Question

2.
For the reaction,
CaCO3(s) = CaO(s) + CO2(9),
partial pressure of co, at 500 K temperature is
10-3 atm. AGº = 25 Kcal. Calculate the value of
free energy change (AG).
(1) 18 Kcal
(2) 28 Kcal
(3) 12 kcal
(4) 15 Kcal​

Answer 1

I think 18 kcal answer

Answer 2

Answer: (1) 18 Kcal

Explanation:

For the reaction :

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

$K=p_{CO_2}=10^{-3}atm$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G=\Delta G^o+RT\ln K_1$

where,

$\Delta G^o$ = Standard Gibbs free energy = 25 kcal

R = Gas constant = $1.98\times 10^{-3}kcal/K mol$

T = temperature =500 K

Putting values in above equation, we get:

$\Delta G=25+(1.98\times 10^{-3}kcal/Kmol)\times 500K\times \ln (10^{-3})\\\\\Delta G=18kcal$

Hence, the Gibbs free energy of the reaction is 18kcal