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Using formation values of -207.4 for HNO3(aq), +90.2 for NO(g), +33.2 for NO2(g) and -285.8 kJ/mol for H2O(l), what is the enthalpy change for 3NO2 + H2O giving 2HNO3 + NO?

A-138.4 kJ
B+138.4 kJ
C-324.6 kJ
D-709.8 kJ
Answer & Solution
Correct answer: A. -138.4 kJ
1. Multiply each formation value by its coefficient in the balanced equation. 2. Products give 2 times minus 207.4 plus 1 times plus 90.2, that is minus 414.8 plus 90.2. 3. The product sum is therefore minus 324.6 kJ. 4. Reactants give 3 times plus 33.2 plus 1 times minus 285.8, that is plus 99.6 minus 285.8. 5. The reactant sum is therefore minus 186.2 kJ. 6. Subtract: minus 324.6 minus minus 186.2 equals minus 324.6 plus 186.2. 7. The reaction enthalpy is minus 138.4 kJ, so the process is exothermic. 8. Stopping at minus 324.6 kJ reports the product sum alone and skips the subtraction. _Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.3 Enthalpy_
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