[VISUAL] Questions 14-17 refer to the following information. Substance | Lewis Diagram | Boiling Poi — Equilibrium Chemistry Question

STEPS:

1. Identify the chemical equation representing the physical process:
The evaporation (vaporization) of liquid methanol to form methanol gas is represented by the phase equilibrium:
$$\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)$$

2. Write the equilibrium constant expression ($K_p$) for this process:
* The equilibrium constant in terms of partial pressures, $K_p$, is calculated using the partial pressures of gaseous products divided by gaseous reactants.
* Since pure liquids and solids have an activity of 1, they are omitted from the equilibrium expression.
* Therefore, the expression for this process simplifies to:
$$K_p = P_{\text{CH}_3\text{OH}}$$
where $P_{\text{CH}_3\text{OH}}$ is the equilibrium partial pressure of methanol vapor.

3. Read the equilibrium vapor pressure from the graph:
* Initially, the container is completely evacuated, so the initial pressure of methanol gas is $0\text{ atm}$.
* As liquid methanol evaporates, the pressure in the vessel rises due to the accumulation of $\text{CH}_3\text{OH}(g)$ molecules in the gas phase.
* Once dynamic equilibrium is achieved (where the rate of evaporation equals the rate of condensation), the pressure stabilizes and reaches a constant plateau.
* According to the provided graph, the pressure levels off at exactly $0.196\text{ atm}$.

4. Calculate $K_p$:
* Substitute the equilibrium partial pressure of the gas into the $K_p$ expression:
$$K_p = P_{\text{CH}_3\text{OH}} = 0.196$$
* Comparing this value to the multiple-choice options, $0.196$ is closest to $0.2$, which identifies Option B as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (0.04): A student might arrive at this value by squaring the equilibrium pressure ($0.196^2 \approx 0.038 \approx 0.04$), perhaps due to a misconception about the stoichiometry or by mistakenly applying a quadratic equilibrium calculation.
• Option C is incorrect (5): This value represents the reciprocal of the equilibrium vapor pressure ($1 / 0.196 \approx 5.1$). A student would calculate this if they incorrectly placed reactants over products in their $K_p$ expression, treating the liquid as a product in the numerator.
• Option D is incorrect (30): This is a distractor that corresponds to the time on the x-axis (approx. $30\text{ seconds}$) at which the system first reaches dynamic equilibrium. It confuses time (a kinetic factor) with the equilibrium constant (a thermodynamic factor).