Question

Consider the following reaction:
$\text{C}(graphite)\text{ + }{{\text{O}}_{2}}(g)\text{ }\to \text{ C}{{\text{O}}_{2}}\text{(g)}$ : H = ${{x}_{1}}$ cal
$\text{C}(diamond)\text{ + }{{\text{O}}_{2}}(g)\text{ }\to \text{ C}{{\text{O}}_{2}}\text{(g)}$ : H = ${{x}_{2}}$ cal
What is the heat of transition of graphite into diamond?

Answer 1

Hint: The two reactions given above must be written in such a way that only graphite and diamond are the terms left in the reaction. The remaining atoms and molecules must cancel out on the products as well as on the reactants side. When we reverse the reaction i.e. products become reactants and vice versa the polarity of heat of reaction is also reversed.


Complete step-by-step answer:
Enthalpy is a property of a thermodynamic system, that is a convenient state function used as a means of measurement in various chemical and biological systems at constant pressure.
A state function is a property such that its value does not depend on the path taken by the physical quantity to reach a specific value.
In contrast to state function, functions that depend on the path taken by the quantity to go between two values are called path functions. Both these functions are often used in thermodynamics.
We will now write the two reactions such that only graphite and diamond terms remain in the question and other compounds get cancelled out.
$\text{C}(graphite)\text{ + }{{\text{O}}_{2}}(g)\text{ }\to \text{ C}{{\text{O}}_{2}}\text{(g)}$ H = ${{x}_{1}}$ cal
$\text{C}{{\text{O}}_{2}}\to \text{ C(diamond) + }{{\text{O}}_{2}}(g)$ H = $-{{x}_{2}}$ cal
Final reaction : $\text{C}(graphite)\text{ }\to \text{ C(}diamond)$ H = ${{x}_{1}}-{{x}_{2}}$ cal
Therefore, the heat of transition of graphite to diamond is ${{x}_{1}}-{{x}_{2}}$ cal.

The correct answer is option (C).

Additional information: Thermodynamics is a branch of physics that deals with the physical quantities heat, work and temperature and their relation to radiation and energy released.
The behaviour of these quantities is in accordance with the laws of thermodynamics. The 3 laws of thermodynamics give a quantitative description using the above physical quantities at a microscopic level.


Note: We can never define the absolute enthalpy of a substance or a reaction. This is the reason why we consider change in enthalpy for a reaction like enthalpy of formation, enthalpy of atomisation etc.