Question

The standed free energy of fromation of NO(g) is 86.6 kj/ mol at 298 K what is the standed free energy of fromation of ` NO_(2) g` at 298 k? `K _(p)= 1.6 xx 10^(12)`
A. `0.5[2xx86.600-R(298)ln(1.6xx10^(12))`
B. `R(298)ln(1.6xx10^(12))-86.600`
C. `86.600+r(298)ln(1.6xx10^(12))`
D. `86.600-(ln(1.6xx10^(12)))/(R(298))`

Answer 1

Correct Answer - A
`DeltaG=DeltaG^@+RT" ln "K_(p)` . . (i)
`DeltaG=0 and DeltaG^(@)=2DeltaG_(f)^(@)NO_(2)-2DeltaG_(f)^(@)NO-2DeltaG_(f)^(@)O_(2)`
`2DeltaG_(f)^(@)NO_(2)=DeltaG^(@)+2DeltaG_(f)^(@)NO` (Since `DeltaG_(f)^(@)O_(2)=0`)
`=[2xx86,600-RT" ln "1.6xx10^(12)]` ltbRgt `DeltaG_(f)^(@)NO_(2)=0.5[2xx86.600-RT" ln "1.6xx10^(-12)]`