Answer the following questions related to H2O. Substance ΔG°f at 298K(kJ/mol) H2O(l) −237.2 H2O(g) −228.4

Question

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.

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Thành Đạt 4 years 2021-08-12T10:28:49+00:00 1 Answers 1054 views 0

Answers ( )

  1. ngochoa
    0
    2021-08-12T10:30:25+00:00
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    Answer:

    1. 8.8kJ/mol

    2. The process is not favorable.

    3. The reason is in the answer.

    4. Process 1.

    Explanation:

    1. ΔG° = ΔG° products – ΔG° reactants

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

    2. The process is not favorable because ΔG > 0

    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)

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