Question
The heat of transition () of graphite into diamond would be, given the following thermochemical equations:
kJ mol
kJ mol
kJ mol
None of these
To determine the heat of transition for the reaction C(graphite) C(diamond), we apply Hess’s Law of Constant Heat Summation. This law states that the total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps .
We are given two combustion reactions: (i) C(graphite) + O(g) CO(g); kJ (ii) C(diamond) + O(g) CO(g); kJ
To find the enthalpy change for the transition from graphite to diamond, we subtract the second equation from the first: [C(graphite) + O(g)] - [C(diamond) + O(g)] [CO(g)] - [CO(g)] C(graphite) - C(diamond) 0 C(graphite) C(diamond)
The enthalpy change for this process () is calculated by performing the same operation on the enthalpy values:
Thus, the heat of transition is kJ mol.
This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.
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