Question

Carbon dioxide reacts with hot carbon in the form of graphite. The equilibrium constant, , for the reaction is 10.0 at 850.°C.
CO2 (g) + C(graphite) -----> 2CO (g)
If 13.5 g of carbon monoxide is placed in a 2.50-L graphite container and heated to 850.°C, what is the mass of carbon dioxide at equilibrium?

Answer 1

Question #60984 – Chemistry – General Chemistry

Question:

Carbon dioxide reacts with hot carbon in the form of graphite. The equilibrium constant, for the reaction is 10.0 at $850^{\circ}\mathrm{C}$. CO2 (g) + C(graphite) ---> 2CO (g)

If $13.5\,\mathrm{g}$ of carbon monoxide is placed in a 2.50-L graphite container and heated to $850^{\circ}\mathrm{C}$, what is the mass of carbon dioxide at equilibrium?

Solution:

$$\mathrm{Kp} = (\mathrm{P}_{\mathrm{CO}})^2 / \mathrm{P}_{\mathrm{CO}_2}$$ $$\mathrm{n}(\mathrm{CO}) = 13.5\,\mathrm{g} / 28\,\mathrm{g/mol} = 0.482\,\mathrm{mol}$$

In equilibrium $\mathrm{n}(\mathrm{CO}) = 0.482 - \mathrm{x}$, $\mathrm{n}(\mathrm{CO}_2) = \mathrm{x}$.

Ideal gas law:

$$\mathrm{PV} = \mathrm{nRT}$$ $$\mathrm{P}(\mathrm{CO}) = \mathrm{n}(\mathrm{CO}) * \mathrm{RT/V}$$ $$\mathrm{Kp} = (\mathrm{RT/V}) * (0.482 - \mathrm{x})^2 / \mathrm{x} = (0.082 * (850 + 273) / 2.5) * (0.482 - \mathrm{x})^2 / \mathrm{x} = 10\,\mathrm{barr}$$ $$\mathrm{x} = 0.231\,\mathrm{mol}$$ $$\mathrm{m}(\mathrm{CO}_2) = 0.231\,\mathrm{mol} * 44\,\mathrm{g/mol} = 10.2\,\mathrm{g}$$

Answer: $\mathrm{m}(\mathrm{CO}_2) = 10.2\,\mathrm{g}$

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