Consider the reaction 2H2O(g) + 2SO2(g)2H2S(g) + 3O2(g) for which H° = 1.036×103

Question

Consider the reaction 2H2O(g) + 2SO2(g)2H2S(g) + 3O2(g) for which H° = 1.036×103 kJ and S° = 152.9 J/K at 298.15 K.

(1) Calculate the entropy change of the UNIVERSE when 2.082 moles of H2O(g) react under standard conditions at 298.15 K. Suniverse = 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? If the reaction is reactant favored, choose 'reactant favored'.

Explanation / Answer

Reaction :- 2H2O(g) + 2SO2(g)2H2S(g) + 3O2(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.082 moles of H2O(g) react at standard conditions.

So's @ 298K: H2S(g)=205.8 J/mol K O2(g)=205.1 J/mol K H2O(l)=69.9 J/mol K SO2(g)=248.2 J/mol K

So I did S(system) = Sum of So (products) - Sum of So (reactants)

= (2(248.2) + 2 (69.9)) - (3(205.1) + 2(205.8))
= -390.7 J/K

So Its 2H2O(g), so divide by two first, then multiply by 2.082 moles.

So the entropy change of the UNIVERSE when 2.082 moles of H2O(g) react = (-390.7 /2) x 2.082 = -406.71 J/K

Let's first calculate the standard state Gibbs free energy change, which will tell us whether this reaction is product or reactant favored at STP:

Go = Ho - To*So = 1036000 J - (298.15K)*(152.9J/K)

Go = 990.42 kJ

The free energy change is positive , so this is a reactant -favored reaction at STP.

The entropy change in positive, so the reaction is not entropically favored, but the enthalpy change is also positive, and sufficiently positive so as to offset the decrease in entropy. This reaction is enthalpy-favored.

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