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NH(3)(g)+3Cl(2)toNCl(3)(g)+3HCl(g),Delta...

`NH_(3)(g)+3Cl_(2)toNCl_(3)(g)+3HCl(g),DeltaH_(1)`
`N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH_(2)`
`H_(2)(g)+Cl_(2)(g)to 2HCl(g),DeltaH_(3)`
The heat of formation of `NCl_(3)`(g) in the terms of
`DeltaH_(1),DeltaH_(2),DeltaH_(3)` is :

A

`DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`

B

`DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`

C

`DeltaH_(f)=DeltaH_(1)-(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
The required equation is
`(1)/(2)N_(2)(g)+(3)/(2)Cl_(2)(g)toNCl_(3)(g)`, `DeltaH_(f)=?`
`(i)+(1)/(2)xx(ii)-(3)/(2)xx(iii)`
`DeltaH_(f)=-DeltaH_(1)+(1)/(2)DeltaH_(2)-(3)/(2)DeltaH_(3)`
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Given NH_(3)(g)+3Cl_(2)(g)toNCl_(3)(g)+3HCl(g)+x_(1) N_(2)(g)+3H_(2)(g)to2NH_(3)(g)+x_(2) H_(2)(g)+Cl_(2)(g)to2HCl(g)-x_(3) The heat of formation of NCl_(3)(g)

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Given : (i) NH_(3) (g) + 3Cl_(2)(g) rarr NCl_(3)(g)+3HCl(g) , Delta H_(1) (ii) N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) , Delta H_(2) (iii) H_(2)(g)+Cl_(2)(g) rarr2HCl(g) , Delta H_(3) Express the enthalpy of formation of NCl_(3)(g)(Delta H_(f)) in terms of DeltaH_(1), Delta H_(2) and Delta H_(3) :

DINESH PUBLICATION-CHEMICAL THERMODYNAMICS AND CHEMICAL ENERGETICS -Exercise
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  12. Enthalpy of an exothermic reaction is always

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