Thermodynamics Question 88
Question: $ {C _{\text{(graphite)}}}+O _{2}(g)\to CO _{2}(g) $
$ \Delta H=-94.05kcalmo{{l}^{-1}} $
$ {C _{\text{(diamond)}}}+O _{2}(g)\to CO _{2}(g); $
$ \Delta H=-94.50kcalmo{{l}^{-1}} $ therefore [DPMT 2001]
Options:
A) $ {C _{\text{(graphite)}}}\to {C _{\text{(diamond)}}} $ ; $ \Delta H _{298K}^{o}=-450calmo{{l}^{-1}} $
B) $ {C _{\text{(diamond)}}}\to {C _{\text{(graphite)}}}; $
$ \Delta H _{298K}^{o}=+450\ \text{cal}\cdot\text{mol}^{-1} $
C) Graphite is the more stable allotrope
D) Diamond is harder than graphite
Show Answer
Answer:
Correct Answer: C
Solution:
Heat energy is also involved when one allotropic form of an element is converted into another. Graphite is the stabler allotrope because the heat of transformation of
$ {C _{(diamond)}}\to {C _{(graphite)}} $ . (i)
$ {C _{(dia)}}+{O _{2(g)}}=C{O _{2(g)}}\Delta H=-94.5kcal $ (ii)
$ {C _{(graphite)}}+{O _{2(g)}}=C{O _{2(g)}}\Delta H=-94.0kcal $
$ \Delta H _{transformation}=-94.5-(-94.0) $
$ = -0.5 kcal$
BETA
