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Author Question: For a particular process, G = H at a given temperature and pressure. Therefore, (Read 48 times)

Capo

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on: Feb 15, 2020

Question 1

For a particular process ΔG is less than ΔH. Therefore
◦ ΔS is positive.
◦ ΔS is zero.
◦ ΔS is negative if ΔH is positive and ΔS is positive if ΔH is negative.
◦ ΔS is negative.

Question 2

For a particular process, ΔG = ΔH at a given temperature and pressure. Therefore,
◦ ΔS is positive if ΔH is negative and negative is ΔH is positive.
◦ ΔS is zero.
◦ ΔS = ΔG/T.
◦ ΔS is negative if ΔG is positive and positive if ΔG is positive.
Chemistry ¦ McMurry, Fay, Robinson ¦ 7th Edition
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Marked as best answer by Capo on Feb 15, 2020

Athena23

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Reply #1 on: Feb 15, 2020
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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Capo

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Reply 2 on: Feb 15, 2020
:D TYSM

emsimon14

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Reply 3 on: Yesterday
Excellent

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