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Question

If 1 mole of aqueous nitric acid is formed, calculate total heat released :

4NH3(g)+5O2(g)4NO(g)+6H2Og

2NOg+O2(g)2NO2(g)

3NO2(g)+H2O(g)2HNO3(aq)+NO(g)

A
-986 KJ
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B
-246.5 KJ
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C
-493 KJ
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D
None of these
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Solution

The correct option is C -493 KJ
First multiply the first equation by 3, second by 6 and third by 4.

12NH3(g)+15O2(g)12NO(g)+18H2Og ΔH = -2721 kJ

12NOg+6O2(g)12NO2(g) ΔH = -678 kJ

12NO2(g)+4H2O(g)8HNO3(aq)+4NO(g) ΔH = -556 kJ

Adding up all the three equations,

12NH3(g)+15O2(g)12NO(g) +6O2(g)+12NO2(g) +4H2O(g) 12NO(g)+18H2Og + 12NO2(g)8HNO3(aq)+4NO(g)

ΔH = -2721 kJ - 678 kJ - 556 kJ

= ΔH = -3955 kJ

1 mole HNO3×3955KJ / 8 mol HNO3
= -493 KJ

Therefore, the correct option is C.

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