# Verifying That Diamond Is More Stable At Certain Conditions

• Bashyboy
In summary, to show which substance is more stable at certain conditions, we need to compare the Gibbs free energy values and choose the one with the lower value. Using the formula for the partial derivative of Gibbs free energy, we can calculate the Gibbs free energy at different pressures. By assuming negligible changes in volume, we can plug in the given values for diamond and graphite to calculate their respective Gibbs free energy values at 15 kbars. The calculations show that graphite has a lower Gibbs free energy, making it less stable at these conditions. Additionally, the reference data provided shows the Gibbs free energy values at a temperature of 293 K and a pressure of 1 bar, which can be used to calculate the Gibbs free energy values at different pressures.
Bashyboy

## Homework Statement

Verify that diamond becomes more stable than graphite at approximately 15 kbars

## The Attempt at a Solution

So, I understand that to show one substance is more stable than another at particular environmental conditions, I just have to show that the one with a lower Gibbs free energy is more stable.

In the section, the formula $\left( \frac{\partial G}{\partial P} \right)_{T,N} = V$. Since the partial derivative is given approximately by $\frac{\partial G}{\partial P} = \frac{\Delta G}{\Delta P}$, I can write $\frac{\Delta G}{\Delta P} = V$.

In the back of the textbook I am using is a table containing the Gibbs free energy of formation of substances at at temperature of T=293 K and a pressure of P = 1 bar, call it $G_1$ Using this information, I figured I could obtain the Gibbs free energy at P = 15 kbar, call it $G_{15}$:

$\frac{\Delta G}{\Delta P} = V \implies$

$\Delta G = \Delta P V \implies$

$G_{15} = \Delta P V + G_1$.

Now, if we assume that the change in volume of the substance is negligible as the pressure changes from P = 1 bar to P = 15 kbar, which is an assumption the author of the textbook uses, then the formula above is applicable. The volume of diamond at T=293 and P=1 bar is 5.31 x 10^-6. Calculating the Gibbs free energy at 15 kbars of the diamond:

$G_{15,D} = (15 \times 10^{3}~bar -1~bar)(5.31 \times 10^{-6} ~m^3)+ 2.900 \times 10^{3} ~J = 2900 ~J$

And for graphite, where the volume is 3.42 x 10^-6

$G_{15,G} = (15 \times 10^{3}~bar -1~bar)(3.42 \times 10^{-6}) + 0 = 0.0513 ~J$.

Clearly the Gibbs free energy is less for graphite. What did I do incorrectly?

Where does the 3.42 come from? You seem to have a smaller volume for the graphite than for the diamond, which is clearly wrong.

Whoops...I read it incorrectly. The diamond should have the volume of 3.42. But this still does not change the fact that the Gibbs free energy of graphite is much smaller than that of diamond's.

ok, so next I question the G1 values. What exactly does it say in the back of the book, and how did you use that to calculate the G1s?

I uploaded a screen shot of the reference data that I am using, and the paragraph that precedes it. Does this answer your question regarding the G1's, or do you require more information?

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Okay, I think I understand now. Number that is closer to being negative is the substance that is less stable. Therefore, the graphite would be less stable at these particular environmental conditions.

How many Pascals to the Bar?

## 1. How do you determine the stability of diamond?

To determine the stability of diamond, we can use thermodynamic calculations and phase diagrams to analyze the conditions under which diamond is the most stable form of carbon.

## 2. What conditions are necessary for diamond to be the most stable form of carbon?

Diamond is most stable at high pressures and temperatures, typically found deep within the Earth's mantle. However, certain chemical reactions or changes in the surrounding environment can also affect its stability.

## 3. What experiments can be performed to verify the stability of diamond?

Experiments such as diamond anvil cell experiments, high-temperature and high-pressure experiments, and vibrational spectroscopy can be used to verify the stability of diamond at different conditions.

## 4. How does the stability of diamond affect its industrial and commercial use?

The stability of diamond at high pressures and temperatures makes it suitable for industrial and commercial use in cutting tools, drilling equipment, and high-pressure experiments. However, it also makes it challenging to create synthetic diamonds for these purposes.

## 5. Can the stability of diamond be altered?

Yes, the stability of diamond can be altered by introducing impurities or subjecting it to extreme conditions. This can result in the formation of different forms of carbon, such as graphite or amorphous carbon, which have different properties and uses.

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