Ch.9 - Thermochemistry: Chemical Energy
Chapter 9, Problem 96
Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ
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Related Practice
Textbook Question
The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?
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Textbook Question
The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ
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Textbook Question
Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the labo- ratory. Set up a Hess's law cycle that will let you calculate ∆H° for the conversion of methane to ethylene: 2 CH4(g) → C2H4(g) + 2 H2(g) You can use the following information: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ∆H° = -3120.8 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H° = -890.3 kJ C2H4(g) + H2(g) → C2H6(g) ∆H° = -136.3 kJ H2O(l) ∆H°f = -285.8 kJ/mol
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Textbook Question
Set up a Hess's law cycle, and use the following information to calculate ΔH°f for aqueous nitiric acid, HNO3(aq). You will need to use fractional coefficients for some equations. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH° = -137.3 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = -116.2 kJ 4 NH3(g) + 5 O2(g) → 4 NO (g) + 6 H2O(l) ΔH° = -1165.2 kJ NH3(g) ΔH°f = -46.1 kJ/mol H2O(l) ΔH°f = -285.8 kJ/mol
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Textbook Question
What is a compound's standard heat of formation?
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Textbook Question
How is the standard state of an element defined? Why do
elements always have ∆H°f = 0?
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