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Question

NH3(g)+3Cl2(g)NCl3(g)+3HCl(g); ΔH1
N2(g)+3H2(g)2NH3(g); ΔH2
N2(g)+Cl2(g)2HCl(g); ΔH3
The enthalpy of formation of NCl3(g) in the terms of ΔH1, ΔH2 and ΔH3 is :

A
ΔHf=ΔH1+ΔH2232ΔH3
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B
ΔHf=ΔH1+ΔH2232ΔH3
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C
ΔHf=ΔH1ΔH2232ΔH3
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D
None
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Solution

The correct option is A ΔHf=ΔH1+ΔH2232ΔH3
Heat of formation of NCl3 is the heat of the reaction of the reaction in which NCl3 is formed from it's pure components- N2,Cl2.
Therefore the Reaction-
(1/2)N2+(3/2)Cl2NCl3,δHf(NCl3) is obtained by-
Reaction(i) + Reaction(ii)/2 - (3/2)Reaction(iii)
So,
ΔHf=ΔH1+ΔH2232ΔH3

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