WA 07_CHE-121-jan18.docx - Name Robert Velat College ID...

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Name: Robert Velat College ID: 0254046 Thomas Edison State University General Chemistry I with Labs (CHE-121) Section no.: Module 7 Semester and year: JUN2018 Written Assignment 7: Energy and Thermochemistry Answer all assigned questions and problems and show all work. All answers are in BOLD. 1. Consider this reaction: 2CH 3 OH( l ) + 3O 2 ( g ) → 4H 2 O( l ) + 2CO 2 ( g ) H = –1452.8 kJ/mol a. Is this reaction endothermic or exothermic? (2 points) Exothermic b. What is the value of ∆ H if the equation is multiplied throughout by 2? (2 points) -2905.6 kJ/mol c. What is the value of ∆ H if the direction of the reaction is reversed so that the products become the reactants and vice versa? (2 points) 1452.8 kJ/mol d. What is the value of ∆ H if water vapor instead of liquid water is formed as the product? (2 points) 2 C H 3 OH ( l ) + 3 O 2 ( g ) 4 H 2 O ( g ) + 2 C O 2 ( g ) Second half is product, first half is reactants so they switch places: [ 4 ( 241.8 ) + 2 ( 393.5 ) ] [ 2 ( 238.7 ) + 3 ( 0 ) ] =− 1276.8 kJ / mol (Reference: Chang 6.24) 2. The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that is, the conversion of ZnS to ZnO by heating: 2ZnS( s ) + 3O 2 ( g ) → 3ZnO( s ) + 2SO 2 ( g ) ∆ H = –879 kJ/mol Calculate the heat evolved (in kJ) per gram of ZnS roasted. (5 points) 879 kJ 2 Mol x 1 Mol 97.46 gZnS =− 4.51 kJ / g (Reference: Chang 6.25) 3. A 6.22 kg piece of copper metal is heated from 20.5 °C to 324.3 °C. Calculate the heat absorbed (in kilojoules) by the metal. Specific heat Cu = 0.385 J/g°C. (5 points) ∆t = 324.3 20.5 = 303.8 ;s = 0.385 J g ;m = 6.22 kg x 1000 g 1 kg = 6220 gCu 1 Copyright © 2017 by Thomas Edison State University . All rights reserved.
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6220 g x 0.385 J g x 303.8 = 727509.86 J x 1 kJ 1000 J = 728 kJ (Reference: Chang 6.33) 4. A sheet of gold weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a
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