Skip to main content
Ch.19 - Free Energy & Thermodynamics
Tro - Chemistry: A Molecular Approach 5th Edition
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoTro 5th EditionCh.19 - Free Energy & ThermodynamicsProblema 70b
Capitolo 19, Problema 70b

Consider the evaporation of methanol at 25.0 °C : CH3OH(l) → CH3OH(g) b. Find ΔGr at 25.0 °C under the following nonstandard conditions: i. PCH3OH = 150.0 mmHg ii. PCH3OH = 100.0 mmHg iii. PCH3OH = 10.0 mmHg

Guida verificata passo dopo passo
1
Identify the standard Gibbs free energy change (ΔG°) for the evaporation of methanol at 25.0 °C, which can be found in tables or calculated if necessary.
Use the equation ΔG = ΔG° + RT ln(Q) to calculate the Gibbs free energy change under nonstandard conditions, where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.
Convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature (25.0 °C + 273.15 = 298.15 K).
Calculate the reaction quotient Q for each given pressure condition using the formula Q = P_CH3OH / P° where P° is the standard pressure (1 atm or 760 mmHg).
Substitute the values of ΔG°, R, T, and Q into the equation ΔG = ΔG° + RT ln(Q) for each pressure condition to find ΔG for each scenario.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
12m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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 crucial for predicting the spontaneity of a reaction; a negative ΔG indicates a spontaneous process, while a positive ΔG suggests non-spontaneity. In the context of evaporation, ΔG can change with varying conditions such as pressure.
Video consigliato:
Percorso guidato
01:51
Gibbs Free Energy of Reactions

Nonstandard Conditions

Nonstandard conditions refer to situations where the concentrations or partial pressures of reactants and products differ from standard state values (1 bar or 1 M). In this question, the partial pressure of methanol vapor is varied, affecting the Gibbs Free Energy of the evaporation process. Understanding how to adjust ΔG for nonstandard conditions is essential for accurate calculations.
Video consigliato:
Percorso guidato
02:13
Ideal Gas Conditions Example

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant of a reaction to the change in temperature and is often used to derive the Gibbs Free Energy under nonstandard conditions. It can be expressed as ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient. This equation is vital for calculating ΔG at different pressures of methanol vapor during evaporation.
Video consigliato:
Percorso guidato
01:40
Van der Waals Equation
Pratica correlata
Domanda del libro di testo

Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) Calculate ΔG for this reaction at 25 °C under the following conditions: i. PCO2 = 0.112 atm ii. PCCl4 = 0.174 atm iii. PCOCl2 = 0.744 atm

2712
views
1
comments
Domanda del libro di testo

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)

2420
views
1
rank
Domanda del libro di testo

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

2543
views
Domanda del libro di testo

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
Domanda del libro di testo

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
Domanda del libro di testo

Consider the sublimation of iodine at 25.0 °C : I2(s) → I2(g) b. Find ΔG°rxn at 25.0 °C under the following nonstandard conditions: i. PI2 = 1.00 mmHg ii. PI2 = 0.100 mmHg

2930
views
2
rank