The value of `DeltaH_("transition")` of C (graphite) `rarr` C (diamond) is 1.9 kJ/mol at `25^(@)C`. Entropy of graphite is higher than entropy of diamond. This implies that :
A
C (diamond) is more thermodynamically stable than C (graphite) at `25^(@)C`
B
C (graphite) is more thermodynamically stable than C (diamond) at `25^(@)C`
C
diamond will provide more heat on complete combustion at `25^(@)C`
D
`DeltaG_("transition")` of C (diamond) `rarr` C (graphite) is - ve
Text Solution
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The correct Answer is:
B, C, D
`C("graphite")rarrC("diamond")` `DeltaG = DeltaH - TDeltaS` `=1.9 -298 (-ve)=+ve` Graphite is more stable than diamond thermodynamically `Delta_(r)H=(Delta_(C)H)_(G)-(Delta_(C)H)_(D),(Delta_(C)H)_(G)=-x(Delta_(C)H)_(D)=-y` `1.9 =-x+yimplies y=1.9 +x` Diamond provides more heat on complete combustion in comparison of graphite.
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The value of DeltaH_("transition") of C (graphite) to C (diamond) is 1.9 kJ/mol at 25^(@) C entropy of graphite is higher than entropy of diamond. This implies that:
For conversion C (graphite) rarrC (diamond) the DeltaS is
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