b How much energy is evolved or absorbed as heat in the oxidation of 12.0 g of NH3? Heat of oxidation = kJ

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Chapter1: Chemical Foundations
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a Use standard enthalpies of formation from the table below to calculate the standard enthalpy change for this reaction.
Species AƒH° (kJ/mol)
NH3(g)
-45.90
NO(g)
+90.29
H₂O(g) -241.83
Standard enthalpy change = -906.22
The standard molar enthalpy of formation is the enthalpy change for the formation of 1 mol of a compound directly from its component elements in their
standard states.
The enthalpy change for the reaction:
4 NH3(g) + 5 O₂ (g) → 4 NO(g) + 6 H₂O(g)
^ƒH˚(kJ/mol) -45.90
0
+90.29
-241.83
ΔΗ°
=
[(4
(4 mol) +90.29
(+9
-[(4 mol) (
kJ/mol-rxn
-45.90
kJ
mol
kJ
mol
−(−183.60 kJ) = −906.22 kJ/mol-rxn
KJ
+ (6 mol) | −241.83
+ (5 mol)(0)
(0)]
Correct
kJ
5)]-
mol
-
=
= (-1089.82 kJ) –
b How much energy is evolved or absorbed as heat in the oxidation of 12.0 g of NH3 ?
Heat of oxidation =
Transcribed Image Text:a Use standard enthalpies of formation from the table below to calculate the standard enthalpy change for this reaction. Species AƒH° (kJ/mol) NH3(g) -45.90 NO(g) +90.29 H₂O(g) -241.83 Standard enthalpy change = -906.22 The standard molar enthalpy of formation is the enthalpy change for the formation of 1 mol of a compound directly from its component elements in their standard states. The enthalpy change for the reaction: 4 NH3(g) + 5 O₂ (g) → 4 NO(g) + 6 H₂O(g) ^ƒH˚(kJ/mol) -45.90 0 +90.29 -241.83 ΔΗ° = [(4 (4 mol) +90.29 (+9 -[(4 mol) ( kJ/mol-rxn -45.90 kJ mol kJ mol −(−183.60 kJ) = −906.22 kJ/mol-rxn KJ + (6 mol) | −241.83 + (5 mol)(0) (0)] Correct kJ 5)]- mol - = = (-1089.82 kJ) – b How much energy is evolved or absorbed as heat in the oxidation of 12.0 g of NH3 ? Heat of oxidation =
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