The following reaction is performed at 298 K.
`2NO(g)+O_(2)(g) hArr 2NO_(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. `86600-("In" (1.6xx10^(12)))/(R(298))`
B. `0.5[2xx86,600-R(298) " In " (1.6xx10^(12))]`
C. R (298) In `(1.6xx10^(12))-86600`
D. 86600+R(298) In `(1.6xx10^(12))`

Answer

B

Explanation

`Delta G_(NO(g))^(@)=86.6kJ//"mol"=86600 J//"mol"`
`G_(NO_(2)(g))^(@)=x J//"mol"`
`T=298, K_(P)= 1.6xx10^(12)`
`DeltaG^(@)= -RT " In " K_(P)`
Given equation,
`2NO(g)+O_(2)(g) hArr 2NO_(2)(g)`
`therefore 2DeltaG_(NO2)^(@)-2Delta G_(NO)^(@)= - R(298) " In" (1.6xx10^(12))`
`2Delta G_(NO2)^(@)-2xx86600= - R (298)"In" (1.6xx10^(12))`
`2Delta G_(NO2)^(@)=2xx86600-R(298) "IN" (1.6xx10^(12))`
`DeltaG_(NO2)^(@)=(1)/(2)[2xx86600-R(298) " In" (1.6xx10^(12))J`
`=0.5[2xx86600-R(298)"In" (1.6xx10^(12))`