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Author Question: Consider the following hypothetical reaction at 310 K. Standard free energies of formation are given ... (Read 109 times)

maegan_martin

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on: Mar 21, 2021

Question 1

Consider the following hypothetical reaction (at 316.1 K). Standard free energies in kJ/mol are given in parentheses.
    A           B     +     C          ΔG° = ?
      (-32.2)            (207.8)            (-237.0)
What is the value of the equilibrium constant for the reaction at 316.1 K?

◦ 0.32
◦ 1.0
◦ 6.7 × 104

◦ 273
◦ 0.42

Question 2

Consider the following hypothetical reaction at 310 K. Standard free energies of formation are given in parentheses.
 
B
     C
ΔG° = –36.2 kJ/mol
 
(?)
(176.4 kJ/mol)
 
Calculate the standard free energy of formation of compound B.

​


◦ 212.6 kJ/mol
◦ –212.6 kJ/mol
◦ 140.2 kJ/mol
◦ –140.2 kJ/mol
◦ None of these are correct.
Chemistry: An Atoms First Approach ¦ Zumdahl, DeCoste ¦ 3rd Edition
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Marked as best answer by maegan_martin on Mar 21, 2021

nekcihc358

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Reply #1 on: Mar 21, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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maegan_martin

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Reply 2 on: Mar 21, 2021
Wow, this really help

cam1229

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Reply 3 on: Yesterday
:D TYSM

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