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

AI Thread Summary
The discussion centers on determining the standard enthalpy (ΔH) and entropy (ΔS) changes for a reaction involving silver and oxygen, based on a provided graph. The standard enthalpy change is suggested to be approximately -31 kJ/mol, while the standard entropy change is calculated to be -35 J/mol K. Participants emphasize the importance of converting temperature from Celsius to Kelvin for accurate calculations. The relationship between ΔG, ΔH, and ΔS is confirmed to be linear over a limited temperature range, allowing for the extraction of these values from the graph. Overall, the approach to solving the problem is validated, with specific attention to the graph's details and temperature conversions.
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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 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!
 

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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 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!
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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