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###### Anuradha Bhogaonkar posted an Question
May 11, 2020 • 11:49 am 30 points
• IIT JAM
• Chemistry (CY)

# The following sequence of reaction occurs in commercial production of aqueous nitric acid. 4nh,(g)+50, (g) 4no(g) + 6h,o0) ah=-904kj.1) 3no, (g)+h,o0)2hno, (aq)

The following sequence of reaction occurs in commercial production of aqueous nitric acid. 4NH,(g)+50, (g) 4NO(g) + 6H,O0) AH=-904kJ.1) 3NO, (g)+H,O0)2HNO, (aq)+NO(g) AH=-140kJ.) Determine the total heat liberated (in kJ/mol) at constant pressure for the production of exactly 1 mole of aqueous nitric acid from NH3 by this process. 2NO(g)+0,(g) 2NO, (g) AH=-112kJ ..(2) a.(A) 986 b.(B) 493 c. (C) 246.5 d.(D) None of these

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• ##### Dinesh khalmaniya 1

yeh option b is correct.. here negative charge showing that it is exothermic reaction

yes sir

ans is 493

• ##### Dinesh khalmaniya 1

Multiply the first equation by 3, the second by 6, and the third by 4: 12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ 12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ 12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ Add up all three equations: 12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) → 12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g) ΔH = −2721 kJ − 678 kJ − 556 kJ Cancel like amounts on opposite sides of the arrow, combine amounts of O2 on the same side of the arrow, and do the arithmetic for ΔH: 12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ (1 mole HNO3(aq)) x (−3955 kJ/8 mol HNO3(aq)) = −494 kJ