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Based on the following reactions,
Q. Heat of formation of NO2 (in kcal) is ........
    Correct answer is '4'. Can you explain this answer?
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    Based on the following reactions,Q. Heat of formation of NO2 (in kcal)...
    N2(g) + O2(g) → 2NO ----- (i) ∆rH° = 43 kcal
    2NO(g) + O2(g) → 2NO2(g) -----(ii)     ∆rH° = -35 kcal
    Enthalpy of formation of NO2 means we need to have 1 mole of NO2 from its constituting elements. 
    We will have (i)/2 + (ii)/2
    ½ N2(g) + O2(g) → NO2(g) ∆fH° = 4 kcal
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    Based on the following reactions,Q. Heat of formation of NO2 (in kcal)...
    N2(g) + O2(g) → 2NO ----- (i) ∆rH° = 43 kcal
    2NO(g) + O2(g) → 2NO2(g) -----(ii)     ∆rH° = -35 kcal
    Enthalpy of formation of NO2 means we need to have 1 mole of NO2 from its constituting elements. 
    We will have (i)/2 + (ii)/2
    ½ N2(g) + O2(g) → NO2(g) ∆fH° = 4 kcal
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    Community Answer
    Based on the following reactions,Q. Heat of formation of NO2 (in kcal)...
    I think -4kcal is the answer .Add the 2 eqns we get
    N2 + 2O2 ___ 2NO2. ∆H = -8k.cal. .this is for 2 moles of NO2
    For 1 mole it would be -4K.Cal
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