WA07_CHE-121-mar14

# WA07_CHE-121-mar14 - Thomas Edison State College General...

• Homework Help
• 6
• 100% (9) 9 out of 9 people found this document helpful

This preview shows pages 1–3. Sign up to view the full content.

Thomas Edison State College General Chemistry I with Labs (CHE-121) Section no.: OL009 Semester and year: May 2015 Written Assignment 7: Energy and Thermochemistry Answer all assigned questions and problems, and show all work. 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) Multiplied by 2 the value is 2905.6kJ/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) Direction reversed the value is 1452.8kJ/mol d. What is the value of ∆ H if water vapor instead of liquid water is formed as the product? (2 points) Water vapor the value is -1276.8kJ/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) -879kJ/mole x 2 moles ZnS x (1mole ZnS/97.5g ZnS) = -9.021kJ/g ZnS (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) 1 Copyright © 2014 by Thomas Edison State College. All rights reserved.

This preview has intentionally blurred sections. Sign up to view the full version.

C = 0.385 J/g°C T = 324.3 °C – 20.5 °C = 303.8 °C 6,220g x 0.385 J/g°C x 303.8 °C = 728,000 J / 1,000 = 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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern