The molar heat of formation of nh4no3 is -367.54 and those of n2o and h2o are+81.46kj
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To calculate the enthalpy of reaction of decomposition of NH4 NO3.
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To calculate the enthalpy of reaction of decomposition of NH4 NO3.
Heat of formation :
NH4 NO3 : -367.54 kJ / mol
N2 O : + 81.46 kJ / mol
H2 O : - 285.8 kJ /mol
The reaction is when Ammonium Nitrate is heated it decomposes into Nitrous Oxide and Water.
NH₄ NO₃ ====> N₂ O + 2 H₂O
Enthalpy of the reaction = sum of enthalpies of products minus ethalpies of formation of reactants.
ΔH = 81.46 kj - 2 * 285.8 kj - (-367.54 kj)
= - 122.6 kJ
Heat of formation :
NH4 NO3 : -367.54 kJ / mol
N2 O : + 81.46 kJ / mol
H2 O : - 285.8 kJ /mol
The reaction is when Ammonium Nitrate is heated it decomposes into Nitrous Oxide and Water.
NH₄ NO₃ ====> N₂ O + 2 H₂O
Enthalpy of the reaction = sum of enthalpies of products minus ethalpies of formation of reactants.
ΔH = 81.46 kj - 2 * 285.8 kj - (-367.54 kj)
= - 122.6 kJ
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