Determining Reaction Spontaneity

Question

Does the reaction P₄(s, white) + 5 O₂(g)  P₄O₁₀(s) occur spontaneously at 298 K and standard state conditions?

Calculate the value of S°_univ.

Answer 1

Yes, it is spontaneous. ΔS°_univ = +9.18 × 10³ J/K (about +9.18 kJ/K)

P₄(s, white) + 5 O₂(g) → P₄O₁₀(s)

The rule is ΔS_univ = ΔS_sys + ΔS_surr, and spontaneity requires the total to be positive. Both pieces are needed — that is the whole point of this problem.

Step 1 — ΔS°_sys from absolute entropies

S°: P₄O₁₀(s) = 228.9, P₄(s) = 41.1, O₂(g) = 205.2 J/(mol·K)

ΔS°_sys = 228.9 − [41.1 + 5(205.2)] = 228.9 − 1067.1 = −838.2 J/K

Strongly negative, and it should be — five moles of gas plus a solid collapse into a single solid. Whenever gas moles fall, expect ΔS_sys < 0. Judging spontaneity on this number alone would tell you the reaction cannot happen, which is plainly wrong: white phosphorus ignites in air.

Step 2 — ΔS°_surr from the enthalpy

ΔH°_rxn = ΔH°_f[P₄O₁₀] − 0 − 0 = −2984 kJ (elements in standard states have ΔH°_f = 0)

ΔS°_surr = −ΔH°_sys / T = −(−2 984 000 J) / 298 K = +10 013 J/K

Watch the two details that sink people here: the minus sign (heat released by the system is absorbed by the surroundings) and the kJ → J conversion, since entropies are tabulated in J/K.

Step 3 — combine

ΔS°_univ = −838 + 10 013 = +9175 J/K ≈ +9.18 kJ/K

Positive, so the reaction is spontaneous at 298 K.

What this problem is really teaching

The system becomes far more ordered, and the reaction runs anyway. The combustion is so violently exothermic that the heat dumped into the surroundings raises their entropy by more than twelve times what the system loses. "Entropy always increases" applies to the universe, never to the system alone — and this is the cleanest example of that you will meet.

Cross-check with free energy

ΔG° = ΔH° − TΔS°_sys = −2984 − (298)(−0.8382) = −2734 kJ

and independently ΔG° = −TΔS°_univ = −(298)(9.175) = −2734 kJ ✓

Those two routes agreeing confirms the arithmetic, and it shows why ΔG is useful at all: it is just ΔS_univ rescaled so you never have to leave the system.

(Tabulated values vary slightly between textbooks — ΔH°_f for P₄O₁₀ is sometimes listed near −3010 kJ/mol. Use your own appendix; the conclusion does not change.)