Final answer:
The question deals with thermochemistry in Chemistry and involves standard enthalpy of formation, energy changes in reactions, and standard entropy calculations at standard conditions of 25 °C and 1 atm, specifically focusing on graphite as the standard state of carbon.
Step-by-step explanation:
The question involves the reaction C(graphite) + 2H₂(g) ⟶ CH₄(g) with a change in enthalpy (ΔH°) of −74.6 kJ/mol at 298 K. This kind of reaction pertains to the study of thermochemistry, which is a branch of chemistry that deals with the energy changes that occur during chemical reactions. The standard enthalpy of formation (ΔH°f) describes the enthalpy change when one mole of a compound is formed from its elements in their standard states. For instance, the enthalpy of formation of CO₂(g) from C(s, graphite) and O₂(g) is −393.5 kJ/mol, indicating that the reaction releases energy. Similarly, the standard enthalpy of formation of NO₂(g) from N₂(g) and O₂(g) is +33.2 kJ/mol, showing that energy is absorbed in forming NO₂(g). The standard state of carbon at 25 °C and 1 atm is graphite, as it is more stable than diamond under these conditions.
The change in entropy (∑S°) for a reaction can be calculated using the standard entropy values of the products and reactants - a principle that is critical in predicting the spontaneity of a reaction. The reaction equation must be balanced correctly to reflect the formation of products from reactants at standard conditions.
It is important to note that standard enthalpy (ΔH reaction equations) are always written for forming 1 mole of a product from its constituent elements. For example, for glucose formation from the elements, the reaction would be 6C(s, graphite) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s) with ΔHⁿ = −1273.3 kJ.