4NH3(g) + 502(g) ---> 4NO(g) + 6H2O(1) 2NO(g) + 02(g) - 3NO2(g) + H20(1) ---> 2HNO3((aq) + NO(g) MH = -907 kJ H = -113 kJ WH = -139 kJ 2NO2(g)

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Chapter10: Energy
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Can I calculate this enthalpy only from last equation. Which is already given that the reaction of formation of nitric acid. By using Hess law. That is H of reaction is H of products minus H of reactants.

4NH3(g) + 502(g) ---> 4NO(g) + 6H2O(1)
2NO(g) + 02(g) -
3NO2(g) + H20(1) ---> 2HNO3((aq) + NO(g)
MH = -907 kJ
H = -113 kJ
WH = -139 kJ
2NO2(g)
Transcribed Image Text:4NH3(g) + 502(g) ---> 4NO(g) + 6H2O(1) 2NO(g) + 02(g) - 3NO2(g) + H20(1) ---> 2HNO3((aq) + NO(g) MH = -907 kJ H = -113 kJ WH = -139 kJ 2NO2(g)
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