Saved Bookmarks
| 1. |
Interpret the following Ellingham diagram. . |
|
Answer» (1) `Si_((s)) + O_(2(g)) rarr 2 Si O_(2 (s))` As temperature increases, `Delta_(f) G^(Ө)` becomes less negative , hence `SiO_(2(s))` becomes less stable, i.e., at `500 K, SiO_(2)` is more stable than `2500 K`. (2) `2Mg_((s)) + O_(2(g)) rarr 2MgO_((s))` Abrupt changes in the plot at `~ 800 K` and `~ 1400 K` denote phase change//transition from `s rarr 1` and `1 rarr g`, respectively and represent melting point `(~800 K)` and boiling point `(~ 1400 K)`. `MgO_((s))` becomes less stable with increase in temperature. At `T_(1)`, lines of `(1)` and `(2)` intersect. One can conclude the following : (a) At `T_(1)`, both `MgO` and `SiO_(2)` are equally stable. (b) At temperature `gt T_(1), MgO` is less stable than `SiO_(2)` Hence, `Si` will reduce `MgO` to `Mg`. `2MgO + Si rarr SiO_(2) + 2Mg` ( c) At temperature `lt T_(1), SiO_(2)` is less stable than `MgO`. Hence, `Mg` will reduce `SiO_(2)` to `Si` `SiO_(2) + 2Mg rarr 2MgO + Si` Relative stability of `MgO` and `SiO_(2)` at given temperature can be predicted with the help of Ellingham diagram. (3) `2Zn_((s)) + O_(2(g)) rarr 2ZnO_((s))` `Delta_(f) G^(Ө) (ZnO) gt Delta_(f) G^(Ө) (MgO)` `Delta_(f) G^(Ө) (ZnO) gt Delta_(f)^(Ө) (SiO_(2))` Hence, both `Mg_((s))` and `Si_((s))` can be used as reducing agent for reducing `ZnO_((s))` to `Zn_((s))`. |
|