The following reaction is performed at 298 K ?
`2NO(g)+O_(2)(g)hArr2NO_(2)(g)`
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of `NO_(2)(g)` at 298 K ?
`(K_(p)=1.6xx10^(12))`

Options

A. `R ( 298) ln ( 6 xx 10^(12)) - 86600`
B. `86600+ R ( 298K) ln ( 1.6 xx 10^(12))`
C. `86600 -(ln ( 1.6 xx 10^(12)))/( R( 298))`
D. `0.5 [ 2 xx 86,600 - R( 298 )ln( 1.6 xx 10^(12)) ]`

Answer

d

Explanation

For the given reaction,
`DeltaG_(r)^(@) = (DeltaG_(f)^(@))_("Products")- ( DeltaG_(f)^(@)) _("Reactants")`
`= 2 DeltaG_(f)^(@) (NO_(2)) - 2DeltaG_(f)^(@)(NO){DeltaG_(f)^(@) (O_(2))=0}`
Also, `DeltaG_(f)^(@) = - RT lnK_(p)`
`:. 2[DeltaG_(f)^(@)( NO_(2)) - 2DeltaG_(f)^(@) (NO)]=- RTlnK_(p) `
`or ` 2[DeltaG_(f)^(@) (NO_(2))- 86600 J ] = -R ( 298 )ln ( 1.6 xx 10^(12))`
or `2 DeltaG_(f)^(@) (NO_(2))= [2 xx 86600 -R ( 298) ln ( 1.6 xx 10^(12))]`
or `DeltaG_(f)^(@) (NO_(2))= (1)/(2)[ 2 xx86600-R ( 298)ln ( 1.6 xx 10^(12))]`
`= 0.5 [ 2 xx 86600-R ( 298) ln ( 1.6 xx 10^(12))]`