The densities of graphite and diamond are
Interpretation:
The pressure required for
Concept introduction:
The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by
Answer to Problem 5.35E
The pressure required for
Explanation of Solution
The density of graphite is
The density of diamond is
Graphite and diamond exists in equilibrium as,
The standard Gibbs free energy change for the above reaction is calculated by the formula,
Where,
•
•
The Gibbs free energy of formation of diamond and graphite is
Substitute the value of Gibbs free energy of formation of diamond and graphite in equation (1).
According to the given equation,
The value of
The above equation can also be written as,
According to equation 5.14, the value of
Where,
•
•
Substitute the value of
The molar volume of diamond is calculated by the formula,
Where,
•
•
Substitute the molar mass and density of diamond in above formula.
The molar volume of graphite is calculated by the formula,
Where,
•
•
Substitute the molar mass and density of graphite in above formula.
Substitute equation (4) and equation (5) in equation (3).
Convert
Therefore, the pressure required for
The pressure required for
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Chapter 5 Solutions
EBK PHYSICAL CHEMISTRY
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