Q20.82 CP - Chemistry: Molecular Nature Of Matter And Change

Question

According to advertisements, "a diamond is forever."

(a) Calculate $\mathit{\Delta }{\mathit{H}}^{\mathbf{°}}\mathbf{,}\mathit{\Delta }{\mathit{S}}^{\mathbf{°}}$ , and  $\mathit{\Delta }{\mathit{G}}^{\mathbf{°}}$ at 298k for the phase change Diamond $\to$graphite.

(b) Given the conditions under which diamond jewelry is normally kept, argue for and against the statement in the ad.

(c) Given the answers in part (a), what would need to be done to make synthetic diamonds from graphite?

(d) Assuming  $\mathit{\Delta }{\mathit{H}}^{\mathbf{°}}$ and  $\mathit{\Delta }{\mathit{S}}^{\mathbf{°}}$ do not change with temperature, can graphite be converted to diamond spontaneously at 1 atm?

Step-by-Step Solution

Answer

a) For the given conditions, $\Delta H_{rxn}^{°}=-1.896\frac{\text{k}.\text{J}}{{\text{mol}}^{\text{'}}},\Delta S_{rxn}^{°}=3.427\frac{\text{kJ}}{\text{mol}\cdot \text{K}}$ and $\Delta G_{rxn}^{°}=-2.917 \text{kJ}$

b) Diamonds do change into the graphite allotrope of carbon, therefore the statement isn't entirely accurate. However, in general, the process takes a long time and is very slow in regular room settings.

c) At any temperature, the reaction is non-spontaneous.

d) As a result, it is clear that temperature has an impact on enthalpy and entropy. Otherwise, at any temperature, the reaction would be spontaneous.

Step 1: Calculate Gibb’s free energy

a) Create a chemical reaction for the phase change first:

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

Both enthalpy and entropy are state functions, which means they solely depend on the system's starting and ultimate states:

$\begin{array}{l}\Delta H_{rxn}^{°}=\Delta H_{\text{products }}^{°}-\Delta H_{\text{reactants }}^{°}\\ \Delta S_{rxn}^{°}=\Delta S_{\text{products }}^{°}-\Delta S_{\text{reactants }}^{°}\end{array}$

Then, using Appendix B, we can write and calculate that for enthalpy y.

$\begin{array}{l}\Delta H_{rxn}^{°}=\left(1\Delta H_{C\left(\text{ graphite }\right(s)}^{°}\right)-\left(1\Delta H_{C\left(\text{ diamond }\right)\left(s\right)}^{⌟}\right)\\ \Delta H_{rxn}^{°}=(1 \text{mol}\cdot \left(0.00\frac{\text{kJ}}{\text{mol}}\right))-(1 \text{mol}\cdot \left(1.896\frac{\text{kJ}}{\text{mol}}\right)\\ \Delta H_{rxn}^{°}=-1.896\frac{\text{kJ}}{\text{mol}}\end{array}$

Similarly, for entropy we can write and calculate that

$\begin{array}{l}\Delta S_{ran}^{°}=\left(1\Delta S_{C\left(\text{ graphite }\right)\left(s\right)}^{°}\right)-\left(1\Delta S_{C\left(\text{ diamond }\right)\left(s\right)}^{°}\right)\\ \Delta S_{rxn}^{°}=(1 \text{mol}\cdot \left(5.686\frac{\text{J}}{\text{mol}\cdot \text{K}}\right))-(1 \text{mol}\cdot \left(2.439\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\\ \Delta S_{rxn}^{°}=3.427\frac{\text{J}}{\text{mol}\cdot \text{K}}\end{array}$

Gibb's free energy change can be calculated at 298k :

$\begin{array}{l}\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}\\ \Delta G_{rxn}^{°}=-1.896\frac{\text{k}J}{ \text{mol}}-298 \text{K}\cdot \left(3.427\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\cdot {10}^3\frac{k\cdot J}{J}\\ \Delta G_{rxn}^{°}=-2.917 \text{kJ} \text{J}\end{array}$

Therefore, at the $298 \text{K }\Delta G_{rxn}^{°}=-2.917 \text{kJ}$

Step 2: Whether reaction is spontaneous or not

b) 

Based on the computed values, the diamonds are held at ambient temperature and pressure (298k and 1 atm).

$\Delta G_{rxn}^{°}=-2.917 \text{kJ}<0$

and

$\Delta S_{rxn}^{°}=3.427\frac{\text{J}}{\text{mol}\cdot \text{K}}$meaning that the reaction is spontaneous.

As a result, there's a probability that a spontaneous reaction will change graphite into diamonds. However, the rate of such a reaction would be very dependent on the other variables in place (like pressure and temperature).

$\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}(<0)-T\cdot \Delta S_{rxn}^{°}(>0)$.

Diamonds do change into the graphite allotrope of carbon, therefore the statement isn't entirely accurate. However, in general, the process takes a long time and is very slow in regular room settings.

Step 3: Calculate Gibb&rsquo;s free energy

c) 

The reaction that occurs when graphite is used to make diamonds is:

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

Enthalpy and entropy are the same thing, but with the opposite sign: For enthalpy, we can write and calculate that

$\begin{array}{c}\Delta H_{rxn}^{°}=-[\left(1\Delta H_{C\left(\text{ graphite }\right)\left(s\right)}^{°}\right)-(1\text{ DeltaH}{H}_{C\left(\text{ diamond }\right)\left(s\right)}^{°})]\\ \Delta H_{rxn}^{°}=1.896\frac{\text{kJ}J}{ \text{mol}}\end{array}$

Similarly, we may write and calculate entropy.

$\begin{array}{l}\Delta S_{rxn}^{°}=-[\left(1\Delta S_{C\left(\text{ graphite }\right)\left(s\right)}^{°}\right)-(1\text{ Delta }{S}_{C\left(\text{ diamond }\right)\left(s\right)}^{°})]\\ \Delta S_{rxn}^{°}=-3.427\frac{\text{kJ}}{\text{mol}\cdot \text{K}}\end{array}$

The change in Gibb's free energy can be estimated as follows:

$\begin{array}{l}\Delta G_{r\ge n}^{\infty }=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}\\ \Delta G_{rxn}^{\infty }=1.896\frac{\text{kJ}}{\text{mol}}-T \text{K}\cdot (-3.427\frac{\text{kJ}}{\text{mol}\cdot \text{K}})\cdot {10}^{-3}\frac{\text{kJ}}{\text{J}}.\end{array}$

Since enthalpy is positive and entropy is negative, the reaction would be non-spontaneous at all T. The reason is that graphite is a more stable carbon allotrope than diamond, so the reaction should be activated somehow.