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)`:

Options

A. `Delta H_f = - Delta H_1 + (Delta H_2)/(2)- 3/2 Delta H_3`
B. `Delta H_f = Delta H_f + (Delta H_2)/(2) - 3/2 Delta H_3`
C. `Delta H_f= Delta H_f - (Delta H_2)/(2)- 3/2 (Delta H_3)`
D. None

Answer

A

Explanation

Find `1//2N_2 + 3//2 Cl_2 rarr NCl_3, Delta H_f= ?`
Multiply (ii) by 1/2, (ii) 3/2 and substract from (i), we get
`Delta H_f = -Delta H_1-[-(DeltaH_2)/(2) +3/2 Delta H_3]`
`=- Delta H_1 +(Delta H_2)/(2)-3/2 Delta H_3`