Question

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(g))hArr2CO_((g))`
Calculate `K_(c)` for the reaction at the above temperature.

Answer 1

Correct Answer - `0.149`
Let the total mass of the gaseous mixture be 100 g.
Mass of `CO = 90.55 g`
And,
mass of `CO_(2) = (100 - 90.55) = 9.45` g
Now, number of moles of CO,
`n_(CO) = (90.55)/(28) = 3.234` mol
Number of moles of `CO_(2)` , `n_(CO_(2)) = (9.45)/(44) = 0.215` mol
Particle pressure of CO,
`p_(CO) = (n_(CO))/(n_(CO) + n_(CO_(2))) xx p_("total")`
`= (3.234)/(3.234 + 0.215) xx 1`
`= 0.938` atm
Particle pressure of `CO_(2)`,
`p_(CO) = (n_(CO))/(n_(CO) + n_(CO_(2))) xx p_("total")`
`= (3.234)/(3.234 + 0.215) xx 1`
`= 0.938` atm
Partial pressure of `CO_(2)`,
`p_(CO_(2)) = (n_(CO_(2)))/(n_(CO) + n_(CO_(2))) xx p_("total")`
`= (0.215)/(3.234 + 0.215) xx 1`
`= 0.062` atm
Therefore, `K_(P) = ([CO]^(2))/([CO_(2)])`
`= ((0.938)^(2))/(0.062)`
`= 14.19`
For the given reaction,
`Deltan = 2 - 1 = 1`
We know that,
`K_(P) = K_(C) (RT)^(Deltan)`
`rArr 14.19 = K_(C) (0.082xx 1127)^(1)`
`rArr K_(C) = (14.19)/(0.082 xx 1127)`
`= 0.154` (approximately)