For a system in which DHvap > TDSvap, what state is favored?
DG = DH - TDS
Would you think if DH > TDS, then DG is positive, the reaction (vaporization) is not favored and the .......(liquid/vapor) state is favored?
To determine the state that is favored in a system where DHvap > TDSvap, we need to consider the Gibbs free energy change (ΔG) of the process. The Gibbs free energy is calculated using the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
When DHvap > TDSvap, it means that the enthalpy change (DHvap) during vaporization is greater than the temperature multiplied by the entropy change (TDSvap).
The sign of ΔG determines the state that is favored:
- If ΔG < 0, it means the process is spontaneous, and the system is energetically favorable.
- If ΔG > 0, it means the process is non-spontaneous, and the system is not energetically favorable.
- If ΔG = 0, it means the system is at equilibrium.
In this case, let's analyze the equation ΔG = ΔH - TΔS:
Since DHvap > TDSvap, we know that the enthalpy change is greater than the temperature multiplied by the entropy change. Therefore, the ΔH term dominates over the TΔS term, resulting in a positive value for ΔG.
When ΔG > 0, it indicates that the process is non-spontaneous and the system is not energetically favored. Hence, in a system where DHvap > TDSvap, the favored state is the liquid state rather than the vapor state.