Problem 89 - Chemistry: An Atoms-Focused Approach

Question

Lime Enormous amounts of lime (CaO) are used in steel industry blast furnaces to remove impurities from iron. Lime is made by heating limestone and other solid forms of \(\mathrm{CaCO}_{3}(s) .\) Why is the standard molar entropy of \(\mathrm{CaCO}_{3}(s)\) higher than that of \(\mathrm{CaO}(s)\) ? At what temperature is the pressure of \(\mathrm{CO}_{2}(g)\) over \(\mathrm{CaCO}_{3}(s)\) equal to 1.0 atm?$$\begin{array}{lcc} & \Delta H_{f}^{\circ}(\mathrm{k} J / \mathrm{mol}) & S^{\circ}[J /(\mathrm{mol} \cdot \mathrm{K})] \\\C a C O_{3}(s) & -1207 & 93 \\\\\hline \mathrm{CaO}(s) & -636 & 40 \\\\\hline \mathrm{CO}_{2}(g) & -394 & 214 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Question: Explain why the standard molar entropy of CaCO3(s) is higher than that of CaO(s) and find the temperature at which the pressure of CO2(g) over CaCO3(s) is equal to 1.0 atm. Answer: The standard molar entropy of CaCO3(s) is higher than that of CaO(s) because CaCO3(s) has a more complex structure due to the presence of carbonate ions, leading to more randomness. The pressure of CO2(g) over CaCO3(s) is equal to 1.0 atm at a temperature of approximately 1099 K.

Step 1: Explain the difference in molar entropy

Standard molar entropy is a measure of the randomness or disorder of a substance. In general, the more complex the molecule, the higher the entropy. Here, \(\mathrm{CaCO}_{3}(s)\) has a more complex structure due to the presence of carbonate ions in its crystal lattice, which requires more vibrational states and leads to more randomness. Hence, its standard molar entropy is higher than that of \(\mathrm{CaO}(s)\).

Step 2: Write the balanced equation for the decomposition of \(\mathrm{CaCO}_{3}(s)\)

The decomposition of \(\mathrm{CaCO}_{3}(s)\) occurs by heating and can be represented by the following balanced chemical equation: \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s) + \mathrm{CO}_{2}(g)\)

Step 3: Calculate the change in Gibbs free energy for the decomposition reaction

We can use the data on enthalpy (\(\Delta H^\circ\)) and standard molar entropy (\(S^\circ\)) to find the change in Gibbs free energy (\(\Delta G^\circ\)) for the reaction. The formula for the change in Gibbs free energy is: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) First, we need to calculate the \(\Delta H^\circ\) and \(\Delta S^\circ\) for the decomposition reaction. \(\Delta H^\circ_{reaction} = [\Delta H^\circ_{CaO(s)} + \Delta H^\circ_{CO_2(g)}] - \Delta H^\circ_{CaCO_3(s)}\) \(\Delta H^\circ_{reaction} = [(-636-394)-(-1207)] \mathrm{kJ/mol}\) \(\Delta H^\circ_{reaction} = 177\,\mathrm{kJ/mol}\) Now, we calculate the change in entropy: \(\Delta S^\circ_{reaction} = [S^\circ_{CaO(s)} + S^\circ_{CO_2(g)}] - S^\circ_{CaCO_3(s)}\) \(\Delta S^\circ_{reaction} = [(40+214)-93] \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}\) \(\Delta S^\circ_{reaction} = 161\, \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}\)

Step 4: Determine the pressure of \(\mathrm{CO}_{2}(g)\)

To find the temperature at which the pressure of \(\mathrm{CO}_{2}(g)\) over \(\mathrm{CaCO}_{3}(s)\) is equal to 1.0 atm, we need to use the relationship between Gibbs free energy change and the equilibrium constant. We can write: \(\Delta G^\circ = -RT \ln{K}\) Here, K is the equilibrium constant. Since we are looking for a pressure of \(\mathrm{CO}_{2}(g)\) equal to 1.0 atm, we can assume that we are at equilibrium for the decomposition reaction. At equilibrium, \(\Delta G = 0\) (equilibrium condition), so: \(\Delta G^\circ = -RT \ln{1} = 0\) Now, rearrange the equation to find the temperature: \(T = \frac{\Delta H^\circ_{reaction}}{\Delta S^\circ_{reaction}}\) \(T = \frac{177\,\mathrm{kJ/mol}}{161\, \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}}\) \(T = \frac{177000\,\mathrm{J/mol}}{161\, \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}}\) \(T \approx 1099\,\mathrm{K}\) Therefore, the pressure of \(\mathrm{CO}_{2}(g)\) over \(\mathrm{CaCO}_{3}(s)\) is equal to 1.0 atm at a temperature of approximately 1099 K.

Key Concepts

Entropy

Entropy is a key concept in thermodynamics that measures the amount of disorder or randomness in a system. In simple terms, the higher the entropy, the more chaotic the system is. Molecules with more complex structures, like \(CaCO_{3}(s)\), tend to have higher entropy because they have more vibrational, rotational, and translational motion.
CaCO₃ contains carbonate ions that increase its structural complexity. The presence of these ions means there are more possible arrangements and movements within the solid, accounting for its higher standard molar entropy compared to \(\text{CaO}(s)\).
- Entropy tends to increase with temperature, as well as with the number of molecules or atoms in a substance.
- When a substance undergoes a change, such as a chemical reaction, the total entropy of the system and the surroundings often changes. This can be used to determine the feasibility of a reaction.
Understanding entropy helps in predicting the direction of natural processes and reactions, playing a crucial role in chemical equilibrium.

Gibbs Free Energy

Gibbs Free Energy, denoted by \(\Delta G\), is a thermodynamic quantity that can predict whether a process will occur spontaneously. The change in Gibbs Free Energy combines the system's enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) changes with temperature (\(T\)) through the equation:
\(\Delta G = \Delta H - T \Delta S\).
- A negative \(\Delta G\) indicates a spontaneous process at constant pressure and temperature.
- A positive \(\Delta G\) suggests that the reaction is non-spontaneous unless conditions are changed.
- If \(\Delta G = 0\), the system is at equilibrium.For the decomposition of \(CaCO_{3}(s)\), calculating \(\Delta G\) allows us to determine the temperature at which the process occurs under given conditions. This is essential for understanding reaction conditions in industrial processes like lime production and material extraction.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of a forward reaction equals the rate of its reverse, and the concentrations of the reactants and products remain unchanged over time. At this stage, the system's free energy is minimized, and \(\Delta G = 0\). In terms of the equilibrium constant (\(K\)), the Gibbs Free Energy change is related as:
\(\Delta G^{\circ} = -RT \ln{K}\),
where \(R\) is the universal gas constant.- At equilibrium, the reaction quotient (i.e., the ratio of the concentrations of products to reactants) equals the equilibrium constant \(K\).
- For the decomposition of lime, the system reaches equilibrium when the pressure of \(CO_{2}(g)\) is 1 atm, indicating a stable and balanced state under these conditions.Understanding equilibrium is crucial for controlling reaction conditions in industrial processes, allowing for the optimization of product yield while minimizing energy consumption.

Decomposition Reaction

Decomposition reactions involve breaking down a compound into simpler substances. These reactions often require energy input, such as heat, in order to proceed. For example, the decomposition of calcium carbonate \(\text{CaCO}_{3}(s)\) is a common reaction in thermodynamics.
- The balanced chemical equation is:\(\text{CaCO}_{3}(s) \rightarrow \text{CaO}(s) + \text{CO}_{2}(g)\).
- Such reactions are essential in industry, as seen in lime production, where \(\text{CaO}(s)\) is derived from limestone by applying heat.Decomposition is a critical process as it often leads to the formation of products that are useful in various chemical processes. Understanding the energy and entropy changes involved helps optimize these reactions for industrial efficiency and cost-effectiveness.