Would 1% of `CO_2` in the air be sufficient to prevent any loss in weight when `Ag_2CO_3` is dried at `120^@C? Ag_2CO_3 (s) Leftrightarrow Ag_2O (s)+CO_(2) (g), k_(p)=0.0095" atm at "120^@C?` How low would the partial pressure of `CO_2` have to be to promote this reaction at `120^@C?`

Explanation

`K_(p)=(p_(Ag_(2)O) xx p_(CO_(2)))/(p_(Ag_(2)CO_(3))=p_(CO_(2))`
`p_(CO_(2))=0.0095" atm =constant at 120"^@C`
Thus, if `Ag_2CO_3` is taken in a closed container, a small amount of it would decompose to give `CO_2` gas until the partial pressure of `CO_2` reaches 0.0095 atm. As this is the equilibrium pressure of `CO_2` the decomposition would then stop.
Now since partial pressure of `CO_2` in air is 0.01 atm `(CO_2" is 1% in air")` which is much greater than 0-0095 atm, the equilibrium would practically shift to the left-hand side completely, or in other words, there would be no loss in weight of `Ag_2CO_2` (by decomposition) if placed in air containing 1% `CO_2`
Further, if the partial pressure of `CO_2` in air is less than the equilibrium pressure of 0.0095 atm, the decomposition of `Ag_2CO_2` would continue till the cor pressure around `Ag_2CO_2` becomes 0.0095 atm.