Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C},\) diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine \(\Delta H\) for C(diamond) \(\longrightarrow\) C (graphite) (1) C(diamond) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-395.4 \mathrm{~kJ}\) (2) \(2 \mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H=566.0 \mathrm{~kJ}\) (3) C(graphite) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-393.5 \mathrm{~kJ}\) (4) \(2 \mathrm{CO}(g) \longrightarrow \mathrm{C}\) ( graphite) \(+\mathrm{CO}_{2}(g) \quad \Delta H=-172.5 \mathrm{~kJ}\)

Carbon is an element that can exist in various crystalline forms, also known as allotropes. The two most well-known crystalline forms of carbon are diamond and graphite. Despite being made of the same element, their properties are significantly different due to their distinct crystal structures.
Diamond has a strong three-dimensional network where each carbon atom is covalently bonded to four other carbon atoms. This structure makes diamond extremely hard and gives it its iconic sparkle. In contrast, graphite consists of layers of carbon atoms arranged in a hexagonal pattern. Each carbon atom in graphite is bonded to three other carbon atoms within the same layer, and the layers are held together by weak van der Waals forces, allowing them to slide over each other easily. This gives graphite its softness and lubricating properties.
Understanding the crystalline forms of carbon is key to solving problems involving transformations between them, as their differing enthalpy values play a crucial role.

The transformation from diamond to graphite involves a change in the arrangement of carbon atoms. Though both are forms of pure carbon, the process of rearranging the atoms in diamond's tight three-dimensional lattice into graphite's layered structure requires an input or release of energy, known as enthalpy change, \( \Delta H \).
This transformation is extremely slow under normal conditions, which means its enthalpy change, \( \Delta H \), cannot be measured directly. Instead, it must be calculated using indirect methods such as Hess's Law. This transformation is significant in thermochemistry because it highlights the concept that different forms of the same element can have different physical and chemical properties due to their structural differences.
In the exercise, the provided chemical equations help us indirectly determine the enthalpy change for the transformation from diamond to graphite by using the known enthalpy changes of related reactions.

Hess's Law is a principle in chemistry that states the total enthalpy change of a reaction is the same, regardless of the path taken to achieve it. This law is especially useful for calculating enthalpy changes for reactions that are difficult to measure directly by using known enthalpies of other reactions.
Mathematically, Hess's Law is expressed as: \( \Delta H_{\text{total}} = \Delta H_{1} + \Delta H_{2} + ... + \Delta H_{n} \), where \( \Delta H_{1}, \Delta H_{2}, ... , \Delta H_{n} \) are the enthalpy changes of the individual steps.
In the provided exercise, Hess's Law is applied by combining and manipulating given chemical equations to calculate the enthalpy change for the transformation of diamond to graphite. By aligning the reactions involving diamond and graphite, reversing necessary reactions, and summing their enthalpy changes, we can determine \( \Delta H \) for the desired reaction.

Thermochemical equations are chemical equations that include the enthalpy change of the reaction. These equations are crucial in determining the energy involved in chemical processes and in solving enthalpy-related problems.
Each thermochemical equation provides two pieces of crucial information: the reactants and products of the reaction, and the enthalpy change \( \Delta H \). For example, the thermochemical equation \( \text{C(graphite) + O}_{2}(g) \rightarrow \text{CO}_{2}(g) \Delta H = -393.5 \text{kJ} \) tells us that the combustion of graphite in oxygen to form carbon dioxide releases 393.5 kJ of energy.
To solve the exercise, it is imperative to understand how to manipulate these equations. By combining and rearranging thermochemical equations properly, we can use the given enthalpy changes to indirectly determine the enthalpy change of the transformation of diamond to graphite.