What Are the Standard Enthalpy and Entropy Changes for the Graphed Reaction?

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SUMMARY

The standard enthalpy change (ΔH) for the reaction 1/2O2 + 2Ag → Ag2O is -31 kJ/mol, while the standard entropy change (ΔS) is -35 J/mol K. The calculations require converting Celsius to Kelvin and utilizing the Gibbs free energy equation, ΔG = ΔH - TΔS, rearranged to a linear form. The slope of the graph represents the entropy change, and the y-intercept corresponds to the enthalpy change. This approach effectively determines the local values of standard changes in enthalpy and entropy.

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  • Understanding of Gibbs free energy equation (ΔG = ΔH - TΔS)
  • Knowledge of temperature conversion from Celsius to Kelvin
  • Ability to interpret graphical data in thermodynamics
  • Familiarity with standard enthalpy and entropy concepts
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  • Study the relationship between Gibbs free energy and equilibrium constants
  • Explore the implications of temperature on reaction spontaneity
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brake4country
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Homework Statement


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

Homework Equations


ΔG = ΔH-TΔS

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!
 
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Multiple choice them may not need calculations.
If (1) and (2) works together what can be the only possibility?
 
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.
 
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?
 
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 celsius, 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!
 

Attachments

  • 20151230_163134[1].jpg
    20151230_163134[1].jpg
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brake4country said:
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 celsius, 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!
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
 
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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