(1) C(s, graphite)+ O2(g) → CO(g) AH =-393.5 kJ/mol (2) H2(g) + ¼O:(g) → H20(g) (3) C(s, graphite) + 2H2(g) → CH4(g) AH = -74.6 kJ/mol %3D AH =-241.8 kJ/mol %3D

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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4. Now use the enthalpy of formation values provided (values for equations 1-3 above) to calculate the enthalpy
of combustion of methane. It should be the same as the number you came up with using Hess's Law.
Transcribed Image Text:4. Now use the enthalpy of formation values provided (values for equations 1-3 above) to calculate the enthalpy of combustion of methane. It should be the same as the number you came up with using Hess's Law.
Calorimetry can be used to measure enthalpy of a reaction and Hess's Law involves adding equations together to
determine an unknown enthalpy of reaction. Another way to determine the enthalpy of a reaction is to use the
standard enthalpy of formation of each component in a reaction. This is the energy absorbed or released when a
compound is formed from elements in their standard states, the state at which the substance is most stable. For
example, carbon dioxide, water, and methane can be formed according to the below reactions:
(1) C(s, graphite) + O2(g) → CO2(g) AH¡
(2) H2(g) + ½O2(g) → H2O(g)
(3) C(s, graphite) + 2H2(g) – CH4(g) AH, =-74.6 kJ/mol
=-393.5 kJ/mol
AH
= -241.8 kJ/mol
ntol form or most stable state, and, by definition,
Transcribed Image Text:Calorimetry can be used to measure enthalpy of a reaction and Hess's Law involves adding equations together to determine an unknown enthalpy of reaction. Another way to determine the enthalpy of a reaction is to use the standard enthalpy of formation of each component in a reaction. This is the energy absorbed or released when a compound is formed from elements in their standard states, the state at which the substance is most stable. For example, carbon dioxide, water, and methane can be formed according to the below reactions: (1) C(s, graphite) + O2(g) → CO2(g) AH¡ (2) H2(g) + ½O2(g) → H2O(g) (3) C(s, graphite) + 2H2(g) – CH4(g) AH, =-74.6 kJ/mol =-393.5 kJ/mol AH = -241.8 kJ/mol ntol form or most stable state, and, by definition,
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