High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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.

Other Questions