Statistical Thermodynamics: Gibbs Free Energy and Chemical Equilibrium

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Flashcards on Gibbs Free Energy and Chemical Equilibrium

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22 Terms

1
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What is the primary focus of today's lecture?

Gibbs free energy, its relation to Boltzmann distributions, and application to chemical equilibrium.

2
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What are the two learning outcomes for this lecture?

Understanding the relationship between Boltzmann distribution and chemical equilibrium, and calculating equilibrium constants using partition functions.

3
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What is the key component needed to calculate Gibbs free energy?

The partition function (Q).

4
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What assumption is made for a monatomic gas when calculating Gibbs free energy?

Only the translational component is considered, assuming an ideal gas with no rotational or vibrational contributions.

5
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Why is calculating Gibbs free energy for helium a good example?

It helps understand the terms in the equation, how to apply them, and how this works for real systems.

6
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What is the first approach to simplify complex calculations?

Define the R term ( gas constant) and break down the equation into individual parts, looking for units along the way.

7
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What must you remember when calculating molar Gibbs energy?

To calculate it per mole, using standard units to match the desired output (joules per mole).

8
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What must you remember to do with pressure for these problems?

Use standard pressure.

9
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What components are included in the partition function for a diatomic gas?

Translational and rotational terms. Vibration is ignored in ideal situations.

10
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What is the value of sigma for symmetric molecules?

Two(2)

11
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What is the first step when the calculation involves frequency and wave numbers?

Convert from wave numbers to frequency

12
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In terms of chemical equilibria what does the Boltzmann distribution describe?

The distribution of molecules occupying different energy levels which changes if there are any energy levels change.

13
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In an endothermic reaction, is the distribution is shifted more towards the reactants or products?

The reactants, indicating a lower partition function value closer to one.

14
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In the calcium carbonate reaction, how is the equilibrium shifted?

Shifted to the right, due to the high entropy and close energy spacing, where the products dominate the space.

15
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How can equilibrium constants be calculated using partition functions?

By using the ratio of partition functions of products to reactants, and take into account the energy difference between them. (Activation energy ignored).

16
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What is the key equation in ionization reactions?

K= (Q products)/(Q reactants)*exp(-delta U/kT)

17
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Why are ionization reactions a good way to calculate?

Because we’re completely removing some of the terms and it simplifies the whole approach. We’re also applying the ideal gas laws at all times.

18
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What contributions must be considered within each partition function in ionization reactions?

Translational motion and spin.

19
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Which terms of a caesium reaction you do not have when atomizing caesium?

Spin terms.

20
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When starting to calculate the problem what should you start with?

Listing all the partition functions to break them out fully so that you can cancel out terms

21
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What is a common question when breaking down the problem?

What terms are associated, which partition functions associated with that species.

22
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What is a general trend to realize when the course transitions?

We have all these interactions, speed between these species and it always gets more complex as we add more things. Remember, breaking it all up is how you solve it.