Given:C(diamond) O2 gives CO2 ∆H =-395kJ C(graphite) O2 gives CO2 ∆H...
Given:C(diamond) O2 gives CO2 ∆H =-395kJ C(graphite) O2 gives CO2 ∆H...
Enthalpy of Formation of Diamond from Graphite
The enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its constituent elements in their standard states. In this case, we are interested in finding the enthalpy of formation of diamond from graphite.
Given:
C(diamond) + O2 → CO2 ∆H = -395 kJ
C(graphite) + O2 → CO2 ∆H = -393 kJ
To find the enthalpy of formation of diamond from graphite, we need to consider the difference in enthalpy between the two reactions.
Key Point:
The enthalpy of formation of a compound is equal to the difference in enthalpy between the product side and the reactant side of the balanced chemical equation.
Calculating the Enthalpy of Formation:
The enthalpy change for the reaction C(diamond) + O2 → CO2 is -395 kJ/mol.
The enthalpy change for the reaction C(graphite) + O2 → CO2 is -393 kJ/mol.
To calculate the enthalpy of formation of diamond from graphite, we need to subtract the enthalpy change for the reaction C(graphite) + O2 → CO2 from the enthalpy change for the reaction C(diamond) + O2 → CO2.
∆H = -395 kJ/mol - (-393 kJ/mol)
∆H = -395 kJ/mol + 393 kJ/mol
∆H = -2 kJ/mol
Answer:
The enthalpy of formation of diamond from graphite is -2 kJ/mol.
Explanation:
The negative sign indicates that the formation of diamond from graphite is an exothermic process, meaning it releases energy. The enthalpy of formation is -2 kJ/mol, indicating that the formation of diamond from graphite is favored thermodynamically.