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Question

Given,
NH3(g)+3Cl2(g)NCl3(g)+3HCl; ΔH1

N2(g)+3H2(g)2NH3(g); ΔH2

H2(g) + Cl2(g) 2HCl(g); ΔH3

The heat of formation of NCl3 in terms of ΔH1,ΔH2 and ΔH3 is:

A

ΔH1+ΔH22+32ΔH3

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B

ΔH1ΔH2232ΔH3

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C

ΔH1ΔH22+32ΔH3

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D

ΔH1+ΔH2232ΔH3

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Solution

The correct option is D

ΔH1+ΔH2232ΔH3


NH3(g)+3Cl2(g)NCl3(g)+3HCl; ΔH1...(i)

N2(g)+3H2(g)2NH3(g); ΔH2...(ii)

H2(g)+Cl2(g)2HCl(g); ΔH3...(iii)

12N2(g)+32Cl2(g)NCl3(g); ΔH4...(iv)
Equation (iv) can be obtained by:
(i)+(ii)×12+(iii)×(32)

ΔH4=ΔH1+ΔH2232ΔH3


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