ACl5(g) → ACl3(g) + Cl2(g) ΔH = ?Given:A4(s) + 6Cl2(g) → 4ACl3(g) ΔH = -2439 kJ4ACl5(g) → A4(s) + 10Cl2(g) ΔH = 3438 kJ 2. AClsle) → ACI3(e) + Clz{e)AH = ?Given:Aa(s) + 6CI2(e) → 4ACI3{g)AH = -2439 kJ4ACIS(e) → A4(s) + 10Cl2le)AH = 3438 kJ

Answered: Image /qna-images/answer/e4b7131d-2ae2-4cb8-8994-56c99cb944ba.jpg

Answered: 2. AClsle) → ACI3(e) + Clz{e) AH = ?…

2. AClsle) → ACI3(e) + Clz{e) AH = ? Given: Aa(s) + 6CI2(e) → 4ACI3{g) AH = -2439 kJ 4ACIS(e) → A4(s) + 10Cl2le) AH = 3438 kJ

World of Chemistry

7th Edition

ISBN:9780618562763

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter10: Energy

Section: Chapter Questions

Problem 38A

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Question

ACl5(g) → ACl3(g) + Cl2(g) ΔH = ?
Given:
A4(s) + 6Cl2(g) → 4ACl3(g) ΔH = -2439 kJ
4ACl5(g) → A4(s) + 10Cl2(g) ΔH = 3438 kJ

2. AClsle) → ACI3(e) + Clz{e)
AH = ?
Given:
Aa(s) + 6CI2(e) → 4ACI3{g)
AH = -2439 kJ
4ACIS(e) → A4(s) + 10Cl2le)
AH = 3438 kJ

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