# Free energy and temperature graphs

1. Dec 29, 2015

### brake4country

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. Dec 30, 2015

### theodoros.mihos

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

3. Dec 30, 2015

### 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.

4. Dec 30, 2015

### Staff: Mentor

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?

5. Dec 30, 2015

### brake4country

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:

• ###### 20151230_163134[1].jpg
File size:
24.1 KB
Views:
45
6. Dec 30, 2015

### Staff: Mentor

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

7. Dec 30, 2015

### brake4country

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!