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Characteristics of Equilibrium Constant

At 1127 K a...

Question

At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $C O$ and $C O_{2}$ in equilibrium with solid carbon has $90.55$ % $C O$ by mass
$C (s) + C O_{2} (g) ⇋ 2 C O (g)$
Calculate $K_{c}$ for this reaction at the above temperature.

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Solution

Let weight of mixture is $100 g$ , then $90.55 g$ of $C O$ & $9.45 g$ of $C O_{2}$ is present.
$\Rightarrow \frac{90.55}{28} = 3.324$ moles of $C O$ & $\frac{9.45}{44} = 0.215$ moles of $C O_{2}$ are present.
Mole fraction of $C O (x_{C O}) = \frac{x_{C O}}{x_{C O_{2}}}$
$\Rightarrow \frac{3.234}{3.234 + 0.215} = 0.938$ .
Mole fraction of $C O_{2} = (1 - X_{C O}) = 1 - 0.938 = 0.062$
Now, partial pressure of $C O = 0.938 \times 1 a t m = 0.938 a t m$
Similarly partial pressure of $C O_{2} = 0.062 \times 1 a t m = 0.062 a t m$
Now, $C (s) + C O_{2} (g) ⟶ 2 C O (g)$
$K_{P} = \frac{P_{[C O]}^{2}}{P_{[C O_{2}]^{1}}} = \frac{(0.938)^{2}}{0.062} = 14.19$
$Δ n_{g} = 2 - 1 = 1$
Now, $K_{P} = K_{C} (R T)^{Δ n_{g}}$
$K_{C} = \frac{K_{P}}{(R T)^{Δ n_{g}}}$
$\Rightarrow \frac{14.19}{(0.0821 \times 1127)^{1}} = 0.1536$
$\Rightarrow K_{C} ≃ 0.153$

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At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by massC(s)+CO2(g)⇋2CO(g)Calculate Kc for this reaction at the above temperature.

At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $C O$ and $C O_{2}$ in equilibrium with solid carbon has $90.55$ % $C O$ by mass
$C (s) + C O_{2} (g) ⇋ 2 C O (g)$
Calculate $K_{c}$ for this reaction at the above temperature.