Differentiate between the enthalpy of formation of H_2 O_(l) and H_2 O_(g). Why

Question

Differentiate between the enthalpy of formation of H_2 O_(l) and H_2 O_(g). Why is it necessary to specify the physical state of water in the following thermochemical equation. 2C_2 H_6(g) + 7O_2(g) rightarrow 4CO_2(g) + 6H_2 O_(l or g) delta H = + or - delta S = + or - delta G = + or - Please use the following terms in your explanation. enthalpy of vaporization, entropy, enthalpy of formation, enthalpy change, entropy, Gibb's free energy, positive, negative, spontaneous, non-spontaneous, combustion, carbon dioxide and water vapor, particles, disorder, work, 1 mole, endothermic, exothermic, chemical potential energy, temperature

Explanation / Answer

This is clearly a combustion, all combustions require fuel + oxidant, produces are always CO2 and H2O, most likely gas.

This will release energy, so the enthalpy change is NEGATIVE. The CO2 (carbon dioxide) is the most oxidized state of carbon, so this implies a large amount of energy released.

The water is typically in vapor state, since the heat released by the reaction is used in order to favour the enthalpy of vaporization.

The enthalpy of formation of C2H6, O2 is much larger than that of CO2 and H2O, which also implies a heat of reaction NEGATIVE.

For the entropy, there is CHAOR created, since 2+7 = 9 moles of reactatns produce 4+6 = 10 mol of reactants.

Not only that, the particles are much more disordered, since they increased in temperature.

The Chemical potential favours a SPONTANEOUS reaction, since the Gibbs free energy is NEGATIVE due to

dG = dH - T*dS

if dH is negatrive and dS is positive, then this is always negative, meaning it is always spotnaneous.

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