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)In(1.6xx10^(12))-86600`
B. `86600+R(298)In(1.6xx10^(12))`
C. `86600-(In(1.6xx10^(12)))/(R(298))`
D. `0.5[2xx86600-R(298)In(1.6xx10^(12))]`

Answer

D

Explanation

`2NO(g) + O_2(g) hArr 2NO_2(g), DeltaG_r`
At equilibrium `DeltaG_r^@ = 0`
So `DeltaG_r = - RT ln K_p`
`DeltaG_r = 2G_f(NO_2)- 2G_f(NO)- G_f (O_2)`
`2G_f(NO_2) = 2G_f (NO) + G_f (O_2) + DeltaG_r`
`G_f (NO_2) = 0.5[2 xx 86600 - R(298) ln(1.6 xx10^12)]`