To calculate the equilibrium constant $K_c$ for the reaction $\text{Q(g)} + \text{X(g)} = 2 \text{M(g)} + \text{N(g)}$, we need to relate it to the other reactions and their equilibrium constants, $K_{c1}$, $K_{c2}$, and $K_{c3}$.

Given reactions:

1. $\text{M(g)} = \text{Z(g)}$, with $K_{c1} = 3.00$
2. $6 \text{R(g)} = 2 \text{N(g)} + 4 \text{Z(g)}$, with $K_{c2} = 0.562$
3. $3 \text{X(g)} + 3 \text{Q(g)} = 9 \text{R(g)}$, with $K_{c3} = 11.1$

We are tasked with finding $K_c$ for the reaction: $\text{Q(g)} + \text{X(g)} = 2 \text{M(g)} + \text{N(g)}$

Strategy:

To solve this, we'll manipulate the given reactions and their equilibrium constants to match the desired reaction.

1. For the reaction $\text{M(g)} = \text{Z(g)}$, rearranging gives $\text{M(g)}$ in terms of $\text{Z(g)}$. This will help us express $\text{M(g)}$ in terms of known quantities. $K_{c1} = 3.00$

2. For the reaction $6 \text{R(g)} = 2 \text{N(g)} + 4 \text{Z(g)}$, we'll divide the entire reaction by 2 to match the coefficients of $\text{N(g)}$ and $\text{Z(g)}$ from the target reaction. $\frac{1}{2} \left( 6 \text{R(g)} = 2 \text{N(g)} + 4 \text{Z(g)} \right) \quad \Rightarrow \quad 3 \text{R(g)} = \text{N(g)} + 2 \text{Z(g)}$ The new equilibrium constant for this halved reaction becomes: $K_{c2}' = \sqrt{K_{c2}} = \sqrt{0.562} = 0.749$

3. For the reaction $3 \text{X(g)} + 3 \text{Q(g)} = 9 \text{R(g)}$, we'll divide this by 3 to match the coefficients of $\text{X(g)}$ and $\text{Q(g)}$ from the target reaction. $\frac{1}{3} \left( 3 \text{X(g)} + 3 \text{Q(g)} = 9 \text{R(g)} \right) \quad \Rightarrow \quad \text{X(g)} + \text{Q(g)} = 3 \text{R(g)}$ The new equilibrium constant for this reduced reaction becomes: [ K_{c3}' = \sqrt[3]{K_{c3}} = \sqrt[3]{11.1} = 2.23 ]

Combining reactions:

Now, we'll combine the three modified reactions to match the target reaction $\text{Q(g)} + \text{X(g)} = 2 \text{M(g)} + \text{N(g)}$.

• Reaction 1 gives us $\text{M(g)} = \text{Z(g)}$ with $K_{c1} = 3.00$.
• Reaction 2 gives us $3 \text{R(g)} = \text{N(g)} + 2 \text{Z(g)}$ with $K_{c2}' = 0.749$.
• Reaction 3 gives us $\text{X(g)} + \text{Q(g)} = 3 \text{R(g)}$ with $K_{c3}' = 2.23$.

Multiplying the equilibrium constants of the combined reactions: $K_c = K_{c1} \times K_{c2}' \times K_{c3}' = 3.00 \times 0.749 \times 2.23 = 5.00$

Thus, the equilibrium constant $K_c$ for the reaction $\text{Q(g)} + \text{X(g)} = 2 \text{M(g)} + \text{N(g)}$ is $\boxed{5.00}$.

Do you have any further questions or want more details on any step?

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Further exploration:

1. What is the relationship between $K_p$ and $K_c$ for gas-phase reactions?
2. How do changes in pressure affect the equilibrium constant of a reaction?
3. Can equilibrium constants be negative? Why or why not?
4. How does temperature influence the value of an equilibrium constant?
5. How is the equilibrium constant expressed for heterogeneous reactions?

Tip: Always check the stoichiometric coefficients carefully when manipulating reactions for calculating equilibrium constants.