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Given that,NH (3 (g)) + 3 Cl (2 (g)) hAr...

Given that,`NH _(3 (g)) + 3 Cl _(2 (g)) hArr NCl _(3 (g)) + 3 HCl _((g)) , - Delta H _(1)`
`N_(2(g)) + 3 H _(2 (g)) hArr 2 NH _(3 (g)) , Delta H _(2)`
`H _(2 (g)) + Cl _(2 (g)) hArr 2 HCl _((g)) , Delta H _(3)`
The enthalpy of formation fo `NCl _(3 (g)) ` in terms of `Delta H _(1) , DeltaH _(2) and Delta H _(3)` is

A

`Delta _(f) H =- Delta H _(i)- ( Delta H _(2))/(2) - (3)/(2) Delta H _(3)`

B

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

C

`Delta _(f) H = Delta H_(i) - ( Delta H _(2))/(2) - (3)/(2) Delta H _(3)`

D

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

Text Solution

Verified by Experts

The given equations are
`NH _(3 (g)) + 3 Cl _(2 (g)) hArr NCl _(3 (g)) + 3 HCl _((g)) : - Delta H _(1)" "(1) `
`N _(2 (g)) + 3 H _(2 (g)) hArr 2 NH _(3 (g)) ,- Delta H _(2) " "(2)`
`H _(2 (g)) + Cl _(2 (g)) hArr 2 HCl _((g)) , Delta H _(3) " "(3)`
We aim at: ` N _(2 (g)) + 3 Cl _(2(g)) hArr 2 NCl _(3 (g)) " " Delta _(f) H = ?`
Addding `2 xx Eq. (1) + Eq. (2) - 3 xx Eq. (3),` we get
`2 Delta _(f) H =- 2 Delta H _(1) + [ - Delta H _(2) - 3 Delta H _(3)]implies Delta _(f) H =- Delta H _(1) - ( Delta H _(2))/(2) - (3)/(2) Delta H _(3)`
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