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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium
Chapter 18, Problem 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.

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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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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 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.
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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.
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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.
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Related Practice
Textbook Question

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

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Textbook Question

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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Textbook Question

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?

316
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Textbook Question

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
Textbook Question

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.

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Textbook Question
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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