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Free energy and temperature graphs

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    My questions are based on a graph. I am trying to figure out the standard ΔH change for the graphed reaction and the standard entropy change.

    (1) What is the standard enthalpy change for the graphed reaction?
    (A) -31 KJ/mol
    (B) 0
    (C) +12 KJ/mol
    (D) +11 KJ/mol

    (2) What is the standard entropy change for the graphed reaction?
    (A) -35 KJ/mol K
    (B) 0
    (C) +12 KJ/mol K
    (D) +35 KJ/mol K

    2. Relevant equations
    ΔG = ΔH-TΔS

    3. The attempt at a solution
    I rearranged the Gibb's free energy equation to resemble a linear function. I know that the enthalpy is the y-intercept and the slope is the entropy but my answers are not matching any of possibilities. I included an attachment graph. I apologize in advance if the format is not standard for this site. Thanks in advance!
     
    Last edited by a moderator: Dec 30, 2015
  2. jcsd
  3. Dec 30, 2015 #2
    Multiple choice them may not need calculations.
    If (1) and (2) works together what can be the only possibility?
     
  4. Dec 30, 2015 #3

    Borek

    User Avatar

    Staff: Mentor

    Can't see any plot in your attachment, just a reaction equation between silver and oxygen, and yes - docx is not a reasonable format. Try just an image - jpg, gif, png.
     
  5. Dec 30, 2015 #4
    I presume your graph shows a plot of the equilibrium constant as a function of temperature (or 1/T). Is that correct? Or, is it a plot of ##\Delta G^0## as a function of temperature?
     
  6. Dec 30, 2015 #5
    All the instructions say is: Refer to the following graph, which shows the temperature dependence of the standard free energy change for the reaction:
    1/2O2+ 2Ag---> Ag2O

    I have attached a jpg of the graph. I also noticed that the x-axis is in Celcius, not Kelvin. Some calculations are probably required to convert. For example, I noticed that the gibb's free energy equation is linear and by rearranging: ΔG = -TΔS + ΔH. So, entropy is slope and enthalpy is y-intercept?

    Please let me all know what you think of this graph and the two questions I posted in the earlier thread. Thanks!
     

    Attached Files:

  7. Dec 30, 2015 #6
    The actual relationship is going to be curved, but this graph describes the behavior over a limited range of temperatures, so it can be used to determine the local values of the standard changes in H and S. Your approach is correct. First write out the equation for ΔG as a linear function of centigrade temperature TC. Then substitute T = TC + 273. Then determine the slope and intercept of the resulting equation to get ΔH and ΔS .

    Chet
     
  8. Dec 30, 2015 #7
    Ok. So for the standard enthalpy change, we have to look at the values in Kelvin. Since at 0K, ΔG = ΔH, we have to look at the value for -273 C on the graph, which is approx. -31 kJ/mol. Similarly, the entropy change is slope, which was a bit tricky because picking two points on the line, the y values have to be converted to kelvin. I get -0.035 kJ/mol K, which converts to the right answer: -35 J/mol K. I hope my rationale is correct here!
     
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