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Q.

Enthalpy of formation of 2 mol of NH3 g is -90 kJ, and ∆HH-H  and ∆HN-H are respectively 435 kJ mol-1 and 390 kJ mol-1. The value of ∆HN≡ N  is

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a

-472.75 KJ

b

-945 KJ

c

472.5 kJ

d

945 kJ mol-1

answer is D.

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Detailed Solution

N2g+3H2g → 2NH3g;    ∆H=-90 kJ Now - 90 =∆HN≡N+ 3∆HH-H-6∆HN-H ∆HN≡N=2340-1305-90 =945
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