Discussion Forum



1)

$\triangle H^{0}_{f}$ values (in $ kJ mol ^{1}$) for graphite , diamond and $C_{60}$  are respectively


A) 0;1.9;38.1

B) 1.8;1.9;38.1

C) 0;0;21.4

D) 1.8;1.9;2.0

Answer:

Option A

Explanation:

$\because$    Graphite is the most stable  state of carbon and its  $\triangle H^{0}_{f}$ is considered as zero ( $\because$  has more van der Waals'  force )  also $\triangle H^{0}_{f}$ for $C_{60}$ > $\triangle H^{0}_{f}$ for diamond

$\triangle H^{0}_{f}$ values of graphite  , diamond and $C_{60}$ are =0,  1.9 and 38.1 k J mol-1 and $\triangle H^{0}_{f}$ values are in order:

Diamond < Fullerene < Graphite

Hence , option (a) is the correct answer.