Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The standard state Gibbs free energies of formation of C(graphite) and \(\mathrm{C}\) (diamond) at \(T=298 \mathrm{~K}\) are \(\Delta_{f} G^{0}[\mathrm{C}\) (graphite) \(]=0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} G^{0}[\mathrm{C}\) (diamond) \(]=2.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite \([\mathrm{C}\) (graphite) \(]\) to diamond \(\left[C\right.\) (diamond)] reduces its volume by \(2 \times 10^{-}\) \({ }^{6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} .\) If \(C\) (graphite) is converted to \(\mathrm{C}\) (diamond) isothermally at \(T\) \(=298 \mathrm{~K}\), the pressure at which \(\mathrm{C}\) (graphite) is in equilibrium with \(\mathrm{C}\) (diamond), is [Useful information : \(\left.1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^{5} \mathrm{~Pa}\right]\) [Adv. 2017] (a) 14501 bar (b) 58001 bar (c) \(1450 \mathrm{bar}\) (d) 29001 bar

Step-by-Step Solution

Answer

The equilibrium pressure is 14501 bar, corresponding to option (a).

Step 1: Identify the Given Information

We are given the Gibbs free energies of formation for C(graphite) and C(diamond). They are \( \Delta_f G^0[\text{C(graphite)}] = 0 \text{kJ/mol} \) and \( \Delta_f G^0[\text{C(diamond)}] = 2.9 \text{kJ/mol} \). We also know the volume change is \( 2 \times 10^{-6} \text{m}^3/\text{mol} \).

Step 2: Understand the Concept of Equilibrium

At equilibrium, the Gibbs free energy change for the reaction \( \Delta G = 0 \). For the reaction \( \text{C(graphite)} \rightarrow \text{C(diamond)} \), the change in Gibbs free energy is the difference in Gibbs free energy of formation of products and reactants.

Step 3: Calculate \( \Delta_r G^0 \) for the Reaction

The reaction considered is \( \text{C(graphite)} \rightarrow \text{C(diamond)} \). Thus, \( \Delta_r G^0 = \Delta_f G^0[\text{C(diamond)}] - \Delta_f G^0[\text{C(graphite)}] = 2.9 \text{kJ/mol} \). Convert this to J/mol: \( 2.9 \times 1000 = 2900 \text{J/mol} \).

Step 4: Use the Gibbs Free Energy and Pressure Relationship

Since \( \Delta G = \Delta G^0 + \Delta P \Delta V \), at equilibrium \( \Delta G = 0 \), this implies \( \Delta P \Delta V = -\Delta G^0 \). Solve for \( \Delta P \): \( \Delta P = -\frac{\Delta G^0}{\Delta V} \).

Step 5: Calculate \( \Delta P \) Using the Volume Change

We have \( \Delta V = 2 \times 10^{-6} \text{m}^3/\text{mol} \). Substituting the values, \( \Delta P = -\frac{2900}{2 \times 10^{-6}} = -1.45 \times 10^9 \text{Pa} \). Convert this pressure to bars: \( 1.45 \times 10^9 \text{Pa} = 14500 \text{bar} \) (since \( 1 \text{bar} = 10^5 \text{Pa} \)).

Step 6: Determine the Equilibrium Pressure

The atmospheric pressure is 1 bar. Therefore, the equilibrium pressure \( P = 14500 + 1 = 14501 \text{bar} \). This considers the addition of atmospheric pressure since it was not included initially.

Key Concepts

Equilibrium Pressure

Equilibrium pressure is an important concept in thermodynamics, especially when dealing with phase changes like the conversion of graphite to diamond. At equilibrium, the Gibbs free energy change (\( \Delta G \)) for the process is zero. This balance allows us to determine the pressure conditions under which two phases coexist, such as graphite and diamond.
For the conversion process from graphite to diamond, the Gibbs free energy change becomes the driving force to find the equilibrium pressure.

• The equation to use is: \( \Delta G = \Delta G^0 + \Delta P \Delta V \), where \( \Delta G^0 \) is the standard Gibbs energy change, \( \Delta P \) is the pressure change, and \( \Delta V \) is the volume change.
• When \( \Delta G = 0 \), we solve for \( \Delta P \) to find equilibrium pressure.
• In our exercise, adjoining 1 bar to \( 14500 \text{ bar} \) gives the final pressure value \( 14501 \text{ bar} \).

This calculation confirms that at \( 14501 \text{ bar} \), graphite can transform into diamond without a change in Gibbs energy.

Volume Change

Volume change (\( \Delta V \)) is a crucial factor in determining the equilibrium pressure when a substance undergoes a phase or structural change. For the conversion of graphite to diamond, a volume decrease is observed, given as \( 2 \times 10^{-6} \text{ m}^3/\text{mol} \).
Understanding how volume affects free energy is key:

• A decrease in volume can lead to an increase in pressure if the external pressure remains constant.
• In this exercise, the smaller volume of diamond compared to graphite impacts the equilibrium pressure.
• This decrease is incorporated into calculations to determine \( \Delta P \) through the formula \( \Delta P = -\frac{\Delta G^0}{\Delta V} \).

The volume change directly influences how much pressure needs to change to achieve equilibrium under given conditions.

Thermodynamic Equilibrium

Thermodynamic equilibrium is the state where a system is balanced, and there are no net changes occurring within its properties. At this point, all macroscopic flows cease, and conditions like temperature and pressure remain constant throughout the system.
For a chemical reaction or phase change, achieving thermodynamic equilibrium involves several factors:

• The Gibbs free energy change must be zero (\( \Delta G = 0 \)).
• Both phases or products involved will have the same chemical potential, allowing them to coexist without conversion in either direction.
• In the context of our problem, achieving equilibrium between graphite and diamond ensures that at \( 14501 \text{ bar} \), there is no net transformation under constant temperature.

Thus, thermodynamic equilibrium maintains the stability and balance necessary for diamond to coexist with graphite at high pressures.