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 cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.