Question

Hydrogen and carbon dioxide gases are mixed in equal molar amounts at 800 K. A reversible
reaction takes place.At equilibrium, the partial pressures of H2 and CO2 are both 10.0 kPa. Kp is 0.288 at 800 K.
What is the partial pressure of CO in the equilibrium mixture?
the equation is H2(g) + CO2(g)-------)H2O(g) + CO(g) the reaction is reversible.

Answer 1

52977 Chemistry, General Chemistry

Hydrogen and carbon dioxide gases are mixed in equal molar amounts at 800 K. A reversible reaction takes place. At equilibrium, the partial pressures of H₂ and CO₂ are both 10.0 kPa. Kp is 0.288 at 800 K.

What is the partial pressure of CO in the equilibrium mixture?

the equation is H₂(g) + CO₂(g) = H₂O(g) + CO(g) the reaction is reversible.

Answer:

$$\mathrm{H}_2(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}) \rightleftharpoons \mathrm{H}_2\mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g})$$ $$\mathrm{H}_2(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}) \rightleftharpoons \mathrm{H}_2\mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g})$$ $$\begin{array}{l} \text{y mol ... y mol} \quad \text{...} \quad 0 \quad \text{...} \quad \text{initial} \\ - x \quad \text{...} \quad - x \quad \text{...} \quad + x \quad \text{...} \quad + x \quad \text{...} \quad \text{change} \\ y - x \quad \text{...} \quad y - x \quad \text{...} \quad x \quad \text{...} \quad x \quad \text{...} \quad \text{equilibrium} \\ \end{array}$$ $$K_p = P(\mathrm{H}_2\mathrm{O}) P(\mathrm{CO}) / (P(\mathrm{H}_2) P(\mathrm{CO}_2))$$ $$K_p = x^2 / (y - x)^2$$ $$V(K_p) = x / (y - x)$$ $$0.288 = x^2 / (10 - x) \quad x^2 = 0.288 \cdot (10 - x) \quad x^2 = 2.88 - 0.288x \quad x^2 + 0.288x - 2.88 = 0 \quad x = 1.56$$

Therefore, the partial pressure of CO in the equilibrium mixture is 1.56 kPa.

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