"A diamond is forever" is one of the most successful advertising slogans of all time. But is it true? For the reaction shown below, calculate the standard free energy change at 298K and determine whether or not a diamond is "forever". C(diamond) $\to$ C(graphite)
Data: $\Delta$ H_f°(diamond) = 1.895 kJ/mol; S°(diamond) = 2.337 J mol^-1 K^-1;
S°(graphite) = 5.740 J mol^-1K^-1.

Question

"A diamond is forever" is one of the most successful advertising slogans of all time. But is it true? For the reaction shown below, calculate the standard free energy change at 298K and determine whether or not a diamond is "forever". C(diamond)

\to

C(graphite)
Data:

\Delta

H_f°(diamond) = 1.895 kJ/mol; S°(diamond) = 2.337 J mol^-1 K^-1;
S°(graphite) = 5.740 J mol^-1K^-1.

Quizplus · Accepted Answer

The standard free energy change at 298K can be calculated using the Gibbs-Helmholtz equation:

ΔG° = ΔH° - TΔS°

ΔH° = [ΣH(products)] - [ΣH(reactants)] 
ΔH° = [0] - [-1.895 kJ/mol] = 1.895 kJ/mol

ΔS° = [ΣS(products)] - [ΣS(reactants)] 
ΔS° = [5.740 J/mol-K] - [2.337 J/mol-K] = 3.403 J/mol-K

Plugging in the values:

ΔG° = 1.895 kJ/mol - (298 K)(3.403 J/mol-K) 
ΔG° = -2.9 kJ/mol

Since the ΔG° is negative, the reaction is spontaneous and favors the formation of graphite. This means that over time, a diamond will eventually turn into graphite, so it is not "forever" as the advertising slogan suggests. Therefore, the best choice is C.

"A Diamond Is Forever" Is One of the Most Successful →\to→

"A Diamond Is Forever" Is One of the Most Successful $\to$