Question

For each pairing below, select the molecule that will have the lower standard free energy of formation.

a) H_2O(g) and H_2O(l)
b) CO_2(g) and CO_2(aq)
c) NaCl(s) and NaCl(aq)
d) HCl(g) and HCl(aq)

Answer 1

H2O(l), CO2(g), NaCl(s), and HCl(g) will have lower standard free energy of formation.

Step-by-step explanation:

The molecule with the lower standard free energy of formation can be determined by comparing the phases of the substances involved. In general, substances in their standard states have a lower standard free energy of formation. Let's analyze each pairing:

a) H2O(g) and H2O(l): The standard state for water is a liquid at room temperature and pressure, so H2O(l) would have a lower standard free energy of formation than H2O(g).

b) CO2(g) and CO2(aq): The standard state for carbon dioxide is a gas, so CO2(g) would have a lower standard free energy of formation than CO2(aq).

c) NaCl(s) and NaCl(aq): The standard state for sodium chloride is a solid, so NaCl(s) would have a lower standard free energy of formation than NaCl(aq).

d) HCl(g) and HCl(aq): The standard state for hydrochloric acid is a gas, so HCl(g) would have a lower standard free energy of formation than HCl(aq).