A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of temperatures is the following process spontaneous? Assume that gases are at a pressure of 1 atm. (Hint: Use the data below to calculate ΔH and ΔS [assumed independent of temperature and equal to ΔH° and ΔS°, respectively] and then use the definition of ΔG.)

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)

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.