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How to solve for delta S surroundings at standard conditions

  1. Feb 1, 2015 #1
    The question asks to solve for delta S surroundings at 298K when 2.49 moles of H2S (g) react.

    reaction:
    2H2S (g) + 3O2 (g)= 2H2O (g) + 2SO2 (g)

    My problem is that I don't know how to calculate delta S surroundings. So to solve the problem I'm making the assumption that -Delta S surroundings = Delta S system and using products minus reactants to find Delta S system.

    Attempt:

    I begin by finding tabulated S values for all compounds in rxn. Then multiply and cancel units to find out how many moles of each will be in the reaction. Then Using products minus reactants with accurate mol coefficients calculate delta S of rxn. I get this to be -210.4 J/K. Then assuming that Delta S surroundings = -Delta S run I get that Delta S surroundings equals 210.4 J/K.

    If anyone could point me in the right direction that would be great. Thanks in advance
     
  2. jcsd
  3. Feb 2, 2015 #2
    I didn't exactly get how you've done this. But the way I know is, find out the heat released during the reaction. Knowing the temperature, use the formula S = ∫dQ/T
     
  4. Feb 2, 2015 #3
    By the way run was supposed to be rxn in the last sentence of attempt. The way I attempted to solve the problem uses the equation DeltaSrxn= Sum(S products)- Sum(S reactants). Then I am assuming that Delta S surroundings= -Delta S rxn, but that doesn't seem to work here The chemistry class I am in does not use caclulus to solve thermodynamics problems so, valid as your response may be, it doesn't seem to help me in this situation. Thank you for your reply!
     
  5. Feb 2, 2015 #4
    The other thing we are working with is Gibbs Free energy so I wonder if that can somehow show us the Delat S surroundings
     
  6. Feb 2, 2015 #5

    Quantum Defect

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    Gold Member

    Siddarth is correct.

    If the reaction is carried out at constant T and P, find out what the enthalpy change of the reaction is (using heats of formation). Whatever heat this is, this is for the system. Take the negative of this enthalpy change of the system to get the enthalpy change of the surroundings. (DH syst + DH surr = 0)
    Divide this enthalpy change by the temperature, and you will get Delta S.

    Remember, this is why Delta G is useful for determining if reactions are spontaneous at constant T and P:

    Delta G (system) = Delta H (system) - T Delta S (system)

    Delta G (system) = - Delta H (surroundings) - T Delta S (system)

    Delta G (system) = -T Delta S (surroundings) - T Delta S (system)

    Delta G (system) = -T [ Delta S (system) + Delta S (surroundings) ]

    Delta G (system) = -T Delta S (Universe) ==> for a spontaneous process Delta S (Universe) > 0 ==> Delta G < 0 for constant T, P
     
  7. Feb 2, 2015 #6
    This is on the right track, but it is not quite correct. Whenever they speak of the changes in the thermodynamics functions in this type of chemical reaction context, they are talking about the changes between the following two thermodynamic equilibrium states:

    Thermodynamic Equilibrium State 1: Pure reactants (i.e., in separate containers) in stoichiometric proportions at temperature T and 1 atm. pressure

    Thermodynamic Equilibirum State 2: Pure products (i.e., in separate containers) in corresponding stoichiometric proportions at temperature T and 1 atm pressure

    Because the pressure is 1 atm., the states are referred to as "standard states" and are given the superscript 0.

    The change in entropy for the reaction is not equal to ΔH0/T because the change in Gibbs free energy from State 1 to State 2 is not generally zero. Remember that ##ΔG^0=ΔH^0-TΔS^0##. To get the change in entropy from State 1 to State 2, you need to calculate:
    $$ΔS^0=\frac{(ΔH^0-ΔG^0)}{T}$$
    You also have to multiply by the number of moles involved, and, as Quantum Defect says, you need to flip the sign.

    I'm assuming that your tables have standard heats of formation and standard free energies of formation of the species participating in the reaction, and you are asked to determine the entropy change.

    Chet
     
  8. Feb 2, 2015 #7
    Awesome, thanks for the reply. So now I understand that one can solve for ΔS using the formula for Gibbs Energy. I'm wondering why you can't solve for ΔS just by using Sigma(Sproducts)-Sigma(Sreactants, in this case? Although in the context of this problem ΔHrxn and T are given so it makes sense to solve using the ΔG formula. I should also clarify that what I'm being asked is for is ΔSSurroundings, and yes there are tables with standard heats of formation and entropies. So I still don't understand how to find ΔS Surroundings. Could someone connect Gibbs Free Energy with ΔSSurroundings for me? Is it safe to assume that ΔSrxn=-ΔSsurroundings?

    Thanks again in advance and for previous replies.
     
  9. Feb 2, 2015 #8

    Yes.

    Well, since use of ΔS implies that the process is reversible, ΔS of the surroundings should be minus ΔS of the system. Are you saying that subtracting the entropy of the reactants from the entropy of the products does not give you the same answer for ΔS as when you calculate ΔS from ΔH and ΔG? It should. The three things are related using the equation I gave.

    Please show us the details of your work.

    Chet
     
  10. Feb 2, 2015 #9
    I just realized that you did show your work in your original post. Do you have a reason to feel that this is not the correct answer?

    Chet
     
  11. Feb 2, 2015 #10
    Okay thanks for clarifying. That makes complete sense. I was thinking that ΔS should be the same either way you calculate it. Okay so I looked at the problem again and apparently it didn't include and ΔH so I am still unclear on how to solve it. It says at STP so 298K and 1 atm.

    The way I solved previously was using standard entropies from tables and ΔSrxn=Sigma(Sproducts)-Sigma(SReactants)=-ΔSsurroundings
    Then after finding that multiply the value in J/k by 2.49molesH2S (g)/ 2molesH2S (g). to get my answer in J/K.
    The values I used for S^0 are as follows
    H2S (g)= 205.8 j/m*k
    O2 (g)= 205.1 j/m*k
    H2O (g)= 188.8 j/m*k
    SO2 (g)= 248.2 j/m*k
    For each entropy value I then multiply the j/m*k by the molecular coefficient to get a value in j/k. Then I can simply use products minus reactants to find ΔS and multiply that by the proportion of 2.49 molesH2S(g)/ 2 moles H2S (g) to get the ΔSrxn.
    Calculations are as follows
    Sigma S products= (2m*188.8j/m*k)+(2m*248.2j/m*k)
    Sigma S reactants= (3m*205.1j/m*k)+(2m*205.8j/m*k)
    then ΔS=(874)-(1026.9)= -152.9 j/k Then that time 2.49/2= 190.36J/k. Somehow different than my original answer haha.

    Well I plugged in 190.36 j/k in for ΔSsurroundings and got the wrong answer. The program says that I was supposed to use ΔSSurroundings= -ΔHrxn/T to solve so I guess that makes sense. Does this make sense to all of you?

    Thanks for all the replies!
     
  12. Feb 2, 2015 #11
    Apparently, in post #6, I misinterpreted how this reaction is being carried out, and, in particular, misinterpreted the initial and final states. Apparently, in this problem, we are dealing with a closed system in which the reactants are loaded into the reactor, and the system is held at fixed pressure (i.e., fixed external force per unit area) while the reaction occurs. In addition, the reactor is kept in contact with a constant temperature reservoir at 298, so that the initial and final temperatures of the contents are at 298. For a process carried out at constant pressure, the heat added from the reservoir is equal to the change in enthalpy of the contents. So, for this setup, Quantum Defect was correct in saying that ΔS = -ΔH/T (for the reservoir). If you don't have H0 in your tables, but you do have G0 and S0, you can get H0 from H0=G0+TS0. This can be used to get ΔH0.

    Chet
     
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