Given N₂ +3H2 =2NH3 and standard enthalpy is -92.4kj/mol
Then find the standard enthalpy of formation of nh3 gas

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Solution

Given N₂ +3H2 =2NH3 and standard enthalpy is -92.4kj/mol [ its not the standard enthalpy that is given but the heat of the reaction ]

By definition, ΔH∘f is the enthalpy change associated with the formation of 1 mole of product from its constituent elements in their standard states at 298K.

now as per the given equation the heat of the equation is for 2 moles of NH3

so dividing the given equation by 2
1/2 N2 + 3/2 H2 ==> NH3 dH = - 92.4 /2 = - 46.2kJ / mol
so the ΔH∘f the enthalpy change associated with the formation = -46.2kJ/mol

Suggest Corrections

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) : △_{r} H^{⊖} = - 92.4 k J m o l^{- 1}

N H_{3}

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