Chapter 5, Problem 5.17E

(a)

Interpretation Introduction

Interpretation:

The temperature dependence of the vapor pressure of phosphoryl chloride difluoride is given, ln p against 1/T has to be plotted.

Concept introduction:

The relationship to plot is In P versus 1/T is given below.

    $\text{In P}=\frac{-\Delta H_{vap}^o}{\text{RT}}+\frac{\Delta S_{vap}^o}{\text{ R}}$

  $\begin{array}{l}\text{Slope}=\frac{-\Delta H_{vap}^o}{\text{R}},\\ \\ Intercept=\frac{\Delta S_{vap}^o}{\text{R}}\end{array}$

Explanation of Solution

Given,

Temperature, T/K	190	228	250	273
Vapour pressure, P/Torr	3.2	68	240	672

Unit conversion is given below,

    $\begin{array}{l}\text{1 Torrr }=\text{ 0}\text{.00131579}atm\\ \\ \text{(1) 3}\text{.22 Torr }=\text{ 0}\text{.0042}atm\\ \ln p=\ln (\text{0}\text{.0042}atm)=-5.47\\ \\ \text{(2)}68\text{ Torr }=\text{ 0}\text{.089}atm\\ \ln p=\ln (\text{0}\text{.089}atm)=-2.41\\ \\ (3)240\text{Torr }=\text{ 0}\text{.316}atm\\ \ln p=\ln (\text{0}\text{.316}atm)=-1.15\\ \\ (4)672\text{Torr }=\text{ 0}\text{.884}atm\\ \ln p=\ln (\text{0}\text{.884}atm)=-0.123\end{array}$

    $\begin{array}{l}\text{(1)Temperature (T)}=190,\\ \left(\frac{1}{T}\right)=0.00526\\ \\ \text{(2)}\text{Temperature (T)}=228,\\ \left(\frac{1}{T}\right)=0.00438\\ \\ (3)\text{Temperature (T)}=250,\\ \left(\frac{1}{T}\right)=0.004\\ \\ (4)\text{Temperature (T)}=273,\\ \left(\frac{1}{T}\right)=0.00366\end{array}$

Therefore,

Temp. $(K)$	$T^{-1}(K^{-1})$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$\text{3}\text{.22}$	$\text{ 0}\text{.0042}atm$	$-5.47$
$228$	$0.00438$	$\text{68}$	$\text{0}\text{.089}atm$	$-2.41$
$250$	$0.004$	$\text{240}$	$\text{0}\text{.316}atm$	$-1.15$
$273$	$0.00366$	$672$	$\text{0}\text{.884}atm$	$-0.123$

The ln p against 1/T plot is given below,

                        Figure 1

From the curve, slope is given below,

  $\begin{array}{l}y=mx+c\\ y=\left(-3358.714\right)x+12.247\end{array}$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

The relationship to plot is In P versus 1/T, this gives a straight line.

From the curve, slope is given below,

  $\begin{array}{l}y=mx+c\\ y=\left(-3358.714\right)x+12.247\end{array}$

    $\begin{array}{l}\text{Slope}=\frac{-\Delta H_{vap}^o}{\text{R}},\\ -3359=\frac{-\Delta H_{vap}^o}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)}\\ \Delta H_{vap}^o=28kJ\cdot mo{l}^{-1}\end{array}$

The enthalpy of the vaporization is $28kJ\cdot mo{l}^{-1}$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

The relationship to plot is In P versus 1/T, this gives a straight line.

From the curve, slope is given below,

  $\begin{array}{l}y=mx+c\\ y=\left(-3358.714\right)x+12.247\end{array}$

    $\begin{array}{l}Intercept=\frac{\Delta S_{vap}^o}{\text{R}}\\ 12.25=\frac{\Delta S_{vap}^o}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)}\\ \Delta S_{vap}^o=101.8J\cdot K^{-1}\cdot mo{l}^{-1}\end{array}$

The entropy of the vaporization is $101.8J\cdot K^{-1}\cdot mo{l}^{-1}$.

(d)

Interpretation Introduction

Interpretation:

The normal boiling point of phosphoryl chloride difluoride has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1atmandln\text{ (1) }=0$.

    Therefore, it will happen when.

  $\begin{array}{l}{T}^{-1}=0.0036\\ T=2.7\times {10}^{2}K\end{array}$

The normal boiling point of phosphoryl chloride difluoride is $2.7\times {10}^{2}K$

(e)

Interpretation Introduction

Interpretation:

If the pressure is 15 torr, temperature of phosphoryl chloride difluoride has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

The temperature of phosphoryl chloride difluoride value is calculated as follows,

  $\begin{array}{l}\text{In P}=\frac{-\Delta H_{vap}^o}{\text{RT}}+\frac{\Delta S_{vap}^o}{\text{ R}}\\ \text{In }\left(\frac{15Torr}{760Torr}\right)=\frac{-(28000J\cdot mo{l}^{-1})}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)\text{T}}+\frac{(100J\cdot K^{-1}\cdot mo{l}^{-1})}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)}\\ -3.93=\frac{-(28000J\cdot mo{l}^{-1})}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)\text{T}}+12.028\\ -15.96=\frac{-(28000J\cdot mo{l}^{-1})}{\left(8.314J\cdot K^{-1}\cdot mo{l}^{-1}\right)\text{T}}\\ T=211K\end{array}$

The temperature of phosphoryl chloride difluoride value is $211K$