Which of the following is thermodynamically most stable allotrope of carbon?
Graphite
Let's analyze the given question which asks about the most thermodynamically stable allotrope of carbon. Allotropes are different structural modifications of an element; the atoms of the element are bonded together in a different manner.
Carbon exists in several allotropic forms. The question lists Graphite, Diamond, Fullerene, and Carbon-black. The thermodynamic stability of an allotrope is related to its standard enthalpy of formation. The allotrope that is considered the most stable at standard conditions (typically 298 K and 1 atm pressure) is assigned an enthalpy of formation of zero. All other allotropes will have positive enthalpies of formation relative to the most stable one, indicating they are less stable under standard conditions.
Based on experimental data and thermodynamic principles, graphite is considered the most stable allotrope of carbon under standard conditions. The standard enthalpy of formation of graphite is defined as 0 kJ/mol. For other allotropes like diamond, the standard enthalpy of formation is positive, indicating they are less stable than graphite under these conditions.
Therefore, among the given options, graphite is the thermodynamically most stable allotrope of carbon under standard conditions.
| Allotrope | Structure | Stability (vs. Graphite) | Key Properties |
|---|---|---|---|
| Graphite | Layered hexagonal | Most stable (standard conditions) | Soft, good conductor of electricity, lubricant |
| Diamond | 3D tetrahedral | Less stable (standard conditions), very stable (high P, T) | Very hard, poor conductor of electricity |
| Fullerene (e.g., \text{C}_{60}) | Spherical/Cage-like | Less stable | Semiconductor, used in materials science |
| Carbon-black | Amorphous particles | Less stable | Used as pigment, filler |
The stability of different allotropes can be compared using their standard free energy of formation (\Delta_f G^\circ). Since \Delta G = \Delta H - T\Delta S, both enthalpy and entropy play a role. However, under standard conditions, enthalpy differences are often dominant when comparing solid phases of the same element. The transformation of one allotrope to another often requires significant activation energy, which is why less stable allotropes like diamond can exist for long periods under standard conditions.
The concept of standard state defines the most stable form of an element under standard pressure (1 bar) and a specified temperature (usually 298.15 K). For carbon, graphite is defined as the standard state, meaning its standard enthalpy and free energy of formation are zero.
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