Chemistry

The standard internal energy change for a reaction can be symbolized as ΔU°rxn or ΔE°rxn.

For the following reaction equations, calculate the energy change of the reaction at 25 °C and 1.00 bar.

Sn(s) + 2Cl2(g) --> SnCl4(l)
ΔH°rxn = -511.3 kJ/mol
ΔU°rxn = ?

H2(g) + Cl2(g) --> 2HCl(g)
ΔH°rxn = -184.6 kJ/mol
ΔU°rxn = ?

I know ΔU°rxn = q + work
q = ΔH°rxn

How do I find work for each reaction?

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  1. For #2.
    dUrxn = q + w
    q = dH = -184.6 for 1 mol or x 9 for 9 mols = -? kJ.
    w = -pdV
    volume = nRT/P = 9mol*0.08206 L*atm*K*298K/0.9869 atm = about 223L

    w = -0.9869*446 = about -44 kJ
    Then dU = -? kJ - ?w = ?

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  2. How come 223L was doubled?
    And do I need to use 9 mols
    so -184.6 x 9 = -1661.4 kJ/mol?

    Can I do this:

    (1 mol*0.08206*298K)/0.9869 = 24.8L
    then

    w = -0.9869 * 24.8 = -2.45 kJ

    then

    dU = -184.6 - 2.45 = 209.1 kJ/mol ?

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  3. Yes and no.
    I goofed on the 9 mols; don't ask where I picked that up because I don't see it anywhere in the problem. However, what is n for H2 and n for Cl2. That isn't clear from the problem.


    (1 mol*0.08206*298K)/0.9869 = 24.8L
    Yes you can use that but subtitute mols HCl for n. If you used 1 mol H2 and 1 mol Cl2 it produces 2 mols HCl and that's what goes in for n.

    then

    w = -0.9869 * 24.8 = -2.45 kJ
    You should note here that the is p in atm and v in L(but probably not 24.8L) so the units are atm*L and not joules. You must multiply that number by 101.325 to change L*atm to joules.
    then

    dU = -184.6 - 2.45 = 209.1 kJ/mol ?
    Remember that the -184.6 kJ/mol must be multiplied by mols HCl also to find total kJ for dH. Of course the 2.45 you have here isn't the final number either due to the above changes.
    Hope this looks ok. I'll check it and make corrections if necessary.

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  4. It looks ok to me. Let me know if you have other questions about the solution. Do you have clarification on n H2 and nCl2.

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  5. That is incorrect

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  6. This answer is wrong

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  7. The second reaction has no change in the number of moles of gas, so w = 0, and ΔU = ΔH.

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  8. w=Δ°nRT
    so just find the change in mols from the equation.

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  9. Equation #1
    The change in moles is -2
    dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = 506.3 kJ/mol

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  10. Equation #1
    The change in moles is -2
    dU = 511.3 kJ/mol - [(-2)*(8.3145*10^3 kJ/K*mol)*(298.15K)] = -506.3 kJ/mol

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  11. 1) -511.3
    and
    2)-184.6

    there is no change in moles. Just completed this homework and the above answers are correct. delta-H=delta-U

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  12. WINDAL HAS RIGHT ANSWER!!!!

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  13. WINDAL IS GOD

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  14. the answer are
    equation #1 = -506.3
    equation #2 = -184.6

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  15. Welp has the right answer. Not Windal.

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  16. I concur that welp is correct, with the answers being
    equation #1 = -506.3
    equation #2 = -184.6
    because windal forgot to convert into K for equation 1

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