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Industrial Preparation of Ammonia

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 $90.55 %$ of $C O$ by mass.
$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$
Calculate $K_{c}$ for the reaction at the above temperature.

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Solution

Let the mixture has 100g as total mass.

The masses of CO and $C O_{2}$ are $90.55$ g and $100 - 90.55 = 9.45$ g respectively.

The number of moles of CO are $\frac{90.55}{28} = 3.234$ .

The number of moles of $C O_{2}$ are $\frac{9.45}{44} = 0.215$ .

The mole fraction of CO is $\frac{3.234}{3.234 + 0.215} = 0.938$ .

The mole fraction of $C O_{2}$ is $1 - 0.938 = 0.062$ .

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is $0.938 \times 1 = 0.938$ atm.

The partial pressure of carbon dioxide is $0.062 \times 1 = 0.042$ atm.

The expression for the equilibrium constant is:

$K_{p} = \frac{P_{C O}^{2}}{P_{C O_{2}}} = \frac{(0.938)^{2}}{0.062} = 14.19$

$Δ n_{g} = 2 - 1 = 1$

$K_{c} = K_{p} (R T)^{- Δ n} = 14.19 \times (0.0821 \times 1127)^{- 1} = 0.153$ .

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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% of CO by mass.C(s)+CO2(g)⇌2CO(g)Calculate Kc for the 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 $90.55 %$ of $C O$ by mass.
$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$
Calculate $K_{c}$ for the reaction at the above temperature.