∆ G r x n 0 of given reaction at 1450 0 C .

Question

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Answer 1

Question

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

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Answer 2

Question

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

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Answer 3

Question

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

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Answer 4

Question

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

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Answer 5

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

Answer 6

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.

3) Given:

Fe2O3s+3CO(g)→2Fe(s)+3CO2g

The reaction is carried out at a temperature T=14500C

	∆Hf0(kJ/mol)	∆Sf0(J/mol.K)
Fe2O3s	-824.2	87.4
CO(g)	-110.5	197.7
Fe(s)	0.0	27.3
CO2g	-393.5	213.8

4) Calculations:

The unit of ∆Sf0 is J/mol.K, therefore we need to convert T from ⁰C to K.

T(K)=14500C+273.15=1723.15K

Calculation for ∆Hrxn0:

∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

∆Hrxn0=

2 mol×∆Hf,Fes0+3 mol×∆Hf,CO2g0-1 mol×∆Hf,Fe2O3s0+3 mol×∆Hf,COg0

=2 mol×0.0kJmol+3 mol×-393.5kJmol-1 mol×-824.2kJmol+3 mol×-110.5kJmol

=[0-1180.5 kJ]-[(-824.2-331.5)]

=(-1180.5 kJ)--1155.7 kJ

=-1180.5+1155.7 kJ

∆Hrxn0=-24.8 kJ

Calculation for ∆Srxn0:

∆Srxn0=

2 mol×∆Sf,Fes0+3 mol×∆Sf,CO2g0-1 mol×∆Sf,Fe2O3s0+3 mol×∆Sf,COg0

=2 mol×27.3Jmol.K+3 mol×213.8Jmol.K-1 mol×87.4Jmol.K+3 mol×197.7Jmol.K

=54.6JK+641.4 JK-87.4JK+593.1 JK

=696JK-680.5JK

∆Srxn0=15.5 JK

All units of ∆Hrxn0, ∆Srxn0 and ∆Grxn0, must be in same unit, so need to convert ∆Srxn0 from JK to kJK.

∆Srxn0=15.5 JK×1 kJ1000 J=0.0155kJK

Calculation for ∆Grxn0:

∆Grxn0=∆Hrxn0-T∆Srxn0

∆Grxn0=-24.8 kJ-1723.15 K×0.0155kJK

∆Grxn0=-24.8 kJ-26.708825 kJ

∆Grxn0=-51.5 kJ

After calculating values of ∆Hrxn0 and ∆Srxn0, we plugged these values in ∆Grxn0 formula and calculated it.

Conclusion:

Based on the relation between ∆G, ∆H, and ∆S, we find that ∆Grxn0 for the given reaction at 14500C is-51.5 kJ.

Answer 7

Chapter 12, Problem 12.71QA

Interpretation Introduction

To find:

∆Grxn0 of given reaction at 1450 0C.

Answer 8

Expert Solution & Answer

Answer to Problem 12.71QA

Solution:

∆Grxn0 of given reaction at 1450 0C is -51.5 kJ.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Formula:

i. ∆Grxn0=∆Hrxn0-T∆Srxn0

ii. ∆Hrxn0=∑nproducts×∆Hf,products0-∑nreactants×∆Hf,reactants0

iii. ∆Srxn0=∑nproducts×∆Sf,products0-∑nreactants×∆Sf,reactants0

Here ∆Grxn0 is the standard Gibb’s free energy of reaction, ∆Hrxn0 the standard enthalpy of the reaction, ∆Srxn0 the standard entropy of the reaction, T is the absolute temperature, and n is the number of moles of each reactant or product in the balanced equation for the reaction.

2) Concept:

We can calculate the change in enthalpy that accompanies a chemical reaction under standard conditions, ∆Hrxn0, from the difference between the standard molar enthalpies of n_reactants, the moles of reactants and the standard molar enthalpies of n_products, the moles of products. This is expressed mathematically as the formula shown above. Same applies to ∆Srxn0.

The relation between Gibbs free energy (∆Grxn0), enthalpy (∆Hrxn0) and entropy (∆Srxn0) is given by the above formula. We can use this formula to calculate ∆Grxn0 of reaction.