Table 1. Thermodynamic data for carbon allotropes A,H° / kJ mol¯1 AfS°/ J K-1 mol-1 Graphite -393.5 (0) Diamond -395.3 -3.25 C60 -25965 1.54 1. Calculate the change in enthalpy, entropy, and Gibbs free energy for the transformation of 1 mole of graphite into diamond and into C60 under ambient conditions. Then, rank graphite, diamond, and C60 in order of decreasing stability in terms of Gibbs free energy. 2. Under ambient conditions, diamond does not transform into graphite, which is energetically more stable than diamond. Explain why this transformation does not occur.

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Table 1. Thermodynamic data for carbon allotropes
A,H° / kJ mol-1
AfS°/ JK-1 mol-1
Graphite
-393.5
(0)
Diamond
-395.3
-3.25
C60
-25965
1.54
1. Calculate the change in enthalpy, entropy, and Gibbs free energy for the transformation of 1 mole of
graphite into diamond and into C6o under ambient conditions. Then, rank graphite, diamond, and C60
in order of decreasing stability in terms of Gibbs free energy.
2. Under ambient conditions, diamond does not transform into graphite, which is energetically more
stable than diamond. Explain why this transformation does not occur.
Transcribed Image Text:Table 1. Thermodynamic data for carbon allotropes A,H° / kJ mol-1 AfS°/ JK-1 mol-1 Graphite -393.5 (0) Diamond -395.3 -3.25 C60 -25965 1.54 1. Calculate the change in enthalpy, entropy, and Gibbs free energy for the transformation of 1 mole of graphite into diamond and into C6o under ambient conditions. Then, rank graphite, diamond, and C60 in order of decreasing stability in terms of Gibbs free energy. 2. Under ambient conditions, diamond does not transform into graphite, which is energetically more stable than diamond. Explain why this transformation does not occur.
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