Chapter 10, Problem 92E

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404

Chapter
Section

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404
Textbook Problem

# From the following data for liquid nitric acid, deter mine its heat of vaporization and normal boiling point. Temperature (°C) Vapor Pressure (mm Hg) 0. 14.4 10. 26.6 20. 47.9 30. 81.3 40. 133 50. 208 60. 670.

Interpretation Introduction

Interpretation: The heat of vaporization and normal boiling point of liquid nitric acid should be determined.

Concept Introduction:

Equilibrium vapor pressure:

• The pressure throw by vapor of the substance is equilibrium with  liquid or solid state of same substance is called the equilibrium vapor pressure, thus,
•                     ln(Pvap1Pvap2)=ΔΗvapR(1T2-1T1)......(1)Pvap1andPvap2arevapor pressuresT1andT2aretemperaturesΔΗvapisenthalpy of vaporizationR is constant

Heat of vaporization:

• The energy required for a quantity of liquid substance converted in to into a gases substance is called heat of vaporization.
• The vaporization process take place at constant temperature is called enthalpy of vaporization.
Explanation

Explanation

• The given temperature and pressure of nitric acid are recorded as shown above.

To draw the plot fro given temperature against pressure of the nitric acid.

The plate of given pressure and temperature is,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

The given temperature is converted as 104T form and plotted.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  lnâ€‰(Pvap)=-â€‰Î”Î—vapR(1T)+C

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â Â yâ€‰â€‰=â€‰lnâ€‰(Pvap)xâ€‰â€‰=1Tmâ€‰=â€‰slopeâ€‰â€‰â€‰â€‰â€‰=â€‰-â€‰Î”Î—vapRâ€‰bâ€‰â€‰=â€‰interceptâ€‰â€‰â€‰â€‰â€‰=â€‰C

• The given temperature and pressure are plotted and from the slope the enthalpy of vaporization is calculated.

To calculated the enthalpy of vaporization of nitric acid.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  â€‰â€‰â€‰slopeâ€‰=â€‰6.6-2.52.80Ã—10-3-3.70Ã—10-3â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰4600K4600Kâ€‰=â€‰Î”Î—vapR4600â€‰K=â€‰-sÎ”Î—vap8.3145J/Kgmolâ€‰â€‰Î”Î—vap=â€‰38000â€‰J/molâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰38â€‰kJ/mol

• The slope of the plot and the gas constant value are plugging in equation 1 to give enthalpy of vaporization of given metal

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started