Question

Calculate \(\Delta H^{\circ}\) in kilojoules for the synthesis of lime \((\mathrm{CaO})\) from limestone \(\left(\mathrm{CaCO}_{3}\right)\), the key step in the manufacture of cement. \(\begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(s)\right]=-1207.6 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CaO}(s)]=-634.9 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CO}_{2}(g)\right]=&-393.5 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\)

Short answer

The enthalpy change \( \Delta H^{\circ} \) is 179.2 kJ/mol.

Step-by-step solution

Understand the Reaction

The reaction presented is the decomposition of calcium carbonate (CaCO₃) into calcium oxide (CaO) and carbon dioxide (CO₂). This is represented as \( \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \). We are asked to calculate the standard enthalpy change \( \Delta H^{\circ} \) for this reaction.

Use the Enthalpy of Formation Values

We are given the standard enthalpy of formation for the substances involved: \( \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(s)\right] = -1207.6 \text{ kJ/mol} \), \( \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CaO}(s)] = -634.9 \text{ kJ/mol} \), and \( \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CO}_{2}(g)] = -393.5 \text{ kJ/mol} \). These values allow us to calculate the enthalpy change for the reaction.

Apply Hess's Law

According to Hess's law, the standard enthalpy change of a reaction can be calculated using the formula: \[ \Delta H^{\circ} = \sum \Delta H_{\mathrm{f}}^{\circ} \text{(products)} - \sum \Delta H_{\mathrm{f}}^{\circ} \text{(reactants)} \]

Calculate the Enthalpy for Products

For the products \( \mathrm{CaO}(s) \) and \( \mathrm{CO}_{2}(g) \), the total enthalpy of formation is \( \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CaO}(s)] + \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CO}_{2}(g)] = -634.9 + (-393.5) = -1028.4 \text{ kJ/mol} \).

Calculate the Enthalpy for Reactants

For the reactant \( \mathrm{CaCO}_{3}(s) \), the total enthalpy of formation is \( \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(s)\right] = -1207.6 \text{ kJ/mol} \).

Calculate the Overall Enthalpy Change

The standard enthalpy change for the reaction is calculated as the difference between the enthalpy of the products and the reactants:\[ \Delta H^{\circ} = -1028.4 - (-1207.6) \text{ kJ/mol} \]

Solve the Enthalpy Change Equation

After simplifying the expression: \[ \Delta H^{\circ} = -1028.4 + 1207.6 \text{ kJ/mol} = 179.2 \text{ kJ/mol} \]

Conclusion

The standard enthalpy change \( \Delta H^{\circ} \) for the synthesis of lime from limestone is \( 179.2 \text{ kJ/mol} \).

Key concepts

Hess's Law

Hess's Law is an important principle in chemistry that allows us to calculate the enthalpy change of a chemical reaction, regardless of the multiple stages it may go through.
It's based on the fact that enthalpy is a state function, which depends only on the initial and final states of a reaction, not on the path taken. This law gives us the flexibility to use standard enthalpy of formation values to determine the total enthalpy change even if the reaction proceeds through several steps.

• This is particularly useful when direct measurement of the reaction enthalpy is difficult.
• Using Hess's Law involves summing the enthalpies of each step to get the total change.
• The formula is given by: \[ \Delta H^{\circ} = \sum \Delta H_{\mathrm{f}}^{\circ} \text{(products)} - \sum \Delta H_{\mathrm{f}}^{\circ} \text{(reactants)} \]

In the exercise about the decomposition of calcium carbonate, Hess's Law helps us find the overall enthalpy change by using the enthalpies of formation for the products and the reactants.

Standard Enthalpy of Formation

The standard enthalpy of formation is a key concept when discussing energy changes in chemical reactions.
It refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states, under standard conditions (usually 298 K temperature and 1 atm pressure).

• This value is a crucial component in calculating the enthalpy change of reactions using Hess's Law.
• The standard enthalpy of formation for an element in its standard state is zero.
• In the exercise, we utilize the given values:
  • For calcium carbonate \( \Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(s)\right] = -1207.6 \text{ kJ/mol} \)
  • For calcium oxide \( \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CaO}(s)] = -634.9 \text{ kJ/mol} \)
  • For carbon dioxide \( \Delta H_{\mathrm{f}}^{\circ}[\mathrm{CO}_{2}(g)] = -393.5 \text{ kJ/mol} \)

These values enable the calculation of the net energy change in chemical transformations, giving insight into the energetic favorability of the process under standard conditions.

Calcium Carbonate Decomposition

Decomposing calcium carbonate (CaCO₃) to form calcium oxide (CaO) and carbon dioxide (CO₂) is a reaction of significant industrial importance.
This process is central in producing lime, which is a primary ingredient in cement manufacturing, making this reaction critical to construction and building industries. It can be represented by the chemical equation:
\( \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \)

• Decomposition reactions, such as this, involve breaking down a compound into its simpler product forms.
• For the immediate goal, we're interested in calculating the enthalpy change using the enthalpies of formation.
• Understanding the energy changes involved helps in optimizing industrial processes such as controlling the energy costs in lime production.

Knowing the enthalpy change provides valuable information regarding the heat requirements for this decomposition, thereby allowing adjustments for economic and environmental efficiency.