At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass: `C_((s))+CO_(2(s))hArr2CO_((g))` Calculate `K_(c)` for the reaction at the above temperature.

At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2

1.	At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass: `C_((s))+CO_(2(s))hArr2CO_((g))` Calculate `K_(c)` for the reaction at the above temperature.
Answer» `{:(,,C_((s)),+,CO_(2(g)),hArr,2CO_((g))),(,"At equilibrium",9.45g ,,90.55g,,),(,,=(9.45)/(44)"mole",,=(90.55)/(28)"mole",,),(,,=0.21 "mole",,=3.23 "mole",,):}` `K_(p)=((n_(CO))^(2))/(n_(CO_(2)))xx[(P)/(sumn)]^(1)` `=(3.23xx3.23)/(0.21)xx(1)/(3.44)` Now, `K_(c)=(K_(p))/((RT)^(Deltan))=(14.44)/([0.0821xx1127]^(1))` `=0.156 mol litre^(-1)`

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At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass: `C_((s))+CO_(2(s))hArr2CO_((g))` Calculate `K_(c)` for the reaction at the above temperature.

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