Problem 48 - Chemistry: An Atoms-Focused Approach

Question

What are the signs of \(\Delta S, \Delta H,\) and \(\Delta G\) for the formation of dew on a cool night?

Step-by-Step Solution

Answer

Answer: The signs of the changes are as follows: ∆S is negative, ∆H is negative, and ∆G is negative.

Step 1: Determine the change in entropy (∆S) for the formation of dew

Entropy is a measure of the disorder in a system. When water vapor molecules in the air condense to form dew on a surface, the disorder of the system decreases. The molecules go from the vapor phase, which has higher entropy, to the liquid phase, which has lower entropy. Therefore, the change in entropy (∆S) for the formation of dew is negative.

Step 2: Determine the change in enthalpy (∆H) for the formation of dew

Enthalpy is a measure of the total energy in a system. When water vapor condenses to form dew, it releases heat energy into the environment. This is because the vapor phase has higher energy content than the liquid phase. Hence, the change in enthalpy (∆H) for the formation of dew is negative since heat is released.

Step 3: Determine the change in Gibbs free energy (∆G) for the formation of dew

Gibbs free energy is a measure of the maximum amount of work that can be extracted from a thermodynamic system. The relationship between Gibbs free energy (∆G), enthalpy (∆H), and entropy (∆S) is given by the equation: ∆G = ∆H - T∆S where T is the temperature in Kelvin. Since we previously determined that both ∆H and ∆S are negative for the formation of dew, the term T∆S is positive. For a process to be spontaneous, ∆G must be negative. In this case, the negative ∆H outweighs the positive T∆S term, so the overall ∆G is also negative. In conclusion, the signs of ∆S, ∆H, and ∆G for the formation of dew on a cool night are: ∆S: negative ∆H: negative ∆G: negative