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)(g) 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

A

Explanation

For, `M_(2)CO_(3)(s) rarr M_(2)O(s)+CO_(2)(g)`
`:' CO_(2)` is `1%` in air,
`P'_(CO_(2))=1/100xxP_("air")=1/100xx1 "atm"`
Also, for equilibrium `K_(P)=P'_(CO_(2))=0.0095 "atm"`
`:'` Given, `P'_(CO_(2))=0.01 "atm"`
Since decomposition is carried out in pressure 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 the reaction is desired, `P'_(CO_(2))` must be lesser than `0.0095` atm as `P'_(CO_(2))` at equilibrium connot be more than `0.0095` atm.
Alternate method:
For `M_(2)CO_(3)(s) hArr M_(2)O(s)+CO_(2)(g)`
`P_(CO_(2))=(1/100+P)`
`:' 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
`:' K_(P)=1/100+P`
or, `0.0095=P+0.01`
or, `P=-0.0005`
The value of pressure comes negative and thus, it may be concluded that `M_(2)CO_(3)` will not dissociate in pressure of `CO_(2)` of pressure `0.01` atm.