Question

Fe2O3(s) + 2Al(s)Al2O3(s) + 2Fe(s) for which H° = -851.5 kJ and S° = -37.00 J/K at 298.15 K.

(1) Calculate the entropy change of the UNIVERSE when 2.413 moles of Fe2O3(s) react under standard conditions at 298.15 K. S universe =    J/K

(2) Is this reaction reactant or product favored under standard conditions? (

3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy?

Answer 1

(1)

$\Delta H$system = - $\Delta H$ surroundings =  - T$\Delta S$surroundings .

Or, -851.5 KJ/mol = - 298.15 × $\Delta S$ surroundings

Or, $\Delta S$ surroundings  

= (851.5/298.15) KJ/mol.K

= 2.856 KJ/mol.K

= 2856 J/mol. K

now,

- T$\Delta S$universe = - T$\Delta S$surroundings - T$\Delta S$system.

Or, $\Delta S$ universe  = $\Delta S$ surroundings + $\Delta S$ system

= (2856 - 37 ) J/mol.K.

= + 2819 J/mol. K.

for, 2.413 moles of Fe2O3 (s) $\Delta S$ universe

= (+ 2819) × 2.413

= + 6802 J/K

(2)

As, $\Delta S$ universe  of the reaction is positive hence the formation of product is favored.

(3)

Entropy change of the system is negative and $\Delta H$ system is also negative hence heat is released to the surroundings.

Thus $\Delta S$ surroundings  is positive .

So, the formation of product is enthalpy favored.