Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Question
Chapter 11, Problem 39IL

(a)

Interpretation Introduction

Interpretation:

The normal boiling point of dichlorodimethylsilane has to be determined

Concept Introduction:

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

(a)

Expert Solution
Check Mark

Answer to Problem 39IL

The normal boiling point of dichlorodimethylsilane is 343.3K.

The temperatures at which liquid have a vapour pressures of 250mmHgand650mmHg are 302.33K and 327.51K respectively.

The molar enthalpy of vaporization of is 33.6kJ/mol.

Explanation of Solution

The normal boiling point of dichlorodimethylsilane is calculated

Given:

Temperature(K)VapourPressure(mmHg)272.640290.5100324.9400343.3760

Normal boiling point is the temperature when the external pressure is 760mmHg

From the given data it is clear that the temperature at which the pressure is 760mmHg is 343.3K

Thus the normal boiling point of dichlorodimethylsilane is 343.3K.

(b)

Interpretation Introduction

Interpretation:

The graph of lnP versus 1/T has to be plotted. The temperatures at which liquid have a vapour pressures of 250mmHg and 650mmHg has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R[1T2-1T1]

(b)

Expert Solution
Check Mark

Answer to Problem 39IL

Using the given data we can plot the graph of lnP versus 1/T

Chemistry & Chemical Reactivity, Chapter 11, Problem 39IL , additional homework tip  1

The temperature at which liquid has a vapour pressure of 250mmHg are 302.33K.

The temperature at which liquid has a vapour pressure 650mmHg are 327.51K.

Explanation of Solution

The temperatures at which liquid have a vapour pressures of 250mmHg and 650mmHg is calculated

Given:

  Temperature(K)VapourPressure(mmHg)272.640290.5100324.9400343.3760

The values of lnP and 1/T is determined

  lnP1/T3.683.668×10-34.603.423×10-35.993.077×10-36.632.9112×10-3

Using the given data we can plot the graph of lnP versus 1/T

Chemistry & Chemical Reactivity, Chapter 11, Problem 39IL , additional homework tip  2

From the slope of the graph we can find the value of ΔvapH0R

  Slope=y2y1x2x1=(5.9)(4.6)(0.003077)(0.003423)=3757.22

Using the equation for the straight line in the plot

lnP=ΔvapH0R×1T+C

C, the constant value can be calculated by substituting any one of the value of pressure and temperature from the table given in the equation.

Substituting the values

  lnP=-3757.22×1T+17.949

From this equation we can calculate the temperature at which the pressures are 250mmHgand650mmHg.

When the pressure is 250mmHg

  ln250=3757.22×1T+17.94912.4275=3757.22×1T3.3076×10-3=1TT=302.33K

The temperature at which the pressures is 250mmHg is 302.33K

When the pressure is 650mmHg

  ln650=3757.22×1T+17.94911.4720=3757.22×1T3.0533×10-3=1TT=327.51K

The temperature at which the pressures is 650mmHg is 327.51K

(c)

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization has to be explained.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R[1T2-1T1]

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

Molar enthalpy of vaporization: The energy required to convert liquid to gas of 1mol of a substance is called molar enthalpy of vaporization

(c)

Expert Solution
Check Mark

Answer to Problem 39IL

The molar enthalpy of vaporization of is 33.6kJ/mol.

Explanation of Solution

Given:

  Temperature(K)VapourPressure(mmHg)272.640290.5100324.9400343.3760

The molar enthalpy of vaporization using the given data is calculated.

P1=1mmHg,P2=10mmHgT1=287K,T2=326.8KΔvapH0=?

Substituting the values

ln[10040]=ΔvapH00.008314kJ/K.mol[1290.5K1272.6K]0.9162=ΔvapH00.008314kJ/K.mol[1290.5K1272.6K]ΔvapH0=33.6kJ/mol

The molar enthalpy of vaporization using the given data is

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