JEE Challenger
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ChemHardAdv 2017 • Paper 2

Equilibrium Pressure for Graphite to Diamond Transformation

At T=298 KT = 298\text{ K}, the standard state Gibbs free energies of formation of C(graphite) and C(diamond) are given as: ΔfG∘[C(graphite)]=0 kJ mol−1\Delta_f G^\circ[\text{C(graphite)}] = 0\text{ kJ mol}^{-1} ΔfG∘[C(diamond)]=2.9 kJ mol−1\Delta_f G^\circ[\text{C(diamond)}] = 2.9\text{ kJ mol}^{-1} The standard state implies a pressure of 1 bar1\text{ bar}, and substances are pure at a given temperature. Transforming graphite [C(graphite)] into diamond [C(diamond)] decreases its molar volume by 2×10−6 m3 mol−12 \times 10^{-6}\text{ m}^3\text{ mol}^{-1}. If C(graphite) is converted to C(diamond) isothermally at T=298 KT = 298\text{ K}, find the pressure at which C(graphite) exists in thermodynamic equilibrium with C(diamond). [Useful data: 1 J=1 kg m2 s−21\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}; 1 Pa=1 kg m−1 s−21\text{ Pa} = 1\text{ kg m}^{-1}\text{ s}^{-2}; 1 bar=105 Pa1\text{ bar} = 10^5\text{ Pa}]

Options

A

14501 bar14501\text{ bar}

Correct
B

58001 bar58001\text{ bar}

C

1450 bar1450\text{ bar}

D

29001 bar29001\text{ bar}

Topics & Concepts

ThermodynamicsGibbs Free EnergyEnthalpyEntropy and Second Law

Step-by-Step Solution

To determine the pressure at which C(graphite)\text{C(graphite)} exists in thermodynamic equilibrium with C(diamond)\text{C(diamond)} at T=298 KT = 298\text{ K}, we consider the transformation reaction:

C(graphite)⟶C(diamond)\text{C(graphite)} \longrightarrow \text{C(diamond)}

Step 1: Calculate the Standard Gibbs Free Energy Change (ΔrG∘\Delta_r G^\circ)

At the standard state pressure of P1=1 barP_1 = 1\text{ bar} and T=298 KT = 298\text{ K}, the standard Gibbs free energy change of the reaction is given by:

ΔrG∘=ΔfG∘[C(diamond)]−ΔfG∘[C(graphite)]\Delta_r G^\circ = \Delta_f G^\circ[\text{C(diamond)}] - \Delta_f G^\circ[\text{C(graphite)}]

ΔrG∘=2.9 kJ mol−1−0 kJ mol−1=2.9 kJ mol−1=2900 J mol−1\Delta_r G^\circ = 2.9\text{ kJ mol}^{-1} - 0\text{ kJ mol}^{-1} = 2.9\text{ kJ mol}^{-1} = 2900\text{ J mol}^{-1}

Step 2: Relate Gibbs Free Energy to Pressure

The change in Gibbs free energy with respect to pressure at constant temperature is given by fundamental thermodynamic relation:

(∂G∂P)T=V\left(\frac{\partial G}{\partial P}\right)_T = V

For the reaction, the variation of ΔrG\Delta_r G with pressure is:

(∂(ΔrG)∂P)T=ΔVm\left(\frac{\partial (\Delta_r G)}{\partial P}\right)_T = \Delta V_m

where ΔVm\Delta V_m is the change in molar volume during the transformation:

ΔVm=Vm[C(diamond)]−Vm[C(graphite)]=−2×10−6 m3 mol−1\Delta V_m = V_m[\text{C(diamond)}] - V_m[\text{C(graphite)}] = -2 \times 10^{-6}\text{ m}^3\text{ mol}^{-1}

Assuming the solid phases are incompressible (i.e., ΔVm\Delta V_m remains constant with changing pressure), we integrate from the standard pressure P1=1 bar=105 PaP_1 = 1\text{ bar} = 10^5\text{ Pa} to the equilibrium pressure P2P_2:

ΔrG(P2)=ΔrG∘+∫P1P2ΔVm dP\Delta_r G(P_2) = \Delta_r G^\circ + \int_{P_1}^{P_2} \Delta V_m \, dP

ΔrG(P2)=ΔrG∘+ΔVm(P2−P1)\Delta_r G(P_2) = \Delta_r G^\circ + \Delta V_m (P_2 - P_1)

Step 3: Apply the Equilibrium Condition

At thermodynamic equilibrium, the Gibbs free energy change for the phase change must be zero:

ΔrG(P2)=0\Delta_r G(P_2) = 0

Substitute the given values into the equilibrium equation:

2900 J mol−1+(−2×10−6 m3 mol−1)×(P2−105 Pa)=02900\text{ J mol}^{-1} + \left(-2 \times 10^{-6}\text{ m}^3\text{ mol}^{-1}\right) \times (P_2 - 10^5\text{ Pa}) = 0

2×10−6⋅(P2−105)=29002 \times 10^{-6} \cdot (P_2 - 10^5) = 2900

P2−105=29002×10−6=1.45×109 PaP_2 - 10^5 = \frac{2900}{2 \times 10^{-6}} = 1.45 \times 10^9\text{ Pa}

Step 4: Convert Pressure to bar

Since 1 bar=105 Pa1\text{ bar} = 10^5\text{ Pa}:

P2−1 bar=1.45×109 Pa105 Pa bar−1=14500 barP_2 - 1\text{ bar} = \frac{1.45 \times 10^9\text{ Pa}}{10^5\text{ Pa bar}^{-1}} = 14500\text{ bar}

P2=14500+1=14501 barP_2 = 14500 + 1 = 14501\text{ bar}

Thus, the equilibrium pressure required for graphite to be in thermodynamic equilibrium with diamond is 14501 bar14501\text{ bar}.

Correct Option: A

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