Consider given reaction, CaCO3 (s)
CaO (s) +
CO2 (g)
For
above reaction , standard Gibbs energy change is calculated by
using formula as shown below.
rG
0 = 
fG
0 ( products ) - 
fG
0 ( Reactants)


f G
0 ( products ) =
fG
0 CaO (s) +
f G
0 CO2 (g)

f G
0 ( products ) = ( - 604.2 k J / mol ) + ( - 394.4 k
J / mol ) = - 998.6 k J / mol

fG
0 ( Reactants) =
fG
0 CaCO3 (s) = -
1128.8 k J / mol

rG
0 = - 998.6 k J / mol - ( - 1128.8 k J / mol) =
130.2 k J / mol
We have
relation,
rG
0 = - 2.303 R T log K , where K is equilibrium
constant.
At
standard conditions, T = 298 K .
130200 J / mol = - 2.303
8.314 J K -1 mol -1
298 K
log K
log K = - ( 130200 J / mol / 2.303
8.314 J K -1 mol -1
298 K )
log K = - 22.82
Taking antilog on both sides, we get K = 1.52
10 -23
ANSWERS 1)
rG
0 = 130.2 k J / mol and 2) K = 1.52
10 -23
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