C -diamond- rightarrow C -graphite-triangle S-300K- - 10 K-1-C -diamond- - O-2- rightarrow CO-2-triangle H - -91 mol-1- 300 KC -graphite- - O-2- rightarrow CO-2-triangle H - X 300 KX is-81 K cal101 K cal-94 K cal88 K

Question

Toppr Admin · Accepted Answer

C (diamond) → C (graphite)△S300K=10calK−1C (diamond) + O2 → CO2△ H = -91 cal mol−1 at 300 KC (graphite) + O2 → CO2△ H = X at 300 KX is:-81 K cal101 K cal-94 K cal88 K cal

C (diamond) $\to$ C (graphite)
$△ S_{300 K} = 10 c a l K^{- 1}$
C (diamond) + $O_{2}$ $\to$ $C O_{2}$
$△$ H = -91 cal $m o l^{- 1}$ at 300 K
C (graphite) + $O_{2}$ $\to$ $C O_{2}$
$△$ H = X at 300 K
X is:
-81 K cal
101 K cal
-94 K cal
88 K cal