Question

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

Answer 1

Answer: (1) 18 Kcal

Explanation:

The chemical reaction for the decomposition of calcium carbonate follows the equation:

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

The expression for $K_p$ for the given reaction follows:

$K_p={p_{CO_2}}$

We are given:

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

$K_p=10^{-3}$

Relation between standard Gibbs free energy and equilibrium constant follows:

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

where,

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

R = Gas constant = $0.002cal/K mol$

T = temperature = 500 K

Putting values in above equation, we get:

$\Delta G=25 kcal/mol+(0.002kcal/Kmol)\times 500K\times \ln (10^{-3})\\\\\Delta G=18kcal/mol$

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