Chapter 19 Solutions

Chapter 19 Solutions - = × × × = ΔS 20 0g CCl4 1 mol...

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Chemistry 112 – Lecture Problem Solutions CHAPTER 19 Example 1 Solution : ( )+ ( )+ ( )→ ( ) 12H2 g 12N2 g 32O2 g HNO3 g Example 2 Solution : a) Spontaneous b) Non-Spontaneous c) Spontaneous d) Spontaneous e) Non-Spontaneous Example 3 Solution : For an isothermal reversible process, the change in entropy is simply the heat divided by the absolute temperature: = ΔS qrevT A phase change is given by the following chemical equation, with the enthalpy of condensation obtained from a data table: ( ) CCl4 g ( ) ⇄CCl4 l =- . / ΔHcondo 32 6 kJ mol Simply plug the correct values into the equation to find the change in entropy:
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Unformatted text preview: = . × . ×-, × . =-. ΔS 20 0g CCl4 1 mol CCl4153 822 g CCl4 32 600 J1 mol CCl4 1349 6 K 12 1 / J K Example 4 Solution : a) Decreased b) Increased c) Decreased d) Increased Example 5 Solution : Simply subtract the total final entropy from the total initial entropy to find the change in entropy: × . ( )-× . ( )-× . 2mol 188 83 Jmol K H2O g 2mol 130 58 Jmol K H2 g 1mol 205 0 Jmol K ( )=-. O2 g 88 5JK...
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This note was uploaded on 03/10/2009 for the course CHEM 112 taught by Professor Vandersluys,lorschmid,kylem during the Spring '07 term at Penn State.

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