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}=+450calmo{{l}^{-1}} $

C) Graphite is the stabler allotrope

D) Diamond is harder than graphite

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Answer:

Correct Answer: C

Solution:

Heat energy is also involved when one allotropic form of an element is converted in to 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) $

$ =-0.5kcal $ .



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