Answer:
Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:
2 SO3(g) → 2 SO2(g) + O2(g)
ΔHf° (kJ mol-1) -395.7 -296.8
S° (J K-1 mol-1) 256.8 248.2 205.1
ΔH° = kJ
S° = J K⁻¹
Explanation:
The method to solve this problem calls for the use of the Gibbs standard free energy change:
ΔG
= ΔrxnH - TΔSrxn
We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.
Now as mentioned in the hint, we need to determine ΔrxnH and ΔSrxn, which are given by
ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants
where ν is the stoichiometric coefficient in the balanced chemical equation.
For ΔS we have likewise
ΔrxnS = ∑ ν x ΔSº products - ∑ ν x ΔSº reactants
Thus,
ΔrxnH(kJmol⁻¹) = 2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ
ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹
So ΔG kJ = 197.8 - T(0.1879)
and the reaction will become spontaneous when the term T(0.1879) becomes greater that 197.8,
0 = 197.8 - 0.1879 T ⇒ T = 1052 K
so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)
=
. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.