Question

The chemical reaction that causes iron to corrode in air is given by

4Fe + 3O2→2Fe2O3 in which at 298 K
ΔHrxn= 1684 kJ
ΔSrxn= 543.7 J/K

Part A. What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K.

Part B. What is the Gibbs free energy for this reaction at 3652 K? Assume that Delta H and Delta S do not change with temperature.

Answer 1

The standard Gibbs free energy for the reaction at 298 K is -2.29 x 10^5 J. The Gibbs free energy for the reaction at 3652 K is -3.88 x 10^6 J.

Step-by-step explanation:

Part A. To find the standard Gibbs free energy (ΔG°) for the reaction at 298 K, we can use the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. Plugging in the given values, we have:

ΔG° = 1684 kJ - (298 K)(543.7 J/K)

Converting kJ to J, we get:

ΔG° = 1684000 J - (298 K)(543.7 J/K)

Calculating the values gives us a standard Gibbs free energy of -2.29 x 10^5 J.

Part B. Assuming that ΔH° and ΔS° do not change with temperature, we can use the same equation to calculate the Gibbs free energy at 3652 K:

ΔG = ΔH - TΔS

Plugging in the values, we have:

ΔG = 1684 kJ - (3652 K)(543.7 J/K)

Calculating the values gives us a Gibbs free energy of -3.88 x 10^6 J.

Answer 2

(A) The value of
`\Delta G^o` at 298 K is, 1521.9 kJ

(B) The value of
`\Delta G^o` at 3652 K is, -301.59 kJ

Explanation :

Part A :

`\Delta G^o=\Delta H^o-T\Delta S^o`

where,

`\Delta G^o` = standard Gibbs free energy = ?

`\Delta H^o` = standard enthalpy = 1684 kJ = 1684000 J

`\Delta S^o` = standard entropy = 543.7 J/K

T = temperature of reaction =
`298K`

Now put all the given values in the above formula, we get:

`\Delta G^o=(1684000J)-(298K* 543.7J/K)`

`\Delta G^o=1521977.4J=1521.9kJ`

Part B :

`\Delta G^o=\Delta H^o-T\Delta S^o`

where,

`\Delta G^o` = standard Gibbs free energy = ?

`\Delta H^o` = standard enthalpy = 1684 kJ = 1684000 J

`\Delta S^o` = standard entropy = 543.7 J/K

T = temperature of reaction =
`3652K`

Now put all the given values in the above formula, we get:

`\Delta G^o=(1684000J)-(3652K* 543.7J/K)`

`\Delta G^o=-301592.4J=-301.59kJ`