Use the given standard enthalpies of formation (in kJ/mol) to determine the enthalpy of reaction of the following reaction :
`NH_(3)(g)+3F_(2)(g)rarrNF_(3)+3HF(g)`
`Delta H_(f)^(@)(NH_(3)) (g)=-46.2`, `Delta H_(f)^(@)(NF_(3))(g)=-113.0` `DeltaH_(f)^(@)(HF) (g)=-269.0`

Options

A. `-335.8 kJ//mol`
B. `-873.8 kJ//mol`
C. `-697.2 kJ//mol`
D. `-890.4 kJ//mol`

Answer

B

Explanation

`Delta_(r )H=(Delta_(f)H)_(P)-(Delta_(f)H)_(R)`
`Delta_(r )H=-113 + 3xx(-269)+46.2 =-873.8 " kJ"`