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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 these

Text Solution

Verified by Experts

The correct Answer is:
C
Find `(1)/(2)N_(2)+(3)/(2)Cl_(2)rarrNCl_(3),DeltaH_(f)`
Multiply Eq. (ii) by 1/2 and add to Eq. (i), Eq. (iii) by 3/2 and subtract from Eq. (i), we get
`DeltaH_(f)=-DeltaH_(l)-(DeltaH_(2))/(2)-[+(3)/(2)DeltaH_(3)]`
`=-DeltaH_(l)-(DeltaH_(2))/(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)

Given a. NH_(3)(g)+3CI(g)rarr NCI_(3)(g),3HCI(g),DeltaH_(1) b. N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g),DeltaH_(2) c. H_(2)(g)+CI_(2)(g)rarr 2HCI(g),DeltaH_(3) Express the enthalpy of formation of NCI_(3)(g)(Delta_(f)H^(Theta)) in terms of DeltaH_(1), DeltaH_(2) ,and DeltaH_(3) .

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) :

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