Question

4FeS(s)+7O2(g)→2Fe2O3(s)+4SO2(g)

Due to the presence of FeS(s) as an impurity, the combustion of some types of coal results in the formation of SO2(g), as represented by the equation above. Also, SO2(g) can react with O3(g) to form SO3(g), as represented by the equation below.

SO2(g)+O3(g)⇄SO3(g)+O2(g)
∆H∘298=−242kJ/molrxn;∆S∘298=−25J/(K⋅molrxn)

Question
Which of the following is most likely to be true about the reaction between SO2(g)
and O3(g) at 298 K?

a) ΔG°<0 and TΔS∘ is smaller in magnitude than ΔH∘

b) ΔG°<0 and TΔS∘ is larger in magnitude than ΔH∘

c) ΔG°>0 and TΔS∘is smaller in magnitude than ΔH∘

d) ΔG°>0 and TΔS∘ is larger in magnitude than ΔH∘

Answer 1

At 298 K for the reaction SO2(g) + O3(g) to form SO3(g) and O2(g), the Gibbs free energy change (ΔG°) is negative, indicating that the reaction is spontaneous. The thermal energy component TΔS° is less than the enthalpy change ΔH°, hence the correct answer is (a).

Step-by-step explanation:

The reaction in question is SO2(g) + O3(g) ⇌ SO3(g) + O2(g) with given thermodynamic parameters ΔH°298 = -242 kJ/mol and ΔS°298 = -25 J/(K⋅mol). To determine the sign of the Gibbs free energy change (ΔG°) at 298 K, we use the Gibbs free energy equation:

ΔG° = ΔH° - TΔS°

Substituting the given values at 298 K, we calculate:

ΔG° = (-242 kJ/mol) - (298 K)(-25 J/(K⋅mol))

ΔG° = (-242,000 J/mol) - (298 K)(-25 J/(K⋅mol))

ΔG° = (-242,000 J/mol) + 7450 J/mol

ΔG° = -234,550 J/mol or -234.55 kJ/mol

Since ΔG° < 0, the reaction is spontaneous at 298 K. The magnitude of TΔS° (7450 J/mol) is smaller than that of ΔH° (242,000 J/mol). Therefore, the correct answer is: a) ΔG°<0 and TΔS° is smaller in magnitude than ΔH°

Answer 2

Answer:B

Step-by-step explanation: