`F_(2)(g) +2HCI(g) rarr 2HF(g) +CI_(2)(g), DeltaH^(Theta) =- 352.18 kJ`
Given heat of formation of `HF, Delta_(f)H^(Theta)underset((HF))" =- 268.3 kJ` The heat of formation of `HCI` will be

Options

A. `-22 kJ mol^(-1)`
B. `88 kJ mol^(-1)`
C. `-92.21 kJ mol^(-1)`
D. `-183.8 kJ mol^(-1)`

Explanation

`DeltaH^(Theta) reaction = 2Delta_(f)H^(Theta) underset((HF))" - 2Delta_(f)H^(Theta)underset((HCI))`
`-352.18 = 2 xx (-268.3)-2x`
`2x = 2 (-268.3) + 352.18`
`x =- 92.21 kJ mol^(-1)`