Question

Consider the following reaction: CACO3(s) CaO (s) + CO2 (g). Estimate AG for this reaction at each of the following temperatures. (Assume that AH° and AS do not change too much within the given temperature range.)

spontaneous or nonspontaneous

spontaneous or nonspontaneous

spontaneous or nonspontaneous

Answer 1

From thermodynamics the entropy, Gibb's free energy, enthalpy and temperature are related by the following equation,

AG = AH-TAS

where $\Delta G^{0}$ is the standard change in Gibb's free energy

$\Delta H^{0}$ is the standard change in enthalpy

$\Delta S^{0}$ is the standard change in entropy

T is the temperature

For calculating the change in energy for a reaction

(i) we need to subtract the enthalpy of reactants from that of products

(ii) do the similar thing for entropy

(iii) substitute the above values at the given temperature in the above equation.

From thermodynamic database the values of $\Delta H^{0}$ and $\Delta S^{0}$ can be obtained. It is given in the table given below.

The enthalpy change for the reaction is given by  

ΔΗ – ΔΗ, - ΔΗ,

= (-393.51 + -635.09) - (-1207)

= 160.4 KJ / mol

The entropy change is given by  

AS0 = AS, - AS,

= (38.1 + 393.51 ) - 92.9

= 338.71 J/ K.mol

The values of enthalpy and entropy change is same in all cases . We need to change the value of temperature in the equation in the three cases.

a) T = 290 K

AG = AH-TAS

= 160.4 - 290 * 338.71 = -98065.5 KJ / mol

b) T= 1095 K

AG = AH-TAS

= 160.4 - 1095 * 338.71

= -3707275.05 KJ /mol

c)T= 1475 K

AG = AH-TAS

=160.4 - 1475 * 338.71

= -499436.85 KJ /mol

Values from thermodynamic database

	$\Delta H^{0}$(KJ/mol)	$\Delta S^{0}$(J/K.mol)
CaCO3 (s)	-1207	92.9
(s)	-635.09	38.1
$CO_{2}$	-393.51	393.51