For which of the following reactions is S° > 0.

Question

For which of the following reactions is S° > 0.

Choose all that apply.

NH₄Cl(s)NH₃(g) + HCl(g)

2NH₃(g) + 3N₂O(g)4N₂(g) + 3H₂O(g)

2CO(g) + O₂(g)2CO₂(g)

2CO₂(g) + N₂(g)2CO(g) + 2NO(g)

2NO(g) + 2H₂(g)N₂(g) + 2H₂O(l)

Answer 1

Entropy represents a change in the disorder of a system.

If dS⁰ > 0 then the entropy of a system increases

If dS⁰ < 0 then the entropy of a system decreases

Phase changes lead to a change in dS.

Solid->Liquid->Gas represents dS > 0

Gas->Liquid->Solid represents dS < 0

A change in the number of molecules also represents a change in dS

If the molecular number of reactants is greater than the molecular number of products then dS < 0

If the molecular number of products is greater than the molecular number of reactants then dS > 0

NH₄Cl(s)NH₃(g) + HCl(g)

We go from 1 solid reactant to 2 gaseous products. (Solid -> gas; # molecules increases) This is an increase in entropy, dS > 0

2NH₃(g) + 3N₂O(g)4N₂(g) + 3H₂O(g)

We go from 5 gaseous reactants to 7 gaseous products. (No. molecules increase). This is an increase in entropy, dS > 0

2CO(g) + O₂(g)2CO₂(g)

We go from 3 gaseous reactants to 2 gaseous products (No. molecules decrease). This is a decrease in entropy, dS < 0

2CO₂(g) + N₂(g)2CO(g) + 2NO(g)

We go from 3 gaseous reactants to 4 gaseous products (No. molecules increases). This is an increase in entropy, dS > 0

2NO(g) + 2H₂(g)N₂(g) + 2H₂O(l)

We go from 4 gaseous reactants to 3 products (1 gas and 2 liquid). Phase change (gas->liquid). # molecules decreases. Both represent a decrease in entropy, dS < 0

Hope this helps!