Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 16, Problem 16B.6P

(a)

Interpretation Introduction

Interpretation:

The drift speed of Li+, Na+, and K+ in water have to be calculated.

Concept introduction:

Conductivity is a property of a material that depends upon the charge carriers in a sample.  Molar conductivity is the conductivity per unit molar concentration.  According to Kohlrausch’s law, the molar conducivities of the dilute solutions containing strong electrolytes depend upon the square root of the concentration.  The conductivity of a solution when the concentration tends to zero is known as the limiting molar conductivity.  Limiting molar conductivity is the sum of contributions from individual ions.

(a)

Expert Solution
Check Mark

Answer to Problem 16B.6P

The drift speed of Li+ is 8.02×103cms1_.

The drift speed of Na+ is 1.038×102cms1_.

The drift speed of K+ is 1.524×102cms1_.

Explanation of Solution

The drift speed is calculated by the formula shown below.

    s=uE        (1)

Where,

  • u is the mobility of ion.
  • E is the electric field strength.

The electric field is given by the formula shown below.

    E=Δϕl        (2)

Where,

  • Δϕ is the potential difference.
  • l is the distance between two plane parallel plates.

The value of potential difference and distance between two plane parallel plates are 100V and 5.00cm respectively.

Substitute the values of potential difference and distance between two plane parallel plates in equation (2).

    E=Δϕl=100V5.00cm=20Vcm1

For Li+ ion, the value of mobility is 4.01×104cm2V1s1.

Substitute the value of mobility of Li+ ion and E in equation (1).

    s=uE=4.01×104cm2V1s1×20Vcm1=8.02×103cms1_

Therefore, the drift speed of Li+ is 8.02×103cms1_.

For Na+ ion, the value of mobility is 5.19×104cm2V1s1.

Substitute the value of mobility of Na+ ion and E in equation (1).

    s=uE=5.19×104cm2V1s1×20Vcm1=1.038×102cms1_

Therefore, the drift speed of Na+ is 1.038×102cms1_.

For K+ ion, the value of mobility is 7.62×104cm2V1s1.

Substitute the value of mobility of K+ ion and E in equation (1).

    s=uE=7.62×104cm2V1s1×20Vcm1=1.524×102cms1_

Therefore, the drift speed of K+ is 1.524×102cms1_.

(b)

Interpretation Introduction

Interpretation:

The time taken by Li+, Na+, and K+ ions to move from one electrode to other has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 16B.6P

The time taken by Li+ ion to move from one electrode to other is 623.44s_.

The time taken by Na+ ion to move from one electrode to other is 481.69s_.

The time taken by K+ ion to move from one electrode to other is 328.08s_.

Explanation of Solution

The time taken by the each ion to move from one electrode to other is calculated by the formula shown below.

    t=ls        (1)

Where,

  • l is the distance between two plane parallel plates.
  • s is the drift speed.

For Li+ ion, the value of s and l is 8.02×103cms1 and 5.00cm respectively.

Substitute the value of s and l in equation (1).

    t=ls=5.00cm8.02×103cms1=623.44s_

Therefore, the time taken by Li+ ion to move from one electrode to other is 623.44s_.

For Na+ ion, the value of s and l is 1.038×102cms1 and 5.00cm respectively.

Substitute the value of s and l in equation (1).

    t=ls=5.00cm1.038×102cms1=481.69s_

Therefore, the time taken by Na+ ion to move from one electrode to other is 481.69s_.

For K+ ion, the value of s and l is 1.524×102cms1 and 5.00cm respectively.

Substitute the value of s and l in equation (1).

    t=ls=5.00cm1.524×102cms1=328.08s_

Therefore, the time taken by K+ ion to move from one electrode to other is 328.08s_.

(c)

Interpretation Introduction

Interpretation:

The displacement of each ion in centimeter and in term of solvent diameter during half cycle has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 16B.6P

The displacement of Li+ ion in centimeter during half cycle is 1.28×106cm_.

The the displacement of Li+ ion in term of solvent diameters during half cycle is 42.66ds_.

The displacement of Na+ ion in centimeter during half cycle is 1.65×106cm_.

The displacement of Na+ ion in term of solvent diameters during half cycle is 55ds_.

The displacement of K+ ion in centimeter during half cycle is 2.43×106cm_.

The displacement of K+ ion in term of solvent diameters during half cycle is 81ds_.

Explanation of Solution

The displacement during half cycle is calculated as shown below.

    d=01/2vsdt=01/2vuEdt(s=uE)=01/2vuE0sin(2πvt)dt(E=E0sin(2πvt))=uE02πv[cos(2πvt)]01/2v

Solve the above equation.

    d=uE02πv[cos(2πvt)]01/2v=uE02πv[cos(2πv12v)cos(0)]=uE02πv[cos(π)cos(0)]=uE02πv[11]

Solve the above equation.

    d=uE02πv[11]=uE0πv

The displacement cannot be negative.  Therefore, the above equation can be written as shown below.

    d=uE0πv

Where,

  • u is the mobility.
  • E0 is the potential difference.
  • v is the frequency.

The value of E0 and v is 20Vcm1 and 2kHz respectively.

For Li+ ion, the value of mobility is 4.01×104cm2V1s1.

Substitute the value of mobility, E0 and v in equation (1).

    d=uE0πv=4.01×104cm2V1s1×20Vcm-13.14×2kHz(103s11kHz)=8.02×1066.28cm=1.28×106cm_

Therefore, the displacement of Li+ ion in centimeter during half cycle is 1.28×106cm_.

The solvent diameter (ds) is given as 300pm.

     ds=300pm=300pm1010cm1pm=3×108cm1cm=ds3×108

Use 1cm=ds3×108 to calculate the the displacement of Li+ ion in term of solvent diameters during half cycle as shown below.

    d=1.28×106cm=1.28×106(ds3×108)=42.66ds_

The the displacement of Li+ ion in term of solvent diameters during half cycle is 42.66ds_.

For Na+ ion, the value of mobility is 5.19×104cm2V1s1.

Substitute the value of mobility, E0 and v in equation (1).

    d=uE0πv=5.19×104cm2V1s1×20Vcm-13.14×2kHz(103s11kHz)=10.38×1066.28cm=1.65×106cm_

Therefore, the displacement of Na+ ion in centimeter during half cycle is 1.65×106cm_.

Use 1cm=ds3×108 to calculate the the displacement of Na+ ion in term of solvent diameters during half cycle as shown below.

    d=1.65×106cm=1.65×106(ds3×108)=55ds_

The displacement of Na+ ion in term of solvent diameters during half cycle is 55ds_

For K+ ion, the value of mobility is 7.62×104cm2V1s1.

Substitute the value of mobility, E0 and v in equation (1).

    d=uE0πv=7.62×104cm2V1s1×20Vcm-13.14×2kHz(103s11kHz)=10.38×1066.28cm=2.43×106cm_

Therefore, the displacement of K+ ion in centimeter during half cycle is 2.43×106cm_.

Use 1cm=ds3×108 to calculate the the displacement of K+ ion in term of solvent diameters during half cycle as shown below.

    d=2.43×106cm=2.43×106(ds3×108)=81ds_

The displacement of K+ ion in term of solvent diameters during half cycle is 81ds_.

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Chapter 16 Solutions

Atkins' Physical Chemistry

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