The mineral graphite is more stable than diamond under standard conditions
CHEMISTRY · Multi-part · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry
Problem
The mineral graphite is more stable than diamond under standard conditions. Thermodynamic tables list the following standard entropy changes at 298 K:
Reaction 1 C(s, diamond) → C(s, graphite) ΔS°₁ = +2.4 J mol⁻¹ K⁻¹
Reaction 2 C(s, graphite) + O₂(g) → CO₂(g) ΔS°₂ = +179.0 J mol⁻¹ K⁻¹
Consider the overall oxidation of diamond:
C(s, diamond) + O₂(g) → CO₂(g) (Reaction 3)
Answer the following questions.
(a) Calculate the standard entropy change ΔS°₃ for Reaction 3 at 298 K.
(b) Provide a qualitative explanation, in terms of the nature of entropy as a state function, for why the value of ΔS°₃ can be obtained by algebraically combining the values of ΔS°₁ and ΔS°₂.
(c) Based solely on the physical states and number of particles in Reactants and Products of Reaction 3, justify the expected sign of ΔS°₃ and comment on whether your calculated value is consistent with that expectation.