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Question

Given the bond energies N N, H H and N H bonds are 945,436 and 391 kJ mole1 respectively, the enthalpy of the following reaction N2(g) + 3H2(g) 2NH3(g) is


A

-93 kJ

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B

102 kJ

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C

90 kJ

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D

105 kJ

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Solution

The correct option is A

-93 kJ


N N + 3H H 2N|H|HH

945 +3 x 436 2 × (3 × 391) = 2346

Energy absorbed Energy released

Net. Energy released = 2346 2253 = 93 kJ

i.e., ΔH = 93kJ


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