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))`

The standed free energy of fromation of NO(g) is 86.6 kj/ mol at 298 K

1.	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» 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)]`

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