Questions: The equilibrium constant, K₀⁺ for the following reaction is 9.52 x 10⁻² at 350 K: CH₄(g) + CCl₄(g) ⇌ 2 CH₂Cl₂(g) [CH₄]=□ M [CCl₄]=□ M [CH₂Cl₂]=□ M

Transcript text: The equilibrium constant, $\mathrm{K}_{0}^{+}$for the following reaction is $9.52 \times 10^{-2}$ at 350 K : \[ \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{CCl}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \] \[ \begin{array}{l} {\left[\mathrm{CH}_{4}\right]=\square \mathrm{M}} \\ {\left[\mathrm{CCl}_{4}\right]=\square \mathrm{M}} \\ {\left[\mathrm{CH}_{2} \mathrm{Cl}_{2}\right]=\square \mathrm{M}} \end{array} \]

Solution

Solution Steps

Step 1: Understanding the Equilibrium Constant Expression

The equilibrium constant expression for the given reaction is: \[ \mathrm{CH}_{4}(\mathrm{~g}) + \mathrm{CCl}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \]

The equilibrium constant $ K_0^+ $ is given by: \[ K_0^+ = \frac{[\mathrm{CH}_2\mathrm{Cl}_2]^2}{[\mathrm{CH}_4][\mathrm{CCl}_4]} \]

Step 2: Substituting the Given Values

We are given: \[ K_0^+ = 9.52 \times 10^{-2} \]

We need to find the concentrations of $\mathrm{CH}_4$, $\mathrm{CCl}_4$, and $\mathrm{CH}_2\mathrm{Cl}_2$ that satisfy this equilibrium constant.

Step 3: Solving for Concentrations

Since the problem does not provide specific concentrations, we can express the concentrations in terms of variables. Let: \[ [\mathrm{CH}_4] = a \quad \text{M} \] \[ [\mathrm{CCl}_4] = b \quad \text{M} \] \[ [\mathrm{CH}_2\mathrm{Cl}_2] = c \quad \text{M} \]

Then, the equilibrium constant expression becomes: \[ 9.52 \times 10^{-2} = \frac{c^2}{ab} \]

Final Answer

The equilibrium constant expression in terms of the concentrations is: \[ \boxed{9.52 \times 10^{-2} = \frac{[\mathrm{CH}_2\mathrm{Cl}_2]^2}{[\mathrm{CH}_4][\mathrm{CCl}_4]}} \]

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