The standard molar entropy values of N e a n d X e has to be compared. Concept Introduction: The entropy of distinguishable particle in terms of partition function can be calculated using following formula. S = U − U ( 0 ) T + N k ln q The entropy of indistinguishable particle in terms of partition function can be calculated using following formula. S = U − U ( 0 ) T + N k ln q − N k ( ln N − 1 ) Where, S = e n t r o p y U − U ( 0 ) = c h a n g e i n int e r n a l e n e r g y k = p l a n k ' s c o n s tan t q = p a r t i t i o n f u n c t i o n So from the above equation it is clear that the entropy depends on partition functions . The expression for the translational partition function is given below. q T = ( 2 π m k T ) 3 2 V h 3 Where, q T = t r a n s l a t i o n a l p a r t i t i o n f u n c t i o n m = m o l e c u l a r m a s s T = t e m p e r a t u r e h = p l a n k ' s c o n s tan t V = v o l u m e a t s tan d a r d c o n d i t i o n s k = b o l t z m a n n c o n s tan t The expression for rotational partition function is given below. q R = k T σ h B Where, q R = r o t a t i o n a l p a r t i t i o n f u n c t i o n k = b o l t z m a n n c o n s tan t T = t e m p e r a t u r e h = p l a n k ' s c o n s tan t σ = s y m m e t r y n u m b e r B = r o t a t i o n a l c o n s tan t The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Question

Want to see the full answer?

Answer 1

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

help

Done 11:14 ⚫ worksheets.beyondlabz.com 5 (a). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 6 5 (b). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 7 6. Are there any differences between the spectra you obtained in Beyond Labz and the predicted spectra? If so, what were the differences? <

2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled** Peak Chemical Shift (d) 5.7 1 Multiplicity multiplate .......... 5.04 double of doublet 2 4.98 double of doublet 3 4.05 doublet of quartet 4 5 LO 3.80 quartet 1.3 doublet 6 Peak Chemical Shift (d) Multiplicity

Answer 2

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 6

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 7

Chapter 12, Problem 12.11P

(a)

Interpretation Introduction

Interpretation:

The standard molar entropy values of Ne and Xe has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 8

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 9

(b)

Interpretation Introduction

Interpretation:

The standard molar entropy values of H2O and D2O has to be compared.

Concept Introduction:

The entropy of distinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q

The entropy of indistinguishable particle in terms of partition function can be calculated using following formula.

S=U−U(0)T+Nk ln q−Nk(lnN−1)

Where,

    S=entropyU−U(0)=change in internal energyk=plank's constantq=partition function

So from the above equation it is clear that the entropy depends on partition functions.

The expression for the translational partition function is given below.

qT=(2πmkT)32 Vh3

Where,

    qT=translational partition functionm=molecular massT=temperatureh= plank's constantV= volume at standard conditionsk=boltzmann constant

The expression for rotational partition function is given below.

qR=kTσhB

Where,

    qR= rotational partition functionk=boltzmann constantT=temperatureh= plank's constantσ=symmetry numberB=rotational constant

The translational partition function increases with increase in molecular mass and in the case of rotational partition function which depends on moment of inertia, also increases with increase in molecular mass.

Answer 10

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 11

(c)

Interpretation Introduction

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 12 - Prob. 12A.1ST Ch. 12 - Prob. 12A.2ST Ch. 12 - Prob. 12B.2ST Ch. 12 - Prob. 12B.3ST Ch. 12 - Prob. 12B.5ST Ch. 12 - Prob. 12C.1ST Ch. 12 - Prob. 12C.3ST Ch. 12 - Prob. 12C.4ST Ch. 12 - Prob. 12A.1E Ch. 12 - Prob. 12A.2E

Videos

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Want to see the full answer?

Chapter 12 Solutions

Videos

Interpretation: The standard molar entropy values of C(diamond) and C(graphite) has to be compared. Concept Introduction: The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.

Want to see the full answer?

Chapter 12 Solutions

Interpretation:

The standard molar entropy values of C(diamond) and C(graphite) has to be compared.

Concept Introduction:

The entropies of solids influenced greatly by the arrangement of atoms in it. Entropy will be less for those solids with more ordered arrangement.