4C(s) + 4H2(g) + O2(g) → CH3CH2OCOCH3(l)  ΔH°=-480.0 kJ CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l)  ΔH°=-492.0 kJ 2C(s) + 3H2(g) + 1/2O2(g) → CH3CH2OH(l)  ΔH°=-278.0 kJ H2(g) + 1/2O2(g) → H2O(l)  ΔH°=-286.0 kJ calculate ΔH° for the reaction: CH3COOH(l) + CH3CH2OH(l) → CH3CH2OCOCH3(l) + H2O(l)

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.50QE
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Given the following data:


4C(s) + 4H2(g) + O2(g) → CH3CH2OCOCH3(l)  ΔH°=-480.0 kJ
CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l)  ΔH°=-492.0 kJ
2C(s) + 3H2(g) + 1/2O2(g) → CH3CH2OH(l)  ΔH°=-278.0 kJ
H2(g) + 1/2O2(g) → H2O(l)  ΔH°=-286.0 kJ


calculate ΔH° for the reaction:

CH3COOH(l) + CH3CH2OH(l) → CH3CH2OCOCH3(l) + H2O(l)

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