The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
Δ

Q: Why is $\Delta G = 0$ at the final pressure $P_2$?

When two phases (like graphite and diamond) are in equilibrium at a specific temperature and pressure, their Gibbs free energies are exactly equal. Therefore, the difference $\Delta G$ between them is zero.

Q: Why do we assume $\Delta V$ is constant during compression?

Solids like graphite and diamond are highly incompressible. Their volumes change very little with pressure compared to gases, so treating $\Delta V$ as a constant is a very good approximation for this calculation.

Q: Why did we convert $\Delta G^\circ$ to Joules instead of keeping it in kJ?

To maintain unit consistency. The volume change is given in cubic meters ($m^3$), and pressure in Pascals ($N/m^2$). Since $1 \text{ Joule} = 1 \text{ N}\cdot\text{m}$, using Joules ensures all units align perfectly.

Q: What does the negative sign in $\Delta V$ signify?

It means the volume decreases during the transformation. Diamond has a more compact, denser crystal lattice than graphite, so converting one mole of graphite to diamond results in a smaller volume.

Q: Could we just use $P_2 - P_1$ directly in bars?

No, because the product of $\Delta V$ (in $m^3$) and $\Delta P$ (in bars) would not yield Joules. You must convert pressure to Pascals ($N/m^2$) so that $m^3 \times (N/m^2) = N\cdot m = \text{Joules}$.

Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T = 298	ext{ K}$ are
$\Delta_fG^\circ [	ext{C(graphite)}] = 0	ext{ kJ mol}^{-1}$
$\Delta_fG^\circ [	ext{C(diamond)}] = 2.9	ext{ kJ mol}^{-1}$
The standard state means that the pressure should be $1	ext{ bar}$, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 	imes 10^{-6}	ext{ m}^3	ext{ mol}^{-1}$. If C(graphite) is converted to C(diamond) isothermally at $T = 298	ext{ K}$, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information : $1	ext{ J} = 1	ext{ kg m}^2	ext{ s}^{-2}$; $1	ext{ Pa} = 1	ext{ kg m}^{-1}	ext{ s}^{-2}$; $1	ext{ bar} = 10^5	ext{ Pa}$]

Options:

(A) 14501 bar

(B) 29001 bar

(C) 58001 bar

(D) 1405 bar

Sigma Percentile · Accepted Answer

### The Allure of Alchemy: Graphite to Diamond

Imagine holding a piece of pencil lead—graphite—and wanting to transform it into a sparkling diamond. Both are just pure carbon, yet their properties are worlds apart. Why doesn't graphite spontaneously turn into diamond in our hands? The answer lies in thermodynamics.

At standard conditions ($298	ext{ K}$ and $1	ext{ bar}$), the Gibbs free energy of formation for diamond is higher than that of graphite. Specifically, the change in standard Gibbs free energy for the conversion is:

$$\Delta G^\circ = \Delta_f G^\circ_{	ext{diamond}} - \Delta_f G^\circ_{	ext{graphite}} = 2.9	ext{ kJ mol}^{-1} - 0 = 2.9	ext{ kJ mol}^{-1}$$

Because $\Delta G^\circ$ is positive, the process is non-spontaneous at $1	ext{ bar}$. To force this transformation, we must alter the conditions. The question asks us to find the pressure $P_2$ at which graphite and diamond exist in equilibrium at $298	ext{ K}$.

### The Master Equation of Thermodynamics

To understand how pressure affects Gibbs free energy, we turn to the fundamental thermodynamic relation:

$$dG = VdP - SdT$$

Since the process is isothermal (constant temperature), $dT = 0$, and the equation simplifies beautifully to:

$$dG = VdP$$

For a phase transition from graphite to diamond, we are interested in the difference in Gibbs free energy between the two phases, $\Delta G$. Thus, we can write:

$$d(\Delta G) = \Delta V dP$$

Here, $\Delta V$ is the change in molar volume during the transition. We are given that the volume reduces by $2 	imes 10^{-6}	ext{ m}^3	ext{ mol}^{-1}$, so $\Delta V = -2 	imes 10^{-6}	ext{ m}^3	ext{ mol}^{-1}$.

### Integrating to Equilibrium

We need to integrate this expression from our initial state at $P_1 = 1	ext{ bar}$ to our final state at $P_2$, where equilibrium is achieved. Assuming solids are highly incompressible, $\Delta V$ remains constant over this pressure range.

$$\int_{\Delta G_1}^{\Delta G_2} d(\Delta G) = \int_{P_1}^{P_2} \Delta V dP$$

$$\Delta G_2 - \Delta G_1 = \Delta V (P_2 - P_1)$$

Now, what does equilibrium mean in the language of thermodynamics? It means the Gibbs free energy of both phases is exactly equal, so $\Delta G_2 = 0$. Our initial $\Delta G_1$ is simply the standard $\Delta G^\circ$ we calculated earlier, but we must convert it to Joules to match the SI units of volume and pressure:

$$\Delta G_1 = 2.9 	imes 10^3	ext{ J mol}^{-1}$$

### The Final Calculation

Let's substitute our known values into the integrated equation:

$$0 - (2.9 	imes 10^3) = (-2 	imes 10^{-6}) (P_2 - P_1)$$

$$-2900 = -2 	imes 10^{-6} (P_2 - P_1)$$

Solving for the pressure difference $(P_2 - P_1)$:

$$P_2 - P_1 = \frac{-2900}{-2 	imes 10^{-6}} = 1.45 	imes 10^9	ext{ Pa}$$

Since our options are given in bars, we convert Pascals to bars using the relation $1	ext{ bar} = 10^5	ext{ Pa}$:

$$P_2 - P_1 = \frac{1.45 	imes 10^9}{10^5}	ext{ bar} = 14500	ext{ bar}$$

Finally, we add the initial pressure $P_1$ to find the equilibrium pressure $P_2$:

$$P_2 = 1	ext{ bar} + 14500	ext{ bar} = \mathbf{14501	ext{ bar}}$$

This staggering pressure—over 14,000 times the atmospheric pressure at sea level—is required to squeeze the carbon atoms in graphite close enough together to form the dense, rigid lattice of a diamond. This perfectly aligns with Le Chatelier's principle: increasing pressure favors the state with the smaller volume.

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$T h er m o d y nami c S t a t es$

$G ibb s F r ee E n er g y C han g e$

$F u n d am e n t a l E q u a t i o n$

$I n t e g r a t in g t h e E q u a t i o n$

$E q u i l ib r i u m C o n d i t i o n$

$S u b s t i t u t i o n$

$S o l v in g f or P r ess u r e D i f f er e n ce$

$C o n v er t in g U ni t s$

$F ina l P r ess u r e C a l c u l a t i o n$

$L e C ha t e l i e r^{'} s P r in c i pl e$

The Sigma Insight: Entropy and Free Energy

The Allure of Alchemy

The Master Equation of Thermodynamics

Integrating to Equilibrium

The Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermodynamic States

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gibbs Free Energy Change

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fundamental Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Pressure Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Converting Units

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Pressure Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Le Chatelier′s Principle