{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW1_Solutions - estrada(gse77 Homework 1 Sutclie(52410 This...

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

View Full Document Right Arrow Icon
estrada (gse77) – Homework 1 – Sutcliffe – (52410) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE: In this assignment I have given val- ues for molar heat capacities of various gases at either constant pressure or volume in terms of a number times R; the gas law constant, 8.314 J/mol/K. 001 (part 1 of 2) 10.0 points Calculate the work for a system that releases 317 . 3 kJ of heat in a process for which the decrease in internal energy is 121 . 4 kJ. Correct answer: 195 . 9 kJ. Explanation: Δ U = - 121 . 4 kJ q = - 317 . 3 kJ To calculate this, we use the relationship Δ U = q + w, which arises from the first law of thermody- namics. Because the system releases heat, q will be a negative number as will Δ U , because it is a decrease in internal energy: w = Δ U - q = - 121 . 4 kJ - ( - 317 . 3 kJ) = 195 . 9 kJ . 002 (part 2 of 2) 10.0 points Is work done on or by the system during this process? 1. on the system correct 2. by the system Explanation: Because work is positive, the surroundings will do work on the system. 003 10.0 points Carbon monoxide reacts with oxygen to form carbon dioxide by the following reaction: 2 CO(g) + O 2 (g) 2 CO 2 (g) Δ H for this reaction is - 135 . 28 kcal. How could one rationalize the work compo- nent of this reaction? 1. Work would be done on the system be- cause the heat of reaction is negative which means it is an exothermic reaction. 2. Work would be done on the system be- cause the moles of gas decrease going from re- actants to products, thus implying a decrease in volume. correct 3. There is not enough information to pre- dict work. 4. Work would be done by the system be- cause this is a standard combustion reaction that leads to an increase in gas volume. 5. Work can be calculated only if both the enthalpy and entropy changes are known. 6. Work would be done by the system be- cause the heat of reaction is negative which means it is an exothermic reaction. Explanation: w = - Δ nRT Δ n = moles gas (final) - moles gas (initial) Δ n = 2 moles gas - 3 moles gas Δ n = - 1 mole gas As Δ n is negative, w will be positive so work is done by the surroundings on the sys- tem. 004 10.0 points A sample of 1 . 95 grams of compound Y is burned completely in a bomb calorimeter which contains 2500 g of water. The tem- perature rises from 26 . 3 C to 26 . 722 C. What is Δ U rxn for the combustion of compound Y? The hardware component of the calorimeter has a heat capacity of 3 . 53 kJ/ C. The spe- cific heat of water is 4.184 J/g · C, and the MW of Y is 128 g/mol. Correct answer: - 387 . 5 kJ / mol of Y. Explanation: SH = 4.184 J/g · C HC = 3 . 53 kJ/ C m Y = 1 . 95 g MW Y = 128 g/mol
Image of page 1

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

View Full Document Right Arrow Icon
estrada (gse77) – Homework 1 – Sutcliffe – (52410) 2 m water = 2500 g Δ T = 26 . 722 C - 26 . 3 C = 0 . 422 C Heat = (SH)(m water )(Δ T ) + (HC)(Δ T ) , Then divide the heat by the number of moles which is mass/MW.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern