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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145당신이 사용하는 게 아니라요?교과서 변경
18장, 문제 143k

Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):
Step 1. CO(g) + 2 H2(g) S CH3OH(l) ΔS° = - 332 J/K
Step 2. CH3OH1l2 → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K
(k) Calculate an overall ΔG°, ΔH°, and ΔS° for the formation of CH4 from CO and H2.

검증된 단계별 안내
1
Identify the given reactions and their respective \( \Delta S^\circ \) values: Step 1: \( \text{CO(g) + 2 H}_2\text{(g) } \rightarrow \text{ CH}_3\text{OH(l) } \Delta S^\circ = -332 \text{ J/K} \), Step 2: \( \text{CH}_3\text{OH(l) } \rightarrow \text{ CH}_4\text{(g) + } \frac{1}{2} \text{O}_2\text{(g) } \Delta S^\circ = 162 \text{ J/K} \).
To find the overall \( \Delta S^\circ \), add the \( \Delta S^\circ \) values of the two steps: \( \Delta S^\circ_{\text{overall}} = \Delta S^\circ_{\text{Step 1}} + \Delta S^\circ_{\text{Step 2}} \).
Use Hess's Law to determine \( \Delta H^\circ \) and \( \Delta G^\circ \) for each step if not given, or assume standard enthalpy and Gibbs free energy changes if provided.
Calculate the overall \( \Delta H^\circ \) and \( \Delta G^\circ \) by summing the values from each step: \( \Delta H^\circ_{\text{overall}} = \Delta H^\circ_{\text{Step 1}} + \Delta H^\circ_{\text{Step 2}} \) and \( \Delta G^\circ_{\text{overall}} = \Delta G^\circ_{\text{Step 1}} + \Delta G^\circ_{\text{Step 2}} \).
Ensure all units are consistent and check if any additional data or assumptions are needed to complete the calculations, such as standard enthalpies or Gibbs free energies of formation.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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 calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the change in enthalpy and ΔS° is the change in entropy. A negative ΔG° indicates that a reaction is spontaneous under standard conditions.
추천 영상:
가이드 코스
01:51
Gibbs Free Energy of Reactions

Enthalpy (ΔH°)

Enthalpy (ΔH°) is a measure of the total heat content of a system and reflects the energy required to break and form bonds during a chemical reaction. It is an important factor in determining the heat absorbed or released in a reaction. The overall ΔH° for a reaction can be calculated by summing the ΔH° values of individual steps, following Hess's law.
추천 영상:
가이드 코스
02:34
Enthalpy of Formation

Entropy (ΔS°)

Entropy (ΔS°) is a measure of the disorder or randomness in a system. In chemical reactions, it reflects the distribution of energy among the particles and the number of ways the system can be arranged. A positive ΔS° indicates an increase in disorder, which often favors spontaneity, while a negative ΔS° suggests a decrease in disorder, which can hinder spontaneity.
추천 영상:
가이드 코스
02:46
Entropy in Thermodynamics
관련 실천
교과서 질문

Consider the unbalanced equation: (b) Use the data in Appendix B and ΔG°f for IO3-(aq)= -128.0 kJ/mol to calculate ΔG° for the reaction at 25 °C.

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교과서 질문

Consider the unbalanced equation: I2(s) → I-(aq) + IO3-(aq) (d) What pH is required for the reaction to be at equilibrium at 25°C when [I-] = 0.10M and [IO3-] = 0.50 M?

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교과서 질문
For a process to be spontaneous, the total entropy of the system and its surroundings must increase; that is ΔStotal = ΔSsystem + ΔSsurr 7 0 for a spontaneous process Furthermore, the entropy change in the surroundings, ΔSsurr, is related to the enthalpy change for the process by the equa- tion ΔSsurr = - ΔH>T. (b) What is the value of ΔSsurr for the photosynthesis of glu- cose from CO2 at 298 K? 6 CO21g2 + 6 H2O1l2 S C6H12O61s2 + 6 O21g2 ΔG° = 2879 kJ ΔS° = - 262 J>K
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교과서 질문

Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):

Step 1. CO(g) + 2 H2(g) S CH3OH(l) ΔS° = - 332 J/K

Step 2. CH3OH1l2 → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(m) If you were designing a production facility, would you plan on carrying out the reactions in separate steps or together? Explain.

516
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교과서 질문

Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):

Step 1. CO(g) + 2 H2(g) S CH3OH(l) ΔS° = - 332 J/K

Step 2. CH3OH1l2 → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(l) Is the overall reaction spontaneous at 298 K?

278
views
교과서 질문

Consider the unbalanced equation: (a) Balance the equation for this reaction in basic solution.

378
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