Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass

\[\ce{C (s) + CO2 (g) ⇌ 2CO (g)}\]

Calculate K_c for this reaction at the above temperature.

Answer 1

Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO₂ = (100 - 90.55) = 9.45 g

Now, number of moles of CO, `"n"_("CO") = 90.55/28` =  3.234 mol

Number of moles of CO₂, `"n"_("CO"_2) = 9.45/44` = 0.215 mol

Partial 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

Partial pressure of CO₂,

`"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,

Δn = 2 – 1 = 1

We know that,

`"K"_"P" = "K"_"C"("RT")^(triangle"n")`

`=> 14.19 = "K"_"C"(0.082 xx 1127)^1`

= 0.154 (approximately)