Chapter 5, Problem 5.35E

The densities of graphite and diamond are $2.25$ and $3.51\text{g}/{\text{cm}}^3$, respectively. Using the expression

${\Delta }_{\text{rxn}}G={\Delta }_{\text{rxn}}G°+RT\ln \frac{a_{\text{dia}}}{a_{\text{gra}}}$ and equation 5.14, estimate the pressure necessary for ${\Delta }_{\text{rxn}}G$ to equal zero. What is the stable high-pressure solid phase of carbon?

Interpretation Introduction

Interpretation:

The pressure required for ${\Delta }_{\text{rxn}}G$ to become zero and the stable high-pressure solid phase is to be predicted.

Concept introduction:

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\Delta }_{\text{rxn}}G°$ whereas ${\Delta }_{\text{rxn}}G$ depends on the reaction condition and extent of reaction.

Answer to Problem 5.35E

The pressure required for ${\Delta }_{\text{rxn}}G$ to become zero is $1.49\times {10}^4\text{atm}$. This value of pressure is the stable high-pressure of solid carbon.

Explanation of Solution

The density of graphite is $2.25\text{g}/{\text{cm}}^3$.

The density of diamond is $3.51\text{g}/{\text{cm}}^3$.

Graphite and diamond exists in equilibrium as,

$\text{C}\left(\text{graphite}\right)\rightleftharpoons \text{C}\left(\text{diamond}\right)$

The standard Gibbs free energy change for the above reaction is calculated by the formula,

${\Delta }_{\text{rxn}}G°={\Delta }_{\text{f}}G\left(\text{diamond}\right)-{\Delta }_{\text{f}}G\left(\text{graphite}\right)$ (1)

Where,

• ${\Delta }_{\text{f}}G\left(\text{diamond}\right)$ is the Gibbs free energy of formation of diamond.

• ${\Delta }_{\text{f}}G\left(\text{graphite}\right)$ is the Gibbs free energy of formation of graphite.

The Gibbs free energy of formation of diamond and graphite is $2.9\text{kJ}{\text{mol}}^{-1}$ and $0\text{kJ}{\text{mol}}^{-1}$, respectively.

Substitute the value of Gibbs free energy of formation of diamond and graphite in equation (1).

$\begin{array}{rll}{\Delta }_{\text{rxn}}G°& =& 2.9\text{kJ}{\text{mol}}^{-1}-0\text{kJ}{\text{mol}}^{-1}\\ & =& 2.9\text{kJ}{\text{mol}}^{-1}\end{array}$

According to the given equation,

${\Delta }_{\text{rxn}}G={\Delta }_{\text{rxn}}G°+RT\ln \frac{a_{\text{dia}}}{a_{\text{gra}}}$

The value of ${\Delta }_{\text{rxn}}G$ is zero if the value of $RT\ln \frac{a_{\text{dia}}}{a_{\text{gra}}}$ is,

$RT\ln \frac{a_{\text{dia}}}{a_{\text{gra}}}=-2.9\text{kJ}{\text{mol}}^{-1}$

The above equation can also be written as,

$RT\left(\ln a_{\text{dia}}-\ln a_{\text{gra}}\right)=-2.9\text{kJ}{\text{mol}}^{-1}$ (2)

According to equation 5.14, the value of $\ln a_{\text{dia}}$ and $\ln a_{\text{gra}}$ is,

$\begin{array}{rll}\ln a_{\text{dia}}& =& \frac{{\overline{V}}_{\text{dia}}}{RT}\left(p-1\right)\\ \ln a_{\text{gra}}& =& \frac{{\overline{V}}_{\text{gra}}}{RT}\left(p-1\right)\end{array}$

Where,

• ${\overline{V}}_{\text{gra}}$ is the molar volume of graphite.

• ${\overline{V}}_{\text{dia}}$ is the molar volume of diamond.

Substitute the value of $\ln a_{\text{dia}}$ and $\ln a_{\text{gra}}$ in equation (2).

$\begin{array}{r}RT\left(\frac{{\overline{V}}_{\text{dia}}}{RT}\left(p-1\right)-\frac{{\overline{V}}_{\text{gra}}}{RT}\left(p-1\right)\right)=-2.9\text{kJ}{\text{mol}}^{-1}\\ \left({\overline{V}}_{\text{dia}}-{\overline{V}}_{\text{gra}}\right)\left(p-1\right)=-2.9\text{kJ}{\text{mol}}^{-1}\end{array}$ (3)

The molar volume of diamond is calculated by the formula,

${\overline{V}}_{\text{dia}}=\frac{M_{\text{dia}}}{d_{\text{dia}}}$

Where,

• $M_{\text{dia}}$ is the molar mass of diamond.

• $d_{\text{dia}}$ is the density of diamond.

Substitute the molar mass and density of diamond in above formula.

$\begin{array}{rll}{\overline{V}}_{\text{dia}}& =& \frac{12\text{g}{\text{mol}}^{-1}}{3.51\text{g}{\text{cm}}^{-3}}\\ & =& 3.42{\text{cm}}^3{\text{mol}}^{-1}\end{array}$ (4)

The molar volume of graphite is calculated by the formula,

${\overline{V}}_{\text{gra}}=\frac{M_{\text{gra}}}{d_{\text{gra}}}$

Where,

• $M_{\text{gra}}$ is the molar mass of graphite.

• $d_{\text{gra}}$ is the density of graphite.

Substitute the molar mass and density of graphite in above formula.

$\begin{array}{rll}{\overline{V}}_{\text{gra}}& =& \frac{12\text{g}{\text{mol}}^{-1}}{2.25\text{g}{\text{cm}}^{-3}}\\ & =& 5.33{\text{cm}}^3{\text{mol}}^{-1}\end{array}$ (5)

Substitute equation (4) and equation (5) in equation (3).

$\begin{array}{rll}\left(3.42{\text{cm}}^3{\text{mol}}^{-1}-5.33{\text{cm}}^3{\text{mol}}^{-1}\right)\left(p-1\right)& =& -2.9\text{kJ}{\text{mol}}^{-1}\\ p& =& \frac{-2.9\text{kJ}{\text{mol}}^{-1}}{-1.91{\text{cm}}^3{\text{mol}}^{-1}}+1\\ & =& 2.52\text{kJ}{\text{cm}}^3\end{array}$

Convert $2.52\text{kJ}{\text{cm}}^3$ to $\text{atm}$.

$\begin{array}{rll}2.52\text{kJ}{\text{cm}}^3& =& 2.52\text{kJ}{\text{cm}}^3\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\left(\frac{1{\text{dm}}^{-3}\text{atm}}{101.32\text{J}}\right)\left(\frac{{10}^{-3}{\text{m}}^{-3}}{1{\text{dm}}^{-3}}\right)\left(\frac{{10}^6{\text{cm}}^3}{1{\text{m}}^{-3}}\right)\\ & =& 1.49\times {10}^4\text{atm}\end{array}$

Therefore, the pressure required for ${\Delta }_{\text{rxn}}G$ to become zero is $1.49\times {10}^4\text{atm}$. This value of pressure is the stable high-pressure of solid carbon.

Conclusion

The pressure required for ${\Delta }_{\text{rxn}}G$ to become zero is $1.49\times {10}^4\text{atm}$. This value of pressure is the stable high-pressure of solid carbon.