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Ch 18 Free Energy and Thermodynamics Module 1

Authored by Laura Forrester

Chemistry

University

Ch 18 Free Energy and Thermodynamics Module 1
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10 questions

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1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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Which of the following is/are always spontaneous? 

I only

II only

both are always spontaneous

neither are always spontaneous

2.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The standard molar entropy of lead(II) bromide (PbBr2) is 161 J/mol ⋅ K. What is the entropy of 2.45 g of PbBr2?

+1.07 J/K

−1.07 J/K

+161 J/K

−161 J/K

0 J/K

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Which of the following processes is predicted to decrease the entropy of a system?

The number of particles increases.

Temperature increases.

Volume increases.

Pressure increases.

All of these increase the entropy of the system.

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Which of the following will have the greatest standard molar entropy (S°)?

Cl2(g)

Ne(g)

Na(s)

CH3OH(l)

Na2CO3(s)

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Which of the following are listed in order of increasing standard molar entropy?

Au(s) < Au(g) < H2O(l) < CH3OH(l) < H2O(g)

Au(s) < H2O(l) < CH3OH(l) < Au(g) < H2O(g)

Au(s) < H2O(l) < CH3OH(l) < H2O(g) < Au(g)


H2O(l) < CH3OH(l) < Au(s) < Au(g) < H2O(g)

H2O(l) < Au(s) < CH3OH(l) < H2O(g) < Au(g)

6.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

During a spontaneous chemical reaction, it is found that ΔSsys is less than 0. This means that

ΔSsurr is less than 0 and its magnitude is less than ΔSsys.

ΔSsurr is less than 0 and its magnitude is greater than ΔSsys.

ΔSsurr is greater than 0 and its magnitude is less than ΔSsys.

ΔSsurr is greater than 0 and its magnitude is greater than ΔSsys.

an error has been made, as Ssys is greater than 0 by necessity for a spontaneous process.

7.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Determine ΔS° for the reaction D given the following information.

Substance S° (J/mol ⋅ K)

Fe2O3(s) 87.4

Fe(s) 27.3

O2(g) 205.0

−144.9 J/mol ⋅ K

−549.4 J/mol ⋅ K

+899.0 J/mol ⋅ K

+549.4 J/mol ⋅ K

+144.9 J/mol ⋅ K

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