Question

Calculate the free energy change per mole for the following reactions.

(i) CaCO3

(s) + CaO(s) + CO2

(g) 298K ij

∆H = 117.9 kJ, ∆S = 160.4JK–1

(ii) 2 NO2

(g) + N2O4

(g), 298K ij

∆H = –57.2 KJ, ∆S = –175.6JK–1​

Answer 1

Answer:

2 NO2

(g) + N2O4

(g), 298K ij

∆H = –57.2 KJ, ∆S = –175.6JK–1

Answer 2

Answer:

The free energy change of the reaction (i) is equal to -130KJ and for the reaction (ii) is equal to -4.371 KJ.

Explanation:

We have given, the chemical reaction:

CaCO₃ (s)  $\longrightarrow$  CaO (s)   + CO₂ (g)

We use the equation to calculate Gibbs free energy of the reaction;

ΔG = ΔH - TΔS                                                            .................(1)

Given, the enthalpy change, ΔH = 117.9 KJ = 177900J

The entropy change, ΔS = 160.4J/K

The temperature of the reaction , T = 298K

Substitute the all values in the equation (1);

ΔG = 177900J - (298K × 160.4 J/K)

ΔG =  177900J - 477992J

ΔG =  -130100.8J

ΔG = - 130.1KJ

(ii) Given, chemical reaction:

 2NO₂ (g)  $\longrightarrow$  N₂O₄ (g)

Given, the enthalpy change , ΔH = -57.2KJ = -57200J

The entropy change, ΔS = -175.6J/K

We know that ΔG = ΔH - TΔS  

ΔG = -57200J - [298× (-175.6J/K)]

ΔG = -57200J + 52328.8J

ΔG = - 4871.2J

ΔG = - 4.871 KJ

Therefore, the free energy change is equal to -4.871KJ.

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