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We have learned that the enthalpy of vaporization of a

General Chemistry: Principles and Modern Applications | 10th Edition | ISBN: 9780132064521 | Authors: Ralph Petrucci ISBN: 9780132064521 175

Solution for problem 120 Chapter 12

General Chemistry: Principles and Modern Applications | 10th Edition

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General Chemistry: Principles and Modern Applications | 10th Edition | ISBN: 9780132064521 | Authors: Ralph Petrucci

General Chemistry: Principles and Modern Applications | 10th Edition

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Problem 120

We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius Clapeyron equation in the form given in the text (that is, equation 12.2). Instead, we must go back to the differential equation upon which the Clausius Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor and liquid and the temperature. Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, Make appropriate substitutions into the above expression, and separate the P and dP terms from the T and dT terms. The appropriate substitution for means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits and on one side and and on the other. (a) Derive an equation for the vapor pressure of as a function of temperature, if (b) Use the equation derived in (a), together with the fact that the vapor pressure of at 120 K is 10.16 Torr, to determine the normal boiling point of ethylene.

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Chapter 2 Water: The Solvent for Biochemical Reactions POLARITY ● Electron​egativity ​= the tendency of an atom to attract electrons to itself ​ (and become negatively charged)​in a chemical bond ● ​Polar bond = the unequal sharing of electrons between two atoms of different electronegativities ○ Part​ ial negative charge ​δ‐)found​on the ​most electronegative atom ■ Shared electrons more likely to be found at this end of the bond ○ Pa ​ rtial positive charge (δ+)​foun​d on the ​ east electronegative atom ○ Covalent ​ (share electrons) ​ ● Dipole = a bond with a positive and a negative end ○ One

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Chapter 12, Problem 120 is Solved
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Textbook: General Chemistry: Principles and Modern Applications
Edition: 10
Author: Ralph Petrucci
ISBN: 9780132064521

The full step-by-step solution to problem: 120 from chapter: 12 was answered by , our top Chemistry solution expert on 12/23/17, 04:52PM. General Chemistry: Principles and Modern Applications was written by and is associated to the ISBN: 9780132064521. The answer to “We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius Clapeyron equation in the form given in the text (that is, equation 12.2). Instead, we must go back to the differential equation upon which the Clausius Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor and liquid and the temperature. Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, Make appropriate substitutions into the above expression, and separate the P and dP terms from the T and dT terms. The appropriate substitution for means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits and on one side and and on the other. (a) Derive an equation for the vapor pressure of as a function of temperature, if (b) Use the equation derived in (a), together with the fact that the vapor pressure of at 120 K is 10.16 Torr, to determine the normal boiling point of ethylene.” is broken down into a number of easy to follow steps, and 258 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 28 chapters, and 3268 solutions. This textbook survival guide was created for the textbook: General Chemistry: Principles and Modern Applications, edition: 10. Since the solution to 120 from 12 chapter was answered, more than 234 students have viewed the full step-by-step answer.

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We have learned that the enthalpy of vaporization of a