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In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.

Short Answer

Expert verified
In the first experiment \(\Delta H\) is calculated by determining the heat change and using \(\Delta H = \frac{-q}{n}\). Thereafter the same process is repeated for the two parts of the second experiment by first calculating the heat change and subsequently \(\Delta H\) for each part. By comparing the two \(\Delta H\), we can verify Hess's law in that the \(\Delta H\) should remain constant, independent of the route in getting to the end reaction itself, within the limits of experimental error.

Step by step solution

01

Determine the heat transfer for each experiment

In the first experiment, heat transfer \(q = mc \Delta T\), where m and c are the mass and specific heat capacity of the substances, which we can consider as water since the densities are similar, and \(\Delta T\) is the change in temperature. Similarly, for the second experiment, the heat transfers of the two parts are calculated separately and then added together.
02

Calculate \(\Delta H\) for each experiment

This can be determined using the formula \(\Delta H = \frac{-q}{n}\), where n is the number of moles and is calculated as \(n = \text{volume} \times \text{concentration}\). The moles of \( \text{NH}_{3}(\mathrm{aq}) \rightarrow \text{NH}_{3}(\mathrm{g}) \) and \( \text{NH}_{3}(\mathrm{g}) \rightarrow \text{NH}_{4}\text{Cl}(\mathrm{aq}) \) should equal the moles used in the first experiment \( \text{NH}_{3}(\mathrm{aq}) \rightarrow \text{NH}_{4}\text{Cl}(\mathrm{aq}) \), thereby proving that the total equations align. The \(\Delta H\) of each process can be calculated.
03

Compare \(\Delta H\) values

\(\Delta H\) for the overall reaction in each experiment is compared. Because \(\Delta H\) of the reaction should be the same regardless of the steps taken to get there (according to Hess's law), both \(\Delta H\) values should be theoretically equal, within the limits of experimental error.

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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 relationships between chemical reactions and energy changes involving heat. It’s foundational for understanding why reactions occur and what changes they bring about in terms of energy. Specifically, it studies the thermal energy transfer associated with chemical reactions.

When substances react chemically, bonds are broken and formed, which involves the absorption or release of energy, typically in the form of heat. This energy exchange can be measured and calculated, giving insights into the energetics of reactions. Such thermochemical calculations are crucial in fields ranging from materials science to biochemistry, where controlling the energy of reactions is fundamental.
Enthalpy Change
The enthalpy change, denoted as \(\Delta H\), is a measurement of the heat change at constant pressure. It’s an essential concept within thermochemistry because it helps quantify the energy changes during a chemical reaction.

In a given reaction, the enthalpy change represents the difference in heat content between the products and the reactants. If \(\Delta H\) is negative, the reaction is exothermic and releases heat into the surroundings. Conversely, if \(\Delta H\) is positive, the reaction is endothermic, and it absorbs heat from the surroundings. Enthalpy changes are central to understanding whether a reaction will be spontaneous or require energy to proceed.
Chemical Calorimetry
Chemical calorimetry is a technique used to measure the amount of heat involved in a chemical reaction, physical change, or heat capacity. Utilizing devices called calorimeters, scientists can determine the heat exchanges associated with a reaction.

In a typical calorimetry experiment, the substance under investigation reacts in a closed vessel, and the resulting temperature change is measured. This change helps calculate the heat transfer (\(q\)) using the formula \(q = mc\Delta T\), where \(m\) stands for the mass, \(c\) for the specific heat capacity, and \(\Delta T\) for the temperature change. The precise measurements of heat transfer are instrumental in understanding reaction enthalpies and applying Hess's Law in real-world situations.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's based on the law of conservation of mass and the stoichiometric coefficients indicated by a balanced chemical equation.

In practice, stoichiometry is the tool that predicts how much reactants are needed or how much product will form under given conditions. Calculations involve moles, mass, volume, and concentration. In the context of thermochemistry, stoichiometry helps determine the specific amount of a substance involved in a reaction, which then allows for accurate calculation of \(\Delta H\). Understanding stoichiometry is fundamental when applying concepts such as Hess's Law because it ensures that energy change calculations are based on equivalent and proportional amounts of reactants and products.

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Most popular questions from this chapter

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

A 125 \(g\) stainless steel ball bearing \((\mathrm{spht}=\) \(0.50 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\) at \(525^{\circ} \mathrm{C}\) is dropped into \(75.0 \mathrm{mL}\) of water at \(28.5^{\circ} \mathrm{C}\) in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches \(100.0^{\circ} \mathrm{C} .\) What mass of water vaporizes while the boiling continues? \(\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

A British thermal unit (Btu) is defined as the quantity of heat required to change the temperature of 1 lb of water by \(1^{\circ}\) F. Assume the specific heat of water to be independent of temperature. How much heat is required to raise the temperature of the water in a 40 gal water heater from 48 to \(145^{\circ} \mathrm{F}\) in \((\mathrm{a}) \mathrm{Btu}\) (b) kcal; (c) kJ?

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) torches are used in welding. How much heat (in kJ) evolves when 5.0 L of \(C_{2} \mathrm{H}_{2}\) \(\left(d=1.0967 \mathrm{kg} / \mathrm{m}^{3}\right)\) is mixed with a stoichiometric amount of oxygen gas? The combustion reaction is $$\begin{array}{r} \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-1299.5 \mathrm{kJ} \end{array}$$

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

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