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. In P = − Δ H v a p o RT + Δ S v a p o R Slope = − Δ H v a p o R , I n t e r c e p t = Δ S v a p o R

Question

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Answer 1

Question

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

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Answer 2

Question

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

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Answer 3

Question

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

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Answer 4

Question

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

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Answer 5

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

Answer 6

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

Answer 7

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.

$In P=−ΔHvapoRT+ ΔSvapo R$

$Slope =−ΔHvapoR,Intercept = ΔSvapoR$

Answer 8

(a)

Expert Solution

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,

$1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123$

$(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366$

Therefore,

Temp. $(K)$	$T−1 (K−1)$	V.P (Torr)	V.P (atm)	lnP
$190$	$0.00526$	$3.22$	$0.0042 atm$	$−5.47$
$228$	$0.00438$	$68$	$0.089 atm$	$−2.41$
$250$	$0.004$	$240$	$0.316 atm$	$−1.15$
$273$	$0.00366$	$672$	$0.884 atm$	$−0.123$

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

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM, Chapter 5, Problem 5.17E

Figure 1

From the curve, slope is given below,

$y = mx + cy = (−3358.714)x + 12.247$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Answer 14

(b)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Slope =−ΔHvapoR,− 3359 =−ΔHvapo(8.314J⋅K−1⋅mol−1)ΔHvapo = 28 kJ⋅mol−1$

The enthalpy of the vaporization is $28 kJ⋅mol−1$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Answer 20

(c)

Expert Solution

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,

$y = mx + cy = (−3358.714)x + 12.247$

$Intercept = ΔSvapoR12.25 =ΔSvapo(8.314J⋅K−1⋅mol−1)ΔSvapo = 101.8 J⋅K−1⋅mol−1$

The entropy of the vaporization is $101.8 J⋅K−1⋅mol−1$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

Answer 25

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Refer to part (a)

Answer 26

(d)

Expert Solution

Explanation of Solution

The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is $1 atm and ln (1) = 0$ .

Therefore, it will happen when.

$T−1 = 0.0036T = 2.7 ×102 K$

The normal boiling point of phosphoryl chloride difluoride is $2.7 ×102 K$

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(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)

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

Answer 31

(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)

Answer 32

(e)

Expert Solution

Explanation of Solution

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

$In P=−ΔHvapoRT+ ΔSvapo RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K$

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

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

Explanation of Solution

Interpretation: The enthalpy of the vaporization has to be calculated. Concept introduction: Refer to part (a)

Explanation of Solution

Interpretation: The entropy of the vaporization has to be calculated. Concept introduction: Refer to part (a)

Explanation of Solution

Interpretation: The normal boiling point of phosphoryl chloride difluoride has to be calculated. Concept introduction: Refer to part (a)

Explanation of Solution

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

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Chapter 5 Solutions

Interpretation:

The enthalpy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Interpretation:

The entropy of the vaporization has to be calculated.

Concept introduction:

Refer to part (a)

Interpretation:

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

Concept introduction:

Refer to part (a)

Interpretation:

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

Concept introduction:

Refer to part (a)