Calculate the standard enegry change for the reaction:
`OF_(2)(g) +H_(2)O(g) rarr O_(2)(g) +2HF(g) at 298K`
The standard enthalpies of formation of `OF_(2)(g), H_(2)O(g)`, and `HF(g)` are `+20, -250`, and `-270 kJ mol^(-1)`, respectively.

Options

A. `-307.52kJ`
B. `315.25kJ`
C. `-320.12kJ`
D. `320.12kJ`

Answer

A

Explanation

`DeltaH_("Reaction")^(@)=sumH_("Products")^(@)-sumH_("Reactants")^(@)`
`=[2xxH_(HF)^(@)(g)+H_(O_(2))^(@)(g)]-[H_(OF_(2))^(@)(g)+H_(H_(2)O)^(@)(g)]`
`:.H^(@)` of free elements =0
`:.H_(O_(2))^(@)=0`
`:.DeltaH_(f)^(@)=H^(@)` i.e. standard heat of formation
= standard heat enthalpy of a compound
Also `DeltaH_(f)^(@)` for `F_(2)O` is +ve, `DeltaH_(f)^(@)` for `H_(2)O` is -ve, because heats of comhustion are exorthermic.
`DeltaH_(R)^(@)=[2xx(-270])-0=[20+(-250])-210kJ`
Now `DeltaH^(@)=DeltaU^(@)+DeltanRT`
`:.DeltaH_(R)^(@)=-310xx10^(3)J, Deltan=3-2=1`
`R=8.314J,T=398K`
`:.-310xx10^(3)=DeltaU^(@)+1xx8.314xx298`
`:.Delta U^(@)=-307522.4` joule `:.DeltaU^(@)=-307.52kJ`