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Author Question: For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Therefore: (Read 120 times)

karateprodigy

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on: Jul 8, 2021

Question 1

Determine the enthalpy change in the following equation:

Cl2(g) + H2O(l) → 2 HCl(g) + 1/2 O2(g)
ΔfH° H2O(l) = -285.8 kJ/mol
ΔfH° HCl(g) = -92.31 kJ/mol

◦ 193.5 kJ/mol
◦ -470.4 kJ/mol
◦ 470.4 kJ/mol
◦ 101.2 kJ/mol
◦ -101.2 kJ/mol

Question 2

For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Therefore:
◦ ΔU = 110 kJ and ΔH = 145 kJ
◦ ΔU = 145 kJ and ΔH = 110 kJ
◦ ΔU = 145 kJ and ΔH = 180 kJ
◦ ΔU = 180 kJ and ΔH = 145 kJ
General Chemistry: Principles and Modern Applications ¦ Petrucci, Herring, Madura, Bissonnette ¦ 11th Edition
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Marked as best answer by karateprodigy on Jul 8, 2021

irishcancer18

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Reply #1 on: Jul 8, 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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karateprodigy

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Reply 2 on: Jul 8, 2021
Great answer, keep it coming :)

Jossy

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Reply 3 on: Yesterday
Wow, this really help

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