Consider the reaction S(s, rhombic) + 2CO(g) rightarrow SO_2 (g) + 2C(s, graphit

Question

Consider the reaction S(s, rhombic) + 2CO(g) rightarrow SO_2 (g) + 2C(s, graphite) for which Delta H degree = -75.80 kJ and Delta S degree = -167.6 J/K at 298.15 K. Calculate the entropy change of the UNIVERSE when 2.408 moles of S(s, rhombic) react under standard conditions at 298.15K. Delta S_universe = J/K Is this reaction reactant or product favored under standard conditions? If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? It the reaction is reactant faovred choose 'reactant favored'.

Explanation / Answer

Ho = -75.80 kJ and So = -167.6J/K at 298.15K

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

Go = Ho - T x So = (-75800 J) - (298.15K)(-167.6J/K)

Go = -25830.06 J

The free energy change is negative, so this is a product-favored reaction at STP. The entropy change in negative, so the reaction is not entropically favored, but the enthalpy change is also negative, and sufficiently negative so as to offset the decrease in entropy. This reaction is enthalpy-favored.

Suniverse = Ssystem + Ssurrounding

Ssurrounding = -H0 / T

Suniverse = S0 + (- H /T)

Suniverse = -167.6 J/K + ( 75800 J/ 298.15 K)

Suniverse = 86.6344 J/K for 1 mole

for 2.408 mole

Suniverse = 86.6344 J/K x 2.408

Suniverse = 208.615 J/K

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