The value of ${ΔH}_{transition}$ of C(diamond) is $1.9 kJ / mol$ at $25 ° C$ . Entropy of graphite is higher than entropy of diamond. This implies that:

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$Δ G_{transition} of C(diamond) ⟶ C(graphite) is$ -ve.

C(graphite) is thermodynamically more stable than C(diamond) at 25° C

C(diamond) is thermodynamically more stable than C(graphite) at

diamond will provide more heat on complete combustion at 25° C

answer is B, C, D.

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Detailed Solution

C (graphite) is more thermodynamically stable than C (graphite).
For the reverse process, ΔH<0, Also ΔS changes very negligible due to this transition.
Hence, ΔG is <0 for the transition from C(diamond) to C (graphite)
More heat is evolved due to combustion of less stable form.
Hence, options B, C, D are the appropriate answers.

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