Question

Calculate free energy change per mole for the reaction and comment on the
spontaneity of the reaction:
CaCO3------CaO + CO2(g) at 298 K if ∆H = + 177.9 KJ and ∆S = 160.4 JK​

Answer 1

The Gibbs free energy change of the reaction is -130.1 kJ and the reaction is spontaneous in nature.

Explanation:

For the given chemical equation:

$CaCO_3\rightarrow CaO+CO_2(g)$

To calculate the Gibbs free energy of the reaction, we use the equation:  

$\Delta G=\Delta H-T\Delta S$

where,

$\Delta H$ = enthalpy change = +177.9 kJ = 177900 J    (Conversion factor: 1 kJ = 1000 J)

T = Temperature = 298 K

$\Delta S$ = entropy of the reaction = -160.4 J/K

Putting values in above equation, we get:

$\Delta G=177900J-(298K\times (-160.4J/K))\\\\\Delta G=-130100.8J=-130.1kJ$

For a reaction to be spontaneous, the Gibbs free energy of the reaction must be negative.

Learn more about Gibbs free energy change:

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