Consider the evaporation of methanol at 25.0 °C : CH 3 OH(l) → CH 3 OH(g) a. Find ΔG° r at 25.0 °C.

Hello everyone. Today we have the falling problem. Consider the evaporation of methanol at 25 °C. And we are being tasked to find the change and give us free energy for the reaction at 25 °C. So we can do that with the following equation, I think gives free energy for the reaction is equal to the sum of the chain is for energy of our products subtracted by that of the reactives. So if we use the standard table of values for given gives free energies, we can note that for methanol which is our product will be negative 166.6 kilojoules per mole, we subtract that from the gives free energy of our methanol and the liquid form which would be negative 1 62.3 kilojoules per mole. And in doing that, we get a gibbs free energy for the reaction that is equal to 4.300 kilojoules Perel as our final answer or answer twice. D overall, I hope is helped. And until next time

Ch.19 - Free Energy & Thermodynamics

All textbooks

Tro 6th Edition

Ch.19 - Free Energy & Thermodynamics

Problem 74a

Chapter 19, Problem 74a

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

Verified step by step guidance

Step 1: The Gibbs free energy change (ΔG°) for a reaction can be calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the enthalpy change, T is the temperature in Kelvin, and ΔS° is the entropy change. However, in this problem, we are not given the values of ΔH° and ΔS°, so we need to find another way to calculate ΔG°.

Step 2: The standard Gibbs free energy change (ΔG°) can also be calculated from the standard free energy of formation (ΔfG°) of the products and reactants using the equation ΔG° = Σ ΔfG°(products) - Σ ΔfG°(reactants).

Step 3: Look up the standard free energy of formation (ΔfG°) for methanol in both the liquid and gas states in a table of thermodynamic data. These values are typically given in kJ/mol.

Step 4: Substitute the values of ΔfG° for methanol in the liquid and gas states into the equation from step 2. Remember that the ΔfG° for the reactants is subtracted from the ΔfG° for the products.

Step 5: Perform the calculation to find the value of ΔG°. Remember that if ΔG° is negative, the reaction is spontaneous at 25.0 °C, and if ΔG° is positive, the reaction is non-spontaneous at 25.0 °C.

Verified Solution

Video duration:

0m:0s

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

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

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is a crucial concept in predicting the spontaneity of a reaction; a negative ΔG indicates that a process can occur spontaneously, while a positive ΔG suggests non-spontaneity.

Standard State Conditions

Standard state conditions refer to a set of specific conditions (usually 1 bar pressure and a specified temperature, often 25 °C) under which thermodynamic measurements are made. These conditions allow for the comparison of thermodynamic data, such as standard Gibbs free energy changes (ΔG°), which are essential for calculating the favorability of reactions.

Phase Change and Vaporization

Phase change refers to the transition of a substance from one state of matter to another, such as from liquid to gas during evaporation. The process of vaporization involves the absorption of energy, which can be quantified using enthalpy changes. Understanding the energy dynamics of phase changes is vital for calculating ΔG° for reactions involving substances in different states.

Recommended video:

Guided course

01:46

Entropy in Phase Changes

Related Practice

Textbook Question

Determine ΔG° for the reaction: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) Use the following reactions with known ΔG°_rxnvalues:

2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔG°_rxn= -742.2 kJ

CO(g) + 12 O₂( g) → CO₂(g) ΔG°_rxn= -257.2 kJ

Consider the sublimation of iodine at 25.0 °C : I₂(s) → I₂(g) b. Find ΔG°_rxnat 25.0 °C under the following nonstandard conditions: i. P_I2 = 1.00 mmHg ii. P_I2 = 0.100 mmHg

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

1136

views

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

2431

views

Textbook Question

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

617

views

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

2712

views

comments

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

Verified Solution

Key Concepts

Gibbs Free Energy (ΔG)

Standard State Conditions

Phase Change and Vaporization

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