All chemical changes in living and non–living systems obey laws of thermodynamics. The equilibrium c — Physical Chemistry — Thermodynamics Chemistry Question
Enthalpy, entropy, and stability
All chemical changes in living and non–living systems obey laws of thermodynamics. The equilibrium constant of a given reaction is determined by changes in Gibbs free energy, which is in turn determined by enthalpy change, entropy change, and the temperature.
Fill in the blanks (a–f) with all that apply from the following:
equilibrium constant, Keq
entropy change, ∆S
enthalpy change, ∆H
free energy change, ∆G
(a) strongly temperature–dependent ( )
(b) closely related to bond strength ( )
(c) measure of change in randomness ( )
(d) related to the quantity of reactants and products ( )
(e) measure of spontaneity of a reaction ( )
(f) measure of heat released or absorbed ( )
Model Answer
(a) Keq and ∆G
(b) ∆H
(c) ∆S
(d) Keq
(e) ∆G
(f) ∆H
The following equilibrium exists in the vapor phase dissociation of molecular addition compounds of donor molecules, D, and boron compounds, BX3.
D⋅BX3(g) ↔ D(g) + BX3(g)
Kp = [D][BX3] / [D⋅BX3]
Dissociation constants (Kp) of the molecular addition compounds Me3N·BMe3 and Me3P·BMe3 at 100 °C are 0.472 and 0.128, respectively. Calculate the standard free energy change of dissociation at 100 °C for both compounds. Which complex is more stable at this temperature?
Model Answer
From ∆G = –RT lnKp,
∆G is 1.52 kcal mol–1 for Me3P·BMe3 and 0.56 kcal mol–1 for Me3N·BMe3.
Me3P·BMe3 is more stable (less likely to dissociate) than Me3N·BMe3 at 100°C.
The standard entropy change of dissociation, ∆S°, is 45.7 cal mol–1 K–1 for Me3N⋅BMe3 and 40.0 cal mol–1 K–1 for Me3P⋅BMe3. Calculate the standard enthalpy change of dissociation for both compounds. Which compound has the stronger central bond? Assume that ∆H and ∆S are temperature–independent.
Model Answer
∆G = ∆H – T ∆S
∆H373 = ∆G373 + 373 ∆S373 ≈ ∆G373 + 373 ∆S°,
Me3N·BMe3 : ∆H = 0.56 kcal mol–1 + (373 K × 45.7 cal mol–1 K–1) = 17.6 kcal mol–1
Me3P·BMe3 : ∆H = 1.52 kcal mol–1 + (373 K × 40.0 cal mol–1 K–1) = 16.4 kcal mol–1
More heat is needed to dissociate Me3N·BMe3. Therefore, the N–B central bond is stronger.
Which is more critical in determining the overall stability of these addition compounds at 100°C, enthalpy term (∆H) or entropy term (T ∆S)?
Model Answer
Me3N·BMe3 :
∆H = 17.6 kcal mol–1 ; –T ∆S = – 373 K × 45.7 cal mol–1 K–1 = –17.05 kcal mol–1
∆G = 0.56 kcal mol–1
Me3P·BMe3 :
∆G = 16.4 kcal mol–1 ; – T ∆S = – 373 K × 40.0 cal mol–1 K–1 = –14.92 kcal mol–1
∆G = 1.52 kcal mol–1
Enthalpy change is larger for Me3N·BMe3, however, larger increase in the entropy term leads to a smaller increase in Gibbs free energy for Me3N·BMe3.
At what temperature does Me3N⋅BMe3 become more thermodynamically stable than Me3P⋅BMe3? Assume that ∆H and ∆S are temperature–independent.
Model Answer
17 600 cal mol–1 – (45.7 cal mol–1K) T > 16 400 cal mol–1 – (40.0 cal mol–1K) T
5.7 (cal mol–1K) T < 1 200 cal mol–1
T < 210K (– 63 °C)