InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The enthalpy of formation of hypothetical `MgCI` is `-125kJ mol^(-1)` and for `MgCI_(2)` is `-642 kJ mol^(-1)`. What is the enthalpy of the disproportionation of `MgCI`.A. `392 kJ mol^(-1)`B. `-392 kJ mol^(-1)`C. `-767 kJ mol^(-1)`D. `-517 kJ mol^(-1)` |
|
Answer» `{:(i.,Mg(s)+CI_(2)(g)rarrMgCI_(2)(s),,DeltaH_(1)=-642),(ii.,Mg(s)+(1)/(2)CI(g)rarrMgCI(s),,DeltaH_(2)=-125),(iii.,2MgCIrarrMgCI_(2)+Mg,,DeltaH =?):}` `DeltaH = DeltaH_(1) - 2DeltaH_(2)` `=- 642 -2 xx 2 xx (-125) =- 392 kJ mol^(-1)` |
|