Question

Write a balanced chemical equation for the formation of \(\mathrm{CaCO}_{3}(\mathrm{s})\) from the elements in their standard states. Find the value for \(\Delta_{f} H^{\circ}\) for \(\mathrm{CaCO}_{3}(\mathrm{s})\) in Appendix L.

Short answer

Balanced equation: \( \mathrm{Ca(s)} + \mathrm{C(s, graphite)} + \mathrm{O}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s) \). \( \Delta_{f} H^{\circ} = -1206.9 \text{ kJ/mol} \).

Step-by-step solution

Identify Elements in Standard States

First, identify the elements that make up \( \mathrm{CaCO}_{3} \). These are Calcium (Ca), Carbon (C), and Oxygen (O). In their standard states, Calcium is a solid \( \mathrm{Ca(s)} \), Carbon is a solid in the form of graphite \( \mathrm{C(s, graphite)} \), and Oxygen is a gas \( \mathrm{O}_{2}(g) \).

Write Unbalanced Reaction Equation

Write the initial unbalanced chemical equation for the formation of \( \mathrm{CaCO}_{3}(\mathrm{s}) \) from its elements in standard states: \[ \mathrm{Ca(s)} + \mathrm{C(s, graphite)} + \mathrm{O}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s) \]

Balance the Chemical Equation

Verify if the equation is balanced. In our case, the equation already has one atom of Calcium, one atom of Carbon, and three atoms of Oxygen on both sides, making it balanced. Thus, the balanced equation is: \[ \mathrm{Ca(s)} + \mathrm{C(s, graphite)} + \mathrm{O}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s) \]

Find the Enthalpy of Formation

Look up the standard enthalpy of formation \( \Delta_{f} H^{\circ} \) for \( \mathrm{CaCO}_{3}(\mathrm{s}) \) in Appendix L of your chemistry textbook or database. This value is typically given as \( -1206.9 \text{ kJ/mol} \). Ensure you use the correct value from your reference.

Key concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the starting materials, called reactants, and the resulting substances, called products. Each substance is represented by its chemical formula. For example, in a chemical equation, calcium carbonate (CaCO\(_3\)) is the product formed from calcium (Ca), carbon (C), and oxygen (O) in their standard states.
Each part of the equation conveys important information:

• The chemical formulas tell us the number and types of atoms involved.
• The arrow (\(\rightarrow\)) indicates the direction of the reaction from reactants to products.
• Coefficients, when used, show the moles of each substance involved.

Writing and understanding chemical equations help us visualize chemical processes like the formation of new compounds.

Enthalpy of Formation

The enthalpy of formation, symbolized as \(\Delta_{f} H^{\circ}\), is a key concept in thermochemistry. It is the heat change or enthalpy change that occurs when one mole of a compound forms from its constituent elements in their standard states. This value is usually reported in kilojoules per mole (kJ/mol) and is considered under standard conditions (25°C and 1 atm pressure).
In our example, \(\Delta_{f} H^{\circ}\) for CaCO\(_3\)(s) is -1206.9 kJ/mol. This negative value means that the formation of calcium carbonate from its elements releases energy, making it an exothermic reaction.
Enthalpy of formation values are essential in calculating other thermodynamic properties and understanding the stability of compounds.

Balancing Chemical Reactions

Balancing chemical reactions involves ensuring that the number of atoms of each element is the same on both sides of the equation. This aligns with the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Every atom on the reactant side must appear unchanged on the product side.
For our chemical equation \[ \mathrm{Ca(s)} + \mathrm{C(s, graphite)} + \mathrm{O}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s) \]
The atoms are already balanced: one Ca atom, one C atom, and three O atoms are present on both sides. Sometimes, trial and error is needed to balance equations, adjusting coefficients as necessary. However, practice makes this process more intuitive.

Standard States

The concept of standard states provides a reference point for thermochemical calculations. A standard state is defined as the physical state of a substance under standard conditions: 25°C (298 K) temperature and 1 atm pressure. These conditions represent a common baseline for comparing energies and enthalpies of different reactions.
The standard state of each element is its most stable form under these conditions. In our exercise, calcium's standard state is as a solid (Ca(s)), carbon is solid in graphite form (C(s, graphite)), and oxygen is a diatomic gas (O\(_2\)(g)). Recognizing these standard states is crucial to correctly assembling reactants for enthalpy calculations and writing accurate chemical equations.