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The value of ΔfHΘfor NH3 is – 91.8 kJ mol–1. Calculate the enthalpy change for the following reaction : 2NH3(g) → N2(g) + 3H2(g) |
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Answer» lpy change of a reaction is calculated as: Σbond enthalpy of reactants- Σbond enthalpy of productso The standard molar enthalpy of FORMATION (for one mole) ΔfHΘ is given for NH3 = – 91.8 kJ mol–1 which is associated with the reverse exothermic reaction N2(G) + 3H2(g) → 2NH3(g) i.e. heat released INFORMATION of AMMONIA is – 91.8 kJ mol –1.Hence, for the decomposition 2NH3(g) →N2(g) + 3H2(g) ΔrHΘ will be =- (– 91.8 kJ mol–1 ) = + 91.8 kJ mol–1 .And in this case for 2 moles of NH3 enthalpy change of the reaction will be ΔrH = (2 X 91.8 ) = 183.6 kJ mol–1 . |
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