Question

What will be the stable state of aggression of carbon at $298K$ and $1$ atm pressure?
A.Graphite
B.Arbon
C.Amorphous carbon
D.None of these

Answer 1

Hint: We know that the element carbon is a non-metal. It is in a solid condition at room temperature. Graphite, diamond, and graphene are all examples of carbon in various forms. Carbon has diverse properties depending on its form. There is only one type of atom in carbon. This denotes that carbon is an element. Carbon is solid at room temperature because its atoms are arranged in a regular pattern.

Complete answer:
We know that carbon is found in the form of graphite and diamond in nature. The change in enthalpy when one mole of a substance in the standard state ($1$ atm of pressure and $298.15K$) is created from its pure elements under the same conditions is known as the standard enthalpy of formation.
The enthalpy difference between graphite and diamond is too large for both of them to have a zero standard enthalpy of formation. The more stable form of carbon is chosen to determine which form is zero. Because this is also the least enthalpy form, graphite has a standard enthalpy of formation of zero.
Thus, Graphite is (slightly) more stable than diamond at standard temperature and pressure. Therefore, the stable state of aggression of carbon at $298K$ and $1atm$ pressure is Graphite.

Hence, Option A is the correct answer.

Note:
We must note that the thermodynamic stability of diverse carbon structures is determined by complex tradeoffs among multiple parameters in the bonding of distinct solids (allotropes) that can be formed using various bonding arrangements. Other than observation or very high-powered calculations concerning the quantum mechanics of the bonding, there is no straightforward way to identify which is the most thermodynamically stable.