Problem 50 - Chemistry and Chemical Reactivity

Question

When calcium carbonate is heated strongly, \(\mathrm{CO}_{2}\) gas is evolved. The equilibrium pressure of the gas is 1.00 bar at \(897^{\circ} \mathrm{C},\) and \(\Delta H_{\mathrm{rxn}}^{\mathrm{e}}\) at \(298 \mathrm{K}=179.0 \mathrm{kJ}\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ Estimate the value of \(\Delta S^{\circ}\) at \(897^{\circ} \mathrm{C}\) for the reaction.

Step-by-Step Solution

Answer

The estimated value of \( \Delta S^{\circ} \) at \(897^{\circ} \mathrm{C}\) is approximately 152.91 J/(mol\cdot K).

Step 1: Convert temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin. The formula to convert from Celsius to Kelvin is: \[ T(K) = T(°C) + 273.15 \]Given \( T = 897°C \), the temperature in Kelvin is:\[ T(K) = 897 + 273.15 = 1170.15 \,K \]

Step 2: Calculate Gibbs Free Energy Change (ΔG)

At equilibrium, the Gibbs Free Energy change for the reaction can be expressed in terms of pressure as:\[ \Delta G^{\circ} = -RT \ln K_p \]Where:- \( R \) is the universal gas constant, \(8.314 \, J/(mol \cdot K)\)- \( T \) is the temperature in Kelvin (1170.15 K)- \( K_p \) is the equilibrium pressure in bar for \( CO_2 \) which is given as 1.00 bar.Since \( \ln(1) = 0 \), we have:\[ \Delta G^{\circ} = -8.314 \times 1170.15 \times 0 = 0 \, J/mol \]

Step 3: Apply the Gibbs Free Energy Equation

Using the relation for Gibbs Free Energy change:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Since \( \Delta G^{\circ} \) is zero at equilibrium as we found, rearrange the equation to solve for \( \Delta S^{\circ} \):\[ 0 = \Delta H^{\circ} - T \Delta S^{\circ} \]\[ \Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} \]

Step 4: Substitute values into the entropy equation

Substitute the given and calculated values into the equation:\[ \Delta S^{\circ} = \frac{179,000 \, J/mol}{1170.15 \, K} \]Calculate \( \Delta S^{\circ} \):\[ \Delta S^{\circ} = 152.91 \, J/(mol \cdot K) \]

Key Concepts

Thermodynamics

Thermodynamics is the study of energy, heat, and their transformations. It can explain the behavior of substances when they undergo changes in temperature, pressure, and other conditions. This branch of science is vital for understanding reactions like the decomposition of calcium carbonate. When calcium carbonate is heated, it breaks down into calcium oxide and carbon dioxide gas, a process described by its thermodynamic properties. The decomposition process is an important concept because it involves energy changes evaluated through enthalpy (\( \Delta H_\text{rxn}^\circ \)) and other thermodynamic variables.
This process is influenced by several factors:

• Temperature: Higher temperatures can increase the rate of reaction.
• Pressure: Affects the equilibrium position of the reaction.
• Energy: Calculated using specific formulas in thermodynamics, such as \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \).

Understanding these concepts is crucial for predicting the behavior of reactions in varying conditions and for industrial applications where this reaction is utilized.

Entropy Change Calculation

Entropy is a measure of disorder or randomness in a system. In chemistry, the change in entropy (\( \Delta S^\circ \)) during a reaction provides insight into the system's energy dispersal. To estimate \( \Delta S^\circ \) for the decomposition of calcium carbonate, we use the relationship: \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \).
At equilibrium, the Gibbs Free Energy change, \( \Delta G^\circ \), is zero, so the equation simplifies to:\[0 = \Delta H^\circ - T \Delta S^\circ \]Rearranging gives us:\[\Delta S^\circ = \frac{\Delta H^\circ}{T}\]This calculation allows us to estimate the entropy change at the given temperature, which reflects the tendency of the reaction to progress and the energy's dispersal.

Equilibrium Pressure

Equilibrium pressure refers to the pressure at which the components of a chemical reaction are in balance, meaning the rate of the forward reaction equals the rate of the reverse reaction. Here, the equilibrium pressure of carbon dioxide evolving from the decomposition of calcium carbonate is 1.00 bar.
This pressure is critical because it defines the state at which the reaction reaches equilibrium. At this point, no further net change occurs in the concentration of reactants and products. In our case, using the equilibrium pressure helps calculate the Gibbs Free Energy change and confirms that \( \Delta G^\circ \) equals zero when the reaction is at equilibrium. Understanding equilibrium pressure is essential to comprehending how reactions stabilize under certain conditions.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G^\circ \)) is a thermodynamic property that helps predict whether a reaction will occur spontaneously. It's defined by the equation:\[\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\]In simple terms, it measures the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. For the decomposition of calcium carbonate, the reaction reaches equilibrium when \( \Delta G^\circ \) is zero.
At equilibrium:

• The forward reaction rate equals the backward reaction rate.
• There is no net energy change in the system.

Thus, the calculation of Gibbs Free Energy provides us with valuable information regarding the feasibility and direction of chemical reactions. It serves as an indicator of reaction spontaneity, guiding chemists in manipulating reaction conditions for desired outcomes.