Question

1. Consider the reaction: H2CO(g) + O2(g)CO2(g) + H2O(l) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.41 moles of H2CO(g) react at standard conditions. S°system =___ J/K

2. For the reaction
3Fe₂O₃(s) + H₂(g)>2Fe₃O₄(s) + H₂O(g)
DeltaH° = -6.0 kJ and DeltaS° = 88.7 J/K
The standard free energy change for the reaction of 1.78 moles of Fe₂O₃(s) at 292 K, 1 atm would be ___ kJ.
This reaction is (reactant, or product)  favored under standard conditions at 292 K.
Assume that DeltaH° and DeltaS° are independent of temperature.

Answer 1

Answer 2

Entropy change for the system:

The balanced reaction is:
${\text{H}}_2\text{CO(g)}+{\text{O}}_2(\text{g})\to {\text{CO}}_2(\text{g})+{\text{H}}_2\text{O(l)}$.
Using standard entropies:
,
,
,
.

Calculate $\mathrm{\Delta }{S}_{\text{system}}^{\circ }$:

$$\mathrm{\Delta }{S}^{\circ }=[S^{\circ }({\text{CO}}_2)+S^{\circ }({\text{H}}_2\text{O})]-[S^{\circ }({\text{H}}_2\text{CO})+S^{\circ }({\text{O}}_2)]=[214+70]-[219+205]=-140$$

For 2.41 moles:

$$\mathrm{\Delta }{S}_{\text{system}}^{\circ }=2.41\times (-140)=-337.4\text{J/K}\mathrm{.}$$

Answer: $-337\text{J/K}$ (rounded).

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2. Standard free energy change and favorability:
   Given:
   $\mathrm{\Delta }{H}^{\circ }=-6.0\text{kJ}$,
   $\mathrm{\Delta }{S}^{\circ }=88.7\text{J/K}$,
   $T=292\text{K}$.

   Calculate $\mathrm{\Delta }{G}^{\circ }$ for the reaction as written:

   $$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }=-6.0\text{kJ}-(292\times 0.0887\text{kJ/K})=-6.0-25.9=-31.9\text{kJ}\mathrm{.}$$

   For 1.78 moles of ${\text{Fe}}_2{\text{O}}_3$:

   $$\mathrm{\Delta }{G}^{\circ }=\left(\frac{-31.9}{3}\right)\times 1.78=-18.9\text{kJ}\mathrm{.}$$

   Since $\mathrm{\Delta }{G}^{\circ }<0$, the reaction is product-favored.

3. 
   

   Answers:

• Free energy change: $-18.9\text{kJ}$.

• Favorability: product