Thermodynamic Signs
Determine the signs of ∆G, ∆H, and ∆S for the following three processes (for example, -+-,++-,+-+).
Water vapor condenses on a cold mirror.
Ethanol burns in air to form carbon dioxide and water vapor.
Dry ice sublimes at room temperature.
I'm interested in your answers/thoughts/rationale. My thoughts for this one.
Water vapor condenses on a cold mirror.
dG = dH - TdS
You know dH for H2O(l) --> H2O(g) is + because heat must be added; therefore, dH must be - for the reverse.
S for gas is greater than S for a liquid because the liquid is more ordered. So dS from gas to liquid must be -.
So for dG = dH - TdS, in which dH is + and - TdS is -T(-S) +, then dG must be - so ---
To determine the signs of ∆G (Gibbs free energy), ∆H (enthalpy), and ∆S (entropy) for the given processes, we can use the following conventions:
- ∆G: Negative (∆G < 0) indicates a spontaneous reaction, positive (∆G > 0) indicates a non-spontaneous reaction, and zero (∆G = 0) indicates the system is at equilibrium.
- ∆H: Negative (∆H < 0) indicates an exothermic reaction (heat is released), positive (∆H > 0) indicates an endothermic reaction (heat is absorbed), and zero (∆H = 0) indicates no exchange of heat with the surroundings.
- ∆S: Positive (∆S > 0) indicates an increase in entropy (disorder), negative (∆S < 0) indicates a decrease in entropy, and zero (∆S = 0) indicates no change in entropy.
Let's analyze each process:
1. Water vapor condenses on a cold mirror:
During condensation, water vapor molecules lose energy and transition from gas to liquid phase. This process is exothermic as heat is released to the surroundings. The randomness of the system decreases as gas molecules become more ordered in the liquid phase. Therefore, the signs are as follows:
∆G: Negative (∆G < 0)
∆H: Negative (∆H < 0)
∆S: Negative (∆S < 0)
2. Ethanol burns in air to form carbon dioxide and water vapor:
The combustion process of ethanol is exothermic and increases the total disorder of the system significantly, resulting in a large positive entropy change. The signs are as follows:
∆G: Negative (∆G < 0)
∆H: Negative (∆H < 0)
∆S: Positive (∆S > 0)
3. Dry ice sublimes at room temperature:
Sublimation refers to the direct transition from the solid phase to the gas phase without passing through the liquid phase. In this case, the solid carbon dioxide (dry ice) undergoes an endothermic process, as heat is absorbed from the surroundings. The randomness of the system increases as solid molecules transfer to the gas phase. Therefore, the signs are as follows:
∆G: Negative (∆G < 0)
∆H: Positive (∆H > 0)
∆S: Positive (∆S > 0)
By understanding the physical and thermodynamic aspects of the given processes, we can determine the signs of ∆G, ∆H, and ∆S for each process.