Calculate the heat of formation of ammonia gas given the following information: 2NH3(l) + 7/2O2(g) → 2 NO2(g) + 3H2O(l) ΔH°= −700kJ ΔHf°NO2 = +33kJ/mol ΔHf°H2O = −286kJ/mol
Expert Answer
Answer
2NH3(l) + 7/2O2(g) → 2 NO2(g) + 3H2O(l) : ΔH°= −700kJ
ΔH°= ΔHf° products – ΔHf° reactants
= [(2xΔHf°NO2(g)) + ( 3 x ΔHf° H2O(l) )] – [(2xΔHf°NH3(l) ) + ( (7/2)xΔHf°O2(g))]
-700 = [(2×33)+(3x(-286))]-[(2xΔHf°NH3(l) ) +((7/2)x0) ]
(2xΔHf°NH3(l) ) = -92 kJ
ΔHf°NH3(l) = -46 kJ
ΔHf°NO2 = +33kJ/mol ΔHf°H2O = −286kJ/mol