Questions: Express the reverse equilibrium constant for the following reaction. 5CH3(g) + 5/2 Cl2(g) <-> 5 CH3Cl(g) K = [CH3Cl]^5 / [CH3]^5 [Cl2]^(5/2)

Express the reverse equilibrium constant for the following reaction. 5CH3(g) + 5/2 Cl2(g) <-> 5 CH3Cl(g) K = [CH3Cl]^5 / [CH3]^5 [Cl2]^(5/2)
Transcript text: Part A Express the reverse equilibrium constant for the following reaction. 5CH3(g) + 5/2 Cl2(g) <-> 5 CH3Cl(g) K = [CH3Cl]^5 --------------- [CH3]^5 [Cl2]^(5/2)
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Solution

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Solution Steps

Step 1: Understanding the Reverse Reaction

To find the reverse equilibrium constant, we need to consider the reverse reaction: \[ 5 \text{CH}_3\text{Cl}(g) \leftrightarrow 5 \text{CH}_3(g) + \frac{5}{2} \text{Cl}_2(g) \]

Step 2: Writing the Equilibrium Expression for the Reverse Reaction

The equilibrium constant for the reverse reaction, \( K_{\text{reverse}} \), is the reciprocal of the equilibrium constant for the forward reaction, \( K \).

Step 3: Expressing the Reverse Equilibrium Constant

Given the equilibrium constant for the forward reaction: \[ K = \frac{[\text{CH}_3\text{Cl}]^5}{[\text{CH}_3]^5 [\text{Cl}_2]^{5/2}} \]

The reverse equilibrium constant is: \[ K_{\text{reverse}} = \frac{1}{K} = \frac{[\text{CH}_3]^5 [\text{Cl}_2]^{5/2}}{[\text{CH}_3\text{Cl}]^5} \]

Final Answer

\[ \boxed{K_{\text{reverse}} = \frac{[\text{CH}_3]^5 [\text{Cl}_2]^{5/2}}{[\text{CH}_3\text{Cl}]^5}} \]

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