Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−380 kJX2+3Y2⟶2XY3ΔH1=−380 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJX2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−220 kJ2Y2+Z2⟶2Y2ZΔH3=−220 kJ Calculate the change in enthalpy for the reaction. 4XY3+7Z2⟶6Y2Z+4XZ2
Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−380 kJX2+3Y2⟶2XY3ΔH1=−380 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJX2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−220 kJ2Y2+Z2⟶2Y2ZΔH3=−220 kJ Calculate the change in enthalpy for the reaction. 4XY3+7Z2⟶6Y2Z+4XZ2
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
Chapter17: Chemcial Thermodynamics
Section: Chapter Questions
Problem 17.8QE
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Question
Given the thermochemical equations
X2+3Y2⟶2XY3ΔH1=−380 kJX2+3Y2⟶2XY3ΔH1=−380 kJ
X2+2Z2⟶2XZ2ΔH2=−120 kJX2+2Z2⟶2XZ2ΔH2=−120 kJ
2Y2+Z2⟶2Y2ZΔH3=−220 kJ2Y2+Z2⟶2Y2ZΔH3=−220 kJ
Calculate the change in enthalpy for the reaction.
4XY3+7Z2⟶6Y2Z+4XZ2
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