Chemistry, asked by itsprince04, 9 months ago

3No2 +2NH3 → 4N2 +3H2O ∆Hf =-879.6 kj/mol if ∆Hf [NH3] = -45.9 kj/mol and ∆Hf [NO2] = -241.8kj/mol then ∆Hf of [No2] will be a)+246Kj b) +82 kj c)-82 kj d) -246kj

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3No2 +2NH3 → 4N2 +3H2O ∆Hf =-879.6 kj/mol if ∆Hf [NH3] = -45.9 kj/mol and ∆Hf [NO2] = -241.8kj/mol then ∆Hf of [No2] will be a)+246Kj b) +82 kj c)-82 kj d) -246kj

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3No2 +2NH3 → 4N2 +3H2O ∆Hf =-879.6 kj/mol if ∆Hf [NH3] = -45.9 kj/mol and ∆Hf [NO2] = -241.8kj/mol then ∆Hf of [No2] will be a)+246Kj b) +82 kj c)-82 kj d) -246kj​

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3No2 +2NH3 → 4N2 +3H2O ∆Hf =-879.6 kj/mol if ∆Hf [NH3] = -45.9 kj/mol and ∆Hf [NO2] = -241.8kj/mol then ∆Hf of [No2] will be a)+246Kj b) +82 kj c)-82 kj d) -246kj