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NH(3)(g) + 3Cl(2)(g) rarr NCl(3)(g) + 3H...

`NH_(3)(g) + 3Cl_(2)(g) rarr NCl_(3)(g) + 3HCl(g), " "DeltaH_(1)`
`N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g), " "Delta H_(2)`
`H_(2)(g) + Cl_(2)(g) rarr 2HCl(g), " "Delta H_(3)`
The heat of formation of NCl3(g) in the terms of `DeltaH_(1), DeltaH_2 and DeltaH_(3)` is :

A

`Delta H_(f)= - Delta H_(1) + (Delta H_(2))/(2) - (3)/(2) Delta H_(3)`

B

`Delta H_(f)= Delta H_(1) + (Delta H_(2))/(2) - (3)/(2) Delta H_(3)`

C

`Delta H_(f)= - Delta H_(1)- (Delta H_(2))/(2)- (3)/(2) Delta H_(3)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
(i) `NH_(3)(g) + 3Cl_(2)(g) hArr NCl_(3)(g) + 3HCl(g), - Delta H_(1)`
(ii) `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), -Delta H_(2)`
(iii) `H_(2)(g) + Cl_(2)(g) hArr 2HCl(g), + Delta H_(3)`
From eq (ii), `Delta_(f) H_(NH_(3)) = -(1)/(2) Delta H_(2)`
From eq (iii) `Delta_(f) H_(HCl) = + (1)/(2) Delta H_(3)`
From eq (i)
`Delta_(r) H= Delta_(f) H_(NCl_(3)) + 3Delta_(f) H_(HCl) - Delta_(f) H_(NH_(3)) - 3 Delta_(f) H_(Cl_(2))- Delta H_(1)= Delta_(f) H_(NCl_(3)) + 3((1)/(2) Delta H_(3))- (- (1)/(2) Delta H_(2)) - 3(0)`
`Delta_(f) H_(NCl_(3)) = - DeltaH_(1) - (1)/(2) Delta H_(2) - (3)/(2) Delta H_(3)`
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