Saved Bookmarks
| 1. |
The equilibrium constant of the reaction : `Zn(s)+2Ag^(+)(aq)toZn(aq)+2Ag(s),E^(@)=1.50V` at 298 K isA. `2.6xx10^(49)`B. `8.7xx10^(51)`C. `6.1xx10^(30)`D. `6.6xx10^(50)` |
|
Answer» Correct Answer - D `nFE^(@)=2.303RT"log" K` `"log"K=(nFE^(@))/(2.303RT)=(2xx96500)/(2.303xx8.3xx298)` `K=6.6xx10^(50)` |
|