Given are the following data at 298 K.ΔfH∘H2O,l=−286kJmol−1 ΔfH∘HNO3,1=−174kJmol−1; ΔfH∘N2O5,g=14.5kJmol−1, The ΔrH∘ for the reaction N2O5(g)+H2O(l)⟶2HNO3(l) is

Given are the following data at 298 K.ΔfHH2O,l=286kJmol1 ΔfHHNO3,1=174kJmol1; ΔfHN2O5,g=14.5kJmol1, The ΔrH for the reaction N2O5(g)+H2O(l)2HNO3(l) is

  1. A

    -76.5 kJ mol-1

  2. B

    76.5 mol-1

  3. C

    97.5 kJ mol-1

  4. D

    -97.5 kJ mol-1

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    Solution:

     (i) H2(g)+12O2(g)H2O(l); ΔfH=286kJmol1  (ii) 12N2(g)+32O2(g)+12H2(g)HNO3(l) ΔfH=174kJmol1  (iii) N2(g)+52O2(g)N2O5(g) ΔH=14.5kJmol1

    The required reaction N2O5(g)+H2O(l)2NHO3(l) is obtained as -Eq.(iii) - Eq. (i) + 2 Eq. (ii) Hence, Δ1H=[14.5(286))+2(174)kJmol1=76.5kJmol1

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