High School

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------------------------------------------------ Calculate the value of the equilibrium constant (K) for the reaction:
CO (g) + H₂O (g) ⇌ CO₂ (g) + H₂ (g)

Given the following amounts at equilibrium in an 8.00 L container at 690°C:
- 1.60 moles CO
- 1.60 moles H₂O
- 4.00 moles CO₂
- 4.00 moles H₂

Answer :

To calculate the equilibrium constant [tex]K[/tex] for the reaction [tex]\text{CO (g)} + \text{H}_2\text{O (g)} \rightleftharpoons \text{CO}_2\text{ (g)} + \text{H}_2\text{ (g)}[/tex], we will use the expression for the equilibrium constant:

[tex]K = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}[/tex]

Step-by-step Calculation:


  1. Calculate the Concentrations of Each Species:

    The concentration of a gas species [tex]A[/tex] in moles per liter [tex][A][/tex] can be found using the formula:

    [tex][A] = \frac{\text{moles of A}}{\text{volume of the container (in liters)}}[/tex]

    Given that the volume of the container is 8.00 L, we can calculate:


    • [tex][\text{CO}] = \frac{1.60 \text{ moles}}{8.00 \text{ L}} = 0.20 \text{ M}[/tex]

    • [tex][\text{H}_2\text{O}] = \frac{1.60 \text{ moles}}{8.00 \text{ L}} = 0.20 \text{ M}[/tex]

    • [tex][\text{CO}_2] = \frac{4.00 \text{ moles}}{8.00 \text{ L}} = 0.50 \text{ M}[/tex]

    • [tex][\text{H}_2] = \frac{4.00 \text{ moles}}{8.00 \text{ L}} = 0.50 \text{ M}[/tex]



  2. Substitute the Concentrations into the Equilibrium Expression:

    Now, substitute the concentrations into the equilibrium constant expression:

    [tex]K = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(0.50)(0.50)}{(0.20)(0.20)}[/tex]


  3. Calculate the Value of [tex]K[/tex]:

    Calculate the numerical value:

    [tex]K = \frac{0.25}{0.04} = 6.25[/tex]



Conclusion:

The equilibrium constant [tex]K[/tex] for this reaction at 690°C is 6.25. This value indicates how the concentrations of the reactants and products compare at equilibrium. A larger [tex]K[/tex] value suggests that the products are favored at equilibrium.