
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

When an unsaturated alcohol $\mathbf{A}$ is restored on a "poisoned" catalyst, the equilibrium constant $K_{c}$ is given by:

$$
K_{c}=\frac{[\Gamma]^{3}}{[Б]^{3}}=\frac{(0.25 \text{ moles/L})^{3}}{(0.15 \text{ moles/L})^{3}} \cong 4.63
$$

The initial concentration of the sole gaseous reactant $\text{B}$ is the sum of its current equilibrium concentration $[\text{B}] = 0.15 \text{ moles/L}$ and the amount consumed during the reaction. As indicated by the reaction, substances B and Г are in a 1:1 ratio, so the amount of B consumed is equal to the equilibrium concentration of Г, which is $[\Gamma] = 0.25 \text{ moles/L}$. Therefore, the initial concentration of $\text{B}$ is $[\text{B}]_{\text{initial}} = 0.15 + 0.25 = 0.4 \text{ moles/L}$. At the start, the gaseous substance Г was absent, so $[\Gamma]_{\text{initial}} = 0 \text{ moles/L}$.

Present the answer in LaTex format: \boxed{Your answer}
user