The problem in your image relates to calculating the overall equilibrium constant $K$ for the combined reactions given their individual equilibrium constants $K_1$ and $K_2$.

The reactions provided are:

1. $2C(s) + O_2(g) \rightleftharpoons 2CO(g)$ with equilibrium constant $K_1$,
2. $\text{Fe}_2O_3(s) + 3CO(g) \rightleftharpoons 2Fe(s) + 3CO_2(g)$ with equilibrium constant $K_2$.

The net reaction is:

$2 \text{Fe}_2O_3(s) + 6C(s) + 3O_2(g) \rightleftharpoons 4Fe(s) + 6CO_2(g)$

To find the overall equilibrium constant $K$, we need to combine the constants $K_1$ and $K_2$. Since the second reaction involves 3 moles of CO, we will need to cube the equilibrium constant $K_1$ to account for that.

Here’s how the combination works:

• The first reaction is doubled (because we need 6 moles of CO), so the new equilibrium constant becomes $K_1^3$ (since the equilibrium constant is raised to the power of the stoichiometric coefficient).
• The second reaction remains unchanged, with equilibrium constant $K_2$.

Now, we multiply the equilibrium constants to get the overall equilibrium constant:

$K = K_1^3 \times K_2$

So, the final expression for $K$ is:

$K = K_1^3 \times K_2$

Would you like more details on how this combination works or have any further questions?

Here are 5 related questions to further expand your understanding:

1. What is the physical significance of equilibrium constants in chemical reactions?
2. How do changes in reaction conditions affect the equilibrium constant?
3. Why is it necessary to cube $K_1$ in this reaction, and what does it represent?
4. Can equilibrium constants be used to predict the direction of a reaction?
5. How are equilibrium constants related to Gibbs free energy in a reaction?

Tip: Remember, when combining equilibrium constants from sequential reactions, the overall constant is the product of the individual constants, adjusting for any changes in the stoichiometry.