Consider the evaporation of methanol at 25.0 °C : CH 3 OH(l) → CH 3 OH(g) c. Explain why methanol spontaneously evaporates in open air at 25.0 °C

Hello. Everyone in this video we're talking about Napoli. So it says to describe why naturally spontaneously sublime at open air at 25°C. Let's recognize that 25°C, that's going to be room temperature and Naftali supply room temperature because there's an equilibrium between the solid and the gas face. So you can stay at room temperature. There's going to be an equilibrium between the solid and the gas face. But it still favors that solid face in equilibrium. And the reason for that is just that the equilibrium concentration in the room is never maintained because it's not a closed system. So there's always going to be an in fluctuation. And because of that, our answer is that the net feeling supply of the room temperature because it's an equilibrium between the solid and the gas phase. And that's going to be answer eight. Let's go ahead and actually highlight that. So a is going to be my final answer for this problem. Thank you all so much for watching

Ch.19 - Free Energy & Thermodynamics

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Tro 5th Edition

Ch.19 - Free Energy & Thermodynamics

Problem 70c

Chapter 19, Problem 70c

Consider the evaporation of methanol at 25.0 °C : CH₃OH(l) → CH₃OH(g) c. Explain why methanol spontaneously evaporates in open air at 25.0 °C

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insert step 1: Understand the concept of spontaneity in thermodynamics. A process is spontaneous if it occurs without needing to be driven by an external force.

insert step 2: Consider the Gibbs free energy change (ΔG) for the evaporation process. A process is spontaneous if ΔG is negative.

insert step 3: Use the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

insert step 4: Recognize that evaporation involves a phase change from liquid to gas, which increases the disorder or entropy (ΔS > 0) of the system.

insert step 5: Even if the enthalpy change (ΔH) is positive (endothermic), the increase in entropy (ΔS) can make the TΔS term large enough to result in a negative ΔG, making the process spontaneous.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Evaporation

Evaporation is the process by which molecules at the surface of a liquid gain enough energy to enter the gas phase. This occurs when some molecules have sufficient kinetic energy to overcome intermolecular forces, allowing them to escape into the air. The rate of evaporation increases with temperature, as higher temperatures provide more molecules with the energy needed to evaporate.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. For methanol at 25.0 °C, the vapor pressure is significant enough that some molecules will escape into the air, leading to spontaneous evaporation. When the vapor pressure exceeds the atmospheric pressure, evaporation occurs more readily.

Thermodynamics and Spontaneity

Thermodynamics helps explain the spontaneity of processes based on changes in energy and entropy. The evaporation of methanol is spontaneous at 25.0 °C because it increases the entropy of the system; as liquid methanol transitions to gas, the disorder increases, favoring the process. This aligns with the second law of thermodynamics, which states that spontaneous processes tend to increase the overall entropy of the universe.

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Spontaneity of Processes

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Textbook Question

Consider the sublimation of iodine at 25.0 °C : I₂(s) → I₂(g) c. Explain why iodine spontaneously sublimes in open air at 25.0 °C

Textbook Question

Consider the evaporation of methanol at 25.0 °C : CH₃OH(l) → CH₃OH(g) a. Find ΔG°_r at 25.0 °C.

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Textbook Question

Consider the evaporation of methanol at 25.0 °C : CH₃OH(l) → CH₃OH(g) b. Find ΔG_r at 25.0 °C under the following nonstandard conditions: i. P_CH3OH = 150.0 mmHg ii. P_CH3OH= 100.0 mmHg iii. P_CH3OH = 10.0 mmHg

Textbook Question

Consider the reaction: CO₂(g) + CCl₄(g) ⇌ 2 COCl₂(g) Calculate ΔG for this reaction at 25 °C under the following conditions: i. P_CO2 = 0.112 atm ii. P_CCl4 = 0.174 atm iii. P_COCl2 = 0.744 atm

Textbook Question

Use data from Appendix IIB to calculate the equilibrium constants at 25 °C for each reaction. a. 2 CO(g) + O₂(g) ⇌ 2 CO₂(g)

Textbook Question

Use data from Appendix IIB to calculate the equilibrium constants at 25 °C for each reaction. b. 2 H₂S(g) ⇌ 2 H₂(g) + S₂(g)

Consider the evaporation of methanol at 25.0 °C : CH3OH(l) → CH3OH(g) c. Explain why methanol spontaneously evaporates in open air at 25.0 °C

Verified video answer for a similar problem:

Key Concepts

Evaporation

Vapor Pressure

Thermodynamics and Spontaneity

Consider the evaporation of methanol at 25.0 °C : CH₃OH(l) → CH₃OH(g) c. Explain why methanol spontaneously evaporates in open air at 25.0 °C