Would `1% CO_(2)` in air be sufficient to prevent any loss in weight when `M_(2)CO_(3)` is heated at `120^(@)C`?
`M_(2)CO_(3)(s) hArr M_(2)O(s)+CO_(2)(g)`
`K_(p)=0.0095` atm at `120^(@)C`. How long would the partial pressure of `CO_(2)` have to be to promote this reaction at `120^(@)C`?

Answer

(No reactions)

Explanation

For, `M_(2)CO_(3)(g)(S) harr M_(2)O(S) + CO_(2)(g)`
`because CO_(2) "is" 1%` in air: `therefore P'_(CO_(2))=(1)/(100)xxP_(air)=(1)/(100)xx1 "atm".=0.01 "atm".`
Also for equilibrium `K_(P)=P'_(CO_(2))=0.0095` atm.
`because "Given", P'_(CO_(2))=0.01` atm.
Since decomposition is carried out in presence of `P'_(CO_(2))` of `0.01` atm and `K_(P)=0.0095` atm, thus, practically no decomposition of `M_(2)CO_(3)`. Thus `1%CO_(2)` is sufficient to prevent any loss in weight.
If at all reacion is desired the `P'_(CO_(2))` must be lesser than `0.0098` atm as `P'_(CO_(2))` at equlibrium cannot be more than `0.0095` atm.
Altrenate solution:
For `M_(2)CO_(3)(S) harr M_(2)O(S) + CO_(2)(S) therefore K_(P)=(1)/(100)+P`
`because K_(P)=P'_(CO_(2)` and the pressure of `CO_(2)` already present in `1//100` atm. Let the decomposition of `M_(2)CO_(3)` produces the `CO_(2)` of pressure P, then
`therefore K_(P)=therefore K_(P)=(1)/(100)+P`
or `0.0095=P+001 "or" P=-0.0005`.
The value of pressure comes negative and thus, it may be concluded that `M_(2)CO_(2)` will not dissociate in pressure of `CO_(2)` of pressure `0.01` atm.