3A+6B→3D AH - 403 kJ/ тol E + 2F→ A AH, = – ΔΗ 105.2 kJ/mol C→E+3D AH3 = +64.8 kJ/mol 6B+4E+6F→C AH = ?

Chemistry for Engineering Students
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ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter9: Energy And Chemistry
Section: Chapter Questions
Problem 9.43PAE
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watch image please(about enthalpy)

ЗА + 6В—3D
AH = - 403kJ/mol
E + 2F→ A
AH, = – 105.2 kJ/mol
C→E + 3D
AH,= +64.8 kJ/mol
ΔΗ
6B+ 4E+6F→C
AH = ?
Transcribed Image Text:ЗА + 6В—3D AH = - 403kJ/mol E + 2F→ A AH, = – 105.2 kJ/mol C→E + 3D AH,= +64.8 kJ/mol ΔΗ 6B+ 4E+6F→C AH = ?
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