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Ch.9 - Thermochemistry: Chemical Energy
Chapter 9, Problem 143

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) ⇌ CH3OH(l) ΔS° = -332 J/K Step 2. CH3OH(l) ⇌ CH4(g) + 1/2 O2(g) ΔS° = 162 J/K. (g) Calculate ΔG° for step 2. (d) Which term is more important, ΔH° or ΔS°? (i) Which term is more important, ΔH° or ΔS°?

Verified step by step guidance
1
To calculate ΔG° for step 2, use the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. You will need the values of ΔH° and ΔS° for step 2, as well as the temperature T in Kelvin.
Since ΔS° for step 2 is given as 162 J/K, ensure that the units are consistent when using the Gibbs free energy equation. If ΔH° is in kJ, convert ΔS° to kJ by dividing by 1000.
Determine the temperature T at which you want to calculate ΔG°. If not specified, standard conditions (298 K) are often used.
Substitute the values of ΔH°, T, and ΔS° into the Gibbs free energy equation to find ΔG° for step 2.
To determine which term is more important, ΔH° or ΔS°, consider the magnitude of each term in the Gibbs free energy equation. Compare the absolute values of TΔS° and ΔH° to see which has a greater impact on ΔG°.

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, while a positive ΔG° suggests non-spontaneity.
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Enthalpy (ΔH°)

Enthalpy (ΔH°) is a measure of the total heat content of a system, reflecting the energy required to create a system at constant pressure. It accounts for both internal energy and the energy associated with pressure and volume. In chemical reactions, a negative ΔH° indicates an exothermic reaction, where heat is released, while a positive ΔH° indicates an endothermic reaction, where heat is absorbed.
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Entropy (ΔS°)

Entropy (ΔS°) is a measure of the disorder or randomness in a system. It quantifies the number of ways a system can be arranged, with higher entropy indicating greater disorder. In chemical reactions, an increase in entropy (positive ΔS°) often favors spontaneity, as systems tend to evolve towards states of higher disorder. The relationship between ΔS° and spontaneity is crucial in determining the overall feasibility of a reaction.
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Related Practice
Textbook Question

Ethyl alcohol has ΔHfusion = 5.02 kJ/mol and melts at - 114.1 °C. What is the value of ΔSfusion for ethyl alcohol?

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Textbook Question

What is the melting point of benzene in kelvin if ΔHfusion = 9.95 kJ/mol and ΔSfusion = 35.7 J/(K mol)?

485
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Textbook Question
Metallic mercury is obtained by heating the mineral cinnabar (HgS) in air:HgS1s2 + O21g2 S Hg1l2 + SO21g2(a) Use the data in Appendix B to calculate ΔH° in kilojoules for the reaction.
719
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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) → CH3OH(l) ΔS° = –332 J/K

Step 2. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(a) Calculate ΔH° in kilojoules for step 1.

463
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. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(e) In what temperature range is step 1 spontaneous?

391
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) → CH3OH(l) ΔS° = –332 J/K

Step 2. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(f) Calculate ΔH° for step 2.

368
views