Calculate the value of the equilibrium constant, Kc, for the reaction Q(g) + X(g) = 2 M(g) + N(g) given that M(g) = Z(g) 6R(g) = 2 N(g) + 4 Z(g) 3X(g) + 3 Q(g) = 9 R(g) Kc1 = 3.00 Kc2 = 0.562 Kc3 = 11.1 Kc =

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.

Calculate the value of the equilibrium constant, Kc, for the reaction Q(g) + X(g) = 2 M(g) + N(g) given that M(g) = Z(g), 6R(g) = 2 N(g) + 4 Z(g), 3X(g) + 3 Q(g) = 9 R(g), Kc1 = 3.00, Kc2 = 0.562, Kc3 = 11.1.

Learn how to calculate the equilibrium constant for the reaction Q(g) + X(g) = 2 M(g) + N(g) using the known constants Kc1 = 3.00, Kc2 = 0.562, and Kc3 = 11.1. This problem explains reaction manipulation and equilibrium constant rules, suitable for advanced high school or undergraduate chemistry students.

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