61 6 x 83145 29815 x ln1061 kj rg29815k

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 19.4 kJ f) 20.4 kJ g) 27.8 kJ h) 31.2 kJ Show your calculations! In this problem, we consider the above chemical reaction under two different conditions 1) at 298.15K and 10.6 bars and 2) at 298.15K and 1.0 bar. We make a cycle based on this observation between: Reactants (298.15K and 10.6 bars) à༎ Products (298.15K and 10.6 bars) (1)â༎ á༎(3) Reactants (298.15K and 1 bar) à༎ Products (298.15K and 1 bar) (2) ΔRG(298.15K and 10.6 bars) = ΔG1(reactants @298.15K from 10.6 bar to 1 bar) + ΔRG2 (reaction @ 298.15K and 1 bar) + ΔG3(products @298.15K from 1 bar to 10.6 bar) ΔG1(reactants @298.15K from 10.6 bar to 1 bar) = n(H2(g)) RT ln (1/10.6) + 5 n(I2(s)) Vm(I2(s)) (1 – 10.6) bar x 10 Pa/bar Note that the change in the free energy for iodine is written as V ΔP because iodine is considered an incompressible solid. Molar volumes of solids and liquids are much smaller than gases, so the term V ΔP is negligible in comparison to the nRT ln(Pf/Pi) term. ΔG1(reactants @298.15K from 10.6 bar to 1 bar) = 3 x 8.3145 x 298.15 x ln (1/10.6) kJ ΔRG2 (reaction @ 298.15K and 1 bar) = 3 mol x (2 x 1.7 kJ/mol) = 10.2 kJ ΔG3(products @298.15K from 1 bar to 10.6 bar) = n(HI(g)) RT ln (10.6/1) = 6 x 8.3145 * 298.15 x ln(10.6/1) kJ ΔRG(298.15K and 10.6 bars) = 27.8 kJ Question 6: (5 points) Consider the following chemical reaction at 1 bar: l2 (s) + H2 (g) à༎ 2HI (g) Using the data in the data section and assuming that the reaction enthalpy change and the rea...
View Full 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