings. Conversely, exothermic reactions will increase the solution’s temperature. Understanding the dissolving process provides valuable insight into enthalpy changes and offers a tangible illustration of solution chemistry in action.
Endothermic and Exothermic Processes
Every chemical reaction tells a thermal tale of energy either being seized or set free. Endothermic and exothermic processes are the two main characters in this thermal narrative. An endothermic process demands energy transfer into the system, resulting in absorbing heat and a cooler environment. An example might be the melting of ice, where energy is needed to overcome molecular bonds.

On the other hand, an exothermic process is like a benevolent benefactor, offering heat to its surroundings, making it warmer. The reaction of ammonium nitrate dissolving in water fits the role of an exothermic process, as indicated by the negative enthalpy change (\(-25.7\textstyle Delta H $kJ/mol\textstyle Delta H\)). This drop in temperature showcases an energy handoff to the surrounding, a signature feature of exothermic chemistry.

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Most popular questions from this chapter

The hydrocarbons acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) have the same empirical formula. Benzene is an "aromatic" hydrocarbon, one that is unusually stable because of its structure. (a) By using data in Appendix \(\mathrm{C}\), determine the standard enthalpy change for the reaction $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$.

Consider the following unbalanced oxidation-reduction reactions in aqueous solution: $$ \begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{Li}(s) & \longrightarrow \mathrm{Ag}(s)+\mathrm{Li}^{+}(a q) \\ \mathrm{Fe}(s)+\mathrm{Na}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Na}(s) \\ \mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{KOH}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ (a) Balance each of the reactions. (b) By using data in Appen\(\operatorname{dix} C,\) calculate \(\Delta H^{\circ}\) for each of the reactions. (c) Based on the values you obtain for \(\Delta H^{\circ}\), which of the reactions would you expect to be thermodynamically favored? (d) Use the activity series to predict which of these reactions should occur. (Section 4.4) Are these results in accord with your conclusion in part (c) of this problem?

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Calculate the fuel value in \(\mathrm{kJ} / \mathrm{g}\) for each of these compounds. \((\mathrm{c})\) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that has long been used as a surgical anesthetic, has the structure \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) The complete combustion of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{~kJ}\). (a) Write a balanced equation for the combustion of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) (b) Write a balanced equation for the formation of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

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