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Limestone stalactites and stalagmites are formed in caves by the following reaction: \(\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow\) \(\mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I)\) If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under \(1 \mathrm{~atm}\) pressure, the reaction performs \(2.47 \mathrm{~kJ}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{CO}_{2}\) forms. At the same time, \(38.95 \mathrm{~kJ}\) of heat is absorbed from the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Short Answer

Expert verified
The values of ΔH and ΔE for this reaction are 43.89 kJ and 41.42 kJ, respectively.

Step by step solution

01

Calculate the internal energy change (ΔE)

We are given the values of heat absorbed (q) and P-V work performed (w). We can use the First Law of Thermodynamics to calculate the internal energy change (ΔE): ΔE = q + w ΔE = 38.95 kJ + 2.47 kJ ΔE = 41.42 kJ
02

Calculate the enthalpy change (ΔH)

Now that we have calculated the internal energy change (ΔE), we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔE) to find ΔH: ΔH = ΔE + PΔV We are given the value of PΔV, which is 2.47 kJ. So, ΔH = 41.42 kJ + 2.47 kJ ΔH = 43.89 kJ The values of ΔH and ΔE for this reaction are 43.89 kJ and 41.42 kJ, respectively.

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Key Concepts

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

Enthalpy Change
Understanding the concept of enthalpy change, denoted as \(\Delta H\), is crucial when studying chemical reactions. Enthalpy change is the heat content that is transferred in a chemical process at constant pressure. To envision this, imagine a reaction taking place in an open container where atmospheric pressure remains constant. If the reaction releases heat, it is exothermic, and \(\Delta H\) is negative. Conversely, if the reaction absorbs heat, it is endothermic, and \(\Delta H\) becomes positive.

In the formation of limestone stalactites and stalagmites, the reaction absorbs \(38.95 \text{kJ}\) from the environment, which indicates an endothermic process. This heat absorption is a critical part of \(\Delta H\), resulting in a positive value. Knowing how to calculate \(\Delta H\) is not only essential in understanding the thermodynamics of the reaction but also in assessing the energy impact on the surrounding environment.

Remember, \(\Delta H\) provides valuable information about the energy changes during a reaction but does not give the complete picture of the system's overall energy change, which is where internal energy change and other thermodynamic parameters come into play.
Internal Energy Change
The internal energy change, signified by \(\Delta E\), relates to how the energy within a system changes. It encompasses all forms of energy present, including kinetic and potential energies at the microscopic level. Unlike enthalpy, which is concerned with heat at constant pressure, internal energy is a more comprehensive measure of energy changes within a system.

According to the first law of thermodynamics, all energy in the universe is conserved. Therefore, the change in internal energy of a system can be determined by the heat added to the system (denoted as \(q\)) and the work done by the system on its surroundings (denoted as \(w\)). The relationship can be succinctly put as \(\Delta E = q + w\).

In the context of limestone stalactite and stalagmite formation, the work done by the reaction to push back the atmosphere, as \(\mathrm{CO}_2\) gas forms, adds to the internal energy. Despite being often smaller than the heat term, this \(P-V\) work is an indispensable component of \(\Delta E\). For students, grasping the concept of internal energy is key to understanding the broader picture of how reactions affect and are affected by their surroundings.
First Law of Thermodynamics
The first law of thermodynamics, often referred to as the law of energy conservation, states that energy can neither be created nor destroyed, only transformed from one form to another. This principle is the foundation of thermochemistry and guides us in understanding how energy transfers during chemical reactions.

When we apply this law to chemical processes, like the formation of stalactites and stalagmites, we say that the total energy of an isolated system is constant. If a reaction system absorbs heat (\(q\)) from its surroundings and does work (\(w\)) on the surroundings, the internal energy change (\(\Delta E\)) equals the sum of the heat added to the system plus the work done by the system on its surroundings, \(\Delta E = q + w\).

For instance, when calculating the internal energy change in the limestone cave reaction, the absorbed heat and the work done on the environment by the expanding gas were both considered to derive the total internal energy change. This concept ensures that all energy transfers are accounted for, maintaining the equilibrium of energy within a given system.
Limestone Stalactite and Stalagmite Formation
Limestone stalactites and stalagmites are fascinating natural structures formed by the deposition of minerals from water dripping in caves. The chemical reaction involved combines aqueous ions of calcium and bicarbonate to form solid calcium carbonate (limestone), carbon dioxide gas, and water.

The formation process is a remarkable example of thermochemistry at work in nature. As the calcium carbonate precipitates out of the water solution, carbon dioxide is released into the cave's atmosphere, creating air pressure that performs work. Meanwhile, heat is absorbed from the cave's environment to drive the reaction forward, indicating an endothermic process.

Understanding the thermochemical principles in this natural process helps clarify how environmental factors, like temperature and air pressure, contribute to the growth of these geological formations. It also provides insights into the cave's ecosystem and energy dynamics. For students, analyzing the formation of stalactites and stalagmites through the lens of enthalpy and internal energy changes solidifies their comprehension of the subtle interplay between chemical reactions and the environment.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is used as a fuel in race cars. (a) Write a balanced equation for the combustion of liquid methanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced by combustion per liter of methanol. Methanol has a density of \(0.791 \mathrm{~g} / \mathrm{mL}\) (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

The sun supplies about \(1.0\) kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{~kW} / \mathrm{m}^{2}\right.\), where a watt \(=1 \mathrm{~J} / \mathrm{s})\). Plants produce the equivalent of about \(0.20 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H &=5645 \mathrm{~kJ} \end{aligned} $$

Consider the combustion of a single molecule of \(\mathrm{CH}_{4}(g)\) forming \(\mathrm{H}_{2} \mathrm{O}(l)\) as a product. (a) How much energy, in J. is produced during this reaction? (b) A typical X-ray photon has an energy of \(8 \mathrm{keV}\). How does the energy of combustion compare to the energy of the X-ray photon?

Suppose you toss a tennis ball upward. (a) Does the kinetic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball, but of twice the mass, how high would it go in comparison to the tennis ball? Explain your answers.

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 Appendix 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? (That is, which would you expect to be spontaneous?) (d) Use the activity series to predict which of these reactions should occur. ono (Section 4.4) Are these results in accord with your conclusion in part (c) of this problem?

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