Question

For the reaction Fe2O3(s) + 3H2(g)2Fe(s) + 3H2O(g) H°rxn = 98.8 kJ and S°rxn = 142.5 J/K The standard free energy change for the reaction of 2.47 moles of Fe2O3(s) at 267 K, 1 atm would be kJ. This reaction is (reactant,product) favored under standard conditions at 267 K. Assume that H°rxn and S°rxn are independent of temperature.

Answer 1

The value of standard Gibbs free energy is, 60.8 kJ

This reaction is reactant favored under standard conditions at 267 K.

Explanation :

As we know that,

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

where,

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

`\Delta H^o` = standard enthalpy = 98.8 kJ = 98800 J

`\Delta S^o` = standard entropy = 142.5 J/K

T = temperature of reaction = 267 K

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

`\Delta G^o=(98800J)-(267K* 142.5J/K)`

`\Delta G^o=60752.5J=60.8kJ`

As we know that:

• A reaction to be spontaneous when
  `\Delta G<0` and reaction will be favored in the forward direction that means favored in products.
• A reaction to be non-spontaneous when
  `\Delta G>0` and reaction will be favored in the backward direction that means favored in reactants.

As, the value of
`\Delta G` is more than zero that means the reaction is non-spontaneous and reaction will be favored in the backward direction that means favored in reactants.