Indicate whether each reaction is spontaneous under standard conditions. drag the appropriate items to their respective bins. 6cl2(g)+2fe2o3(s)→4 fecl3(s)+3o2(g)6 cl subscript 2 ( g) + 2 f e subscript 2 o subscript 3 ( s) rightarrow 4 f e c l subscript3 ( s) + 3 o subscript2 ( g)so2(g)+2h2(g)→s(s)+2h2o(g)s o subscript2 ( g) + 2 h subscript2 ( g) rightarrow s ( s) + 2 h subscript2 o ( g)2so2(g)+o2(g)→2so3(g)2 s o subscript2 ( g) + o subscript2 ( g) rightarrow 2 s o subscript3 ( g)no2(g)+n2o(g)→3no(g)

ΔG° = 108 kJ, Nonspontaneous.

ΔG° = -158 kJ, Spontaneous.

ΔG° = 105 kJ, Nonspontaneous.

Explanation:

The sponteneity of a reaction depends on the standard Gibbs free energy (ΔG°).

If ΔG° < 0, the reaction is spontaneous.

If ΔG° > 0, the reaction is nonspontaneous.

We can calculate ΔG° using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs energy of formation

p: products

r: reactants

6 Cl₂(g) + 2 Fe₂O₃(s) → 4 FeCl₃(s) + 3 O₂(g)

ΔG° = 4 mol . ΔG°f(FeCl₃(s)) + 3 mol . ΔG°f(O₂(g)) - 6 mol . ΔG°f(Cl₂(g)) - 2 mol . ΔG°f(Fe₂O₃(s))

ΔG° = 4 mol . (-344 kJ/mol) + 3 mol . 0 - 6 mol . 0 - 2 mol . (-742 kJ/mol)

ΔG° = 108 kJ

ΔG° > 0 so the reaction is nonspontaneous.

SO₂(g) + 2 H₂(g) → S(s) + 2 H₂O(g)

ΔG° = 1 mol . ΔG°f(S(s)) + 2 mol . ΔG°f(H₂O(g)) - 1 mol . ΔG°f(SO₂(g)) - 2 mol . ΔG°f(H₂(g))

ΔG° = 1 mol . 0 + 2 mol . (-229 kJ/mol) - 1 mol . (-300 kJ/mol) - 2 mol . 0

ΔG° = -158 kJ

ΔG° < 0 so the reaction is spontaneous.

NO₂(g) + N₂O(g) → 3 NO(g)

ΔG° = 3 mol . ΔG°f(NO(g)) - 1 mol . ΔG°f(NO₂(g)) - 1 mol . ΔG°f(N₂O(g))

ΔG° = 3 mol . (86.6 kJ/mol) - 1 mol . (51.3 kJ/mol) - 1 mol . (104 kJ/mol)

ΔG° = 105 kJ

ΔG° > 0 so the reaction is nonspontaneous.

the answer would be a