Answer the following questions related to H2O.

Substance ΔG°f at 298K(kJ/mol)
H2O(l) −237.2
H2O(g) −228.4

(a) Using the information in the table above, determine the value of ΔG° at 298K for the process represented by the equation H2O(l)⇄H2O(g).

Question 2
(b) Considering your answer to part (a), indicate whether the process is thermodynamically favorable at 298K. Justify your answer.

Question 3
(c) Considering your answer to part (b), explain why H2O(l) has a measurable equilibrium vapor pressure at 298K.

Question 4
Water vapor can be produced in two different processes, as represented below.

Process 1 Process 2
H2O(s)⇄H2O(g) H2O(l)⇄H2O(g)
(d) In terms of concepts of entropy and the particle-level structure of the different phases of water, explain why the change in entropy, ΔS, is greater for process 1 than for process 2.

3. As the reaction is an equilibrium, a little amount of water will change from liquid to gas. This is the reason because of water has a measurable equilibrium vapor pressure.

4. Entropy is greater in process 1 because the structure of H2O(s) is more organized than H2O(l) (S of H2O(s) is lower than H2O(l)).

ΔS of H2O(s)⇄H2O(g) is greater than H2O(l)⇄H2O(g)

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-03-12T13:19:36+01:00

Given reaction is not thermodynamically favorable as value of ΔG is positive, water has measurable equilibrium vapor pressure because very little amount of water changes and entropy of 1st process is more due to solid structure of water.

What is Gibb's free energy?

Gibb's free energy of any reaction tells about the amount of free energy available to complete the reaction.

(1). Given chemical reaction is:

H₂O(l) ⇄ H₂O(g), in this reaction water changes its state from liquid to gas. Also given the value of standard free energy:

ΔG° for H₂O (l) = -237.2 kJ/mole

ΔG° for H₂O (g) = -228.4 kJ/mole

ΔG° = ΔG°(product) - ΔG°(reactant)

ΔG° = -228.4kJ/mol - (-237.2kJ/mol) = 8.8kJ/mol

(2). As the calculated value of ΔG° is positive, so the reaction is not thermodynamically favorable, for the spontaneous process value of ΔG° should be negative.

(3). Water has a measurable equilibrium vapor pressure because very little amount of water will change from liquid to gas state in the equilibrium.

(4). Change in entropy of the reaction can be calculated as:

ΔS = ΔS(product) - ΔS(reactant)

In the first process reactant is in solid state which have value near to zero, so we get the higher value of ΔS for 1st process as compare to 2nd.

Hence, answers are: (1). ΔG° = 8.8kJ/mol, (2). not thermodynamically favorable, (3). because very little water changes and (4). due to solid structure of water.

To know more about Gibb's free energy, visit the below link:

https://brainly.com/question/10012881