Calculate ΔH° for the oxidation of nitric oxide to nitrogen dioxide,
2 NO(g) + O2(g) ---> 2 NO2(g),
from the following data:
ΔH°f[NO(g)] = +90.25 kJ/mol
ΔH°f[NO2(g)] = +33.18 kJ/mol
+16.93 kJ/mol
-218.71 kJ/mol
-114.14 kJ/mol
-172.15 kJ/mol
+69.54 kJ/mol
Given:
Hof(NO(g)) = 90.25 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol
Hof(NO2(g)) = 33.18 KJ/mol
Balanced chemical equation is:
2 NO(g) + O2(g) ---> 2 NO2(g)
ΔHo rxn = 2*Hof(NO2(g)) - 2*Hof( NO(g)) - 1*Hof(O2(g))
ΔHo rxn = 2*(33.18) - 2*(90.25) - 1*(0.0)
ΔHo rxn = -114.14 KJ/mol
Answer: -114.14 KJ/mol
Calculate ΔH° for the oxidation of nitric oxide to nitrogen dioxide,
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