InterviewSolution
Saved Bookmarks
InterviewSolution
| 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. `R(298) ln (1.6xx10^(12))-86600`B. `86600+R(298) ln (1.6xx10^(12))`C. `86600-(ln (1.6xx10^(12)))/(R(298))`D. `0.5[2xx86, 600-R(298) ln (1.6xx10^(12))` |
|
Answer» Correct Answer - D `DeltaG_(r)^(@)=Sigma DeltaG_(g)^(@)("Products")-Sigma DeltaG_(f)^(@)("reactants")=-RT ln K_(P)` `2DeltaG_(f(NO_(2)))^(@)-[2DeltaG_(f(NO))^(@)+DeltaG_(f(O_(2)))^(@)]` `=-RT ln K_(P)` `2 DeltaG_(f(NO_(2)))^(@)-[2xx86,600+0]` `=-RT ln K_(P)` `DeltaG_(f(NO_(2)))^(@)= 1/2[2xx86600-Rxx298xxln(1.6xx10^(12))]` |
|