Problem 97 - Chemistry The Central Science

Question

Limestone stalactites and stalagmites are formed in caves by the following reaction: $$ \mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under 1 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?

Step-by-Step Solution

Answer

The enthalpy change (ΔH) for the given reaction is \(41.42 \: \text{kJ}\) and the internal energy change (ΔE) is \(36.48 \: \text{kJ}\).

Step 1: Calculate ΔH

The enthalpy change (ΔH) is given by the equation: ΔH = q + PΔV, where q is the heat absorbed and PΔV is the work done on the system. We are given the values of q (38.95 kJ) and PΔV (2.47 kJ). Plug these values into the equation and calculate ΔH: ΔH = 38.95 kJ + 2.47 kJ = 41.42 kJ So, the enthalpy change for this reaction is 41.42 kJ.

Step 2: Calculate ΔE

The internal energy change (ΔE) is given by the equation: ΔE = q + w, where w is the work done on the system. In this case, the work done (w) is -PΔV (as the work is done by the system, not on the system). We are given the values of q (38.95 kJ) and PΔV (2.47 kJ). Plug these values into the equation and calculate ΔE: ΔE = 38.95 kJ - 2.47 kJ = 36.48 kJ So, the internal energy change for this reaction is 36.48 kJ. In summary, the enthalpy change (ΔH) for the given reaction is 41.42 kJ and the internal energy change (ΔE) is 36.48 kJ.

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), measures the total heat content exchanged in a reaction. It helps us understand if a reaction absorbs or releases energy.

In this problem, we consider the formation of limestone stalactites and stalagmites through a chemical reaction in caves. The main elements are calcium, bicarbonate ions, and water forming calcium carbonate, carbon dioxide, and water.

To calculate \( \Delta H \), we use the formula: \[ \Delta H = q + P \Delta V \]where \( q \) is the heat absorbed, and \( P \Delta V \) is the work done on the system, often due to gas expansion. The values given are \( q = 38.95 \text{ kJ} \) and \( P \Delta V = 2.47 \text{ kJ} \). By plugging these into the equation, we get:

• \( \Delta H = 38.95 \text{ kJ} + 2.47 \text{ kJ} = 41.42 \text{ kJ} \)

This means the reaction is endothermic, absorbing heat from the environment.

Internal Energy

Internal energy change, symbolized by \( \Delta E \), shows the total energy change within a system, accounting for heat absorbed and work done.

In this reaction involving the formation of limestone structures in caves, we need to determine the energy change with respect to internal adjustments. The formula we use here is: \[ \Delta E = q + w \]where \( q \) is the heat absorbed and \( w \) is the work. But remember, in this scenario, work is done by the system because of gas expansion, thus \( w = -P \Delta V \).

Substituting the known values, \( q = 38.95 \text{ kJ} \) and \( P \Delta V = 2.47 \text{ kJ} \), we calculate:

• \( \Delta E = 38.95 \text{ kJ} - 2.47 \text{ kJ} = 36.48 \text{ kJ} \)

This essentially indicates the system has absorbed heat but also done work by expanding, resulting in an overall energy uptake.

Limestone Stalactites and Stalagmites

Stalactites and stalagmites are fascinating natural structures often found in caves, formed by dripstone deposits. They create unique landscapes, exciting for both scientists and explorers.

The process of their formation is a vivid example of a chemical reaction in nature, involving the dissolution of calcium carbonate. When water, rich in calcium ions \( \mathrm{Ca}^{2+}(aq) \) and bicarbonate \( \mathrm{HCO}_{3}^{-}(aq) \), seeps through limestone, calcium carbonate precipitates, forming solid crystals that lengthen over time.

These reactions are crucial because:

• They show how chemical processes shape natural formations.
• They involve both energy release and absorption, displayed by changes in enthalpy and internal energy.
• They demonstrate principles of thermochemistry in real-world settings, with gas expansion playing a role in cave atmospheres.

Each deposited layer builds over many years, reflecting both time and environmental changes affecting the cave system.