Thermodynamics of Sulfuric Acid Formation from Elemental Sulfur: 21-Mat-A1 Thermodynamics, May 2017, Q4

Question 4: Thermodynamics of Sulfuric Acid Formation from Elemental Sulfur (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $T=298$ K (25 °C). Standard enthalpies of formation and standard entropies (all per mole, kJ or J K⁻¹):

Given data — 298 K standard-state properties
Compound	ΔH°_f (kJ mol⁻¹)	S° (J K⁻¹ mol⁻¹)
S(s)	0	32
O₂(g)	0	205
SO₂(g)	−297	248
SO₃(g)	−396	257
H₂O(l)	−286	70
H₂SO₄(l)	−814	157

Find. $\Delta H^\circ$, $\Delta S^\circ$, $\Delta G^\circ$ and $K$ at 298 K for reactions (1)–(3); the balanced overall reaction $S+O_2+H_2O\to H_2SO_4$ and its $\Delta H^\circ$, $\Delta S^\circ$, $\Delta G^\circ$.

Approach. Apply Hess's law ($\Delta H^\circ=\sum\nu_i\Delta H^\circ_{f,i}$, $\Delta S^\circ=\sum\nu_iS^\circ_i$) to each reaction as written, form $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ and $K=\exp(-\Delta G^\circ/RT)$, then combine reactions (1)–(3) stoichiometrically for the overall path.

(a) Enthalpies of reaction. $$\Delta H^\circ_1=\Delta H^\circ_f(SO_2)=\boxed{-297\ \text{kJ}},\quad \Delta H^\circ_2=2\Delta H^\circ_f(SO_3)-2\Delta H^\circ_f(SO_2)=2(-396)-2(-297)=\boxed{-198\ \text{kJ}},$$ $$\Delta H^\circ_3=\Delta H^\circ_f(H_2SO_4)-\Delta H^\circ_f(SO_3)-\Delta H^\circ_f(H_2O)=-814-(-396)-(-286)=\boxed{-132\ \text{kJ}}.$$
(b) Entropies of reaction. $$\Delta S^\circ_1=S^\circ(SO_2)-S^\circ(S)-S^\circ(O_2)=248-32-205=\boxed{+11\ \text{J}\,\text{K}^{-1}},$$ $$\Delta S^\circ_2=2S^\circ(SO_3)-2S^\circ(SO_2)-S^\circ(O_2)=2(257)-2(248)-205=\boxed{-187\ \text{J}\,\text{K}^{-1}},$$ $$\Delta S^\circ_3=S^\circ(H_2SO_4)-S^\circ(SO_3)-S^\circ(H_2O)=157-257-70=\boxed{-170\ \text{J}\,\text{K}^{-1}}.$$
(c) Free energies of reaction ($T=298$ K). $$\Delta G^\circ_1=\Delta H^\circ_1-T\Delta S^\circ_1=-297{,}000-(298)(11)=\boxed{-300.3\ \text{kJ}},$$ $$\Delta G^\circ_2=-198{,}000-(298)(-187)=\boxed{-142.2\ \text{kJ}},\qquad \Delta G^\circ_3=-132{,}000-(298)(-170)=\boxed{-81.3\ \text{kJ}}.$$
(d) Equilibrium constants. $K=\exp(-\Delta G^\circ/RT)$, $RT=(8.314)(298.15)=2478.8$ J: $$\log_{10}K_1=\dfrac{-\Delta G^\circ_1}{2.303\,RT}=\boxed{52.6},\qquad \log_{10}K_2=\boxed{24.9},\qquad \log_{10}K_3=\boxed{14.2}.$$ All three reactions run essentially to completion at 298 K — consistent with strongly negative $\Delta G^\circ$ values in every step.
(e) Balanced overall reaction. Doubling (1) so it supplies the 2 mol SO₂ that (2) consumes, adding (2), then doubling (3) to consume the 2 mol SO₃ produced, and halving the sum: $$2\times(1):\ 2S+2O_2=2SO_2,\qquad (2):\ 2SO_2+O_2=2SO_3,\qquad 2\times(3):\ 2SO_3+2H_2O=2H_2SO_4.$$ Summing and dividing by 2: $$\boxed{S(s)+\tfrac32O_2(g)+H_2O(l)\ \longrightarrow\ H_2SO_4(l)}.$$
(f)–(h) Thermodynamics of the balanced reaction. Either sum $\bigl(2\Delta X_1+\Delta X_2+2\Delta X_3\bigr)/2$ for $X=H,S$, or apply Hess's law directly from formation data (both agree): $$\Delta H^\circ_e=\Delta H^\circ_f(H_2SO_4)-\Delta H^\circ_f(S)-\tfrac32\Delta H^\circ_f(O_2)-\Delta H^\circ_f(H_2O)=-814-0-0-(-286)=\boxed{-528\ \text{kJ}},$$ $$\Delta S^\circ_e=S^\circ(H_2SO_4)-S^\circ(S)-\tfrac32S^\circ(O_2)-S^\circ(H_2O)=157-32-307.5-70=\boxed{-252.5\ \text{J}\,\text{K}^{-1}},$$ $$\Delta G^\circ_e=\Delta H^\circ_e-T\Delta S^\circ_e=-528{,}000-(298)(-252.5)=\boxed{-452.7\ \text{kJ}}.$$

Reaction	ΔH° (kJ)	ΔS° (J/K)	ΔG° (kJ)	log₁₀K
(1) S+O₂=SO₂	−297	+11	−300.3	52.6
(2) 2SO₂+O₂=2SO₃	−198	−187	−142.2	24.9
(3) SO₃+H₂O=H₂SO₄	−132	−170	−81.3	14.2
(e–h) S+⅓O₂+H₂O=H₂SO₄	−528	−252.5	−452.7	—

Concept & Theory

Topic. Hess's law applied to a three-step industrial oxidation/hydration pathway (the contact process for sulfuric acid).

Key relations. $\Delta H^\circ=\sum\nu_i\Delta H^\circ_{f,i}$; $\Delta S^\circ=\sum\nu_iS^\circ_i$; $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-RT\ln K$.

Why this works. Enthalpy, entropy and Gibbs energy are all state functions, so the overall reaction's $\Delta H^\circ$, $\Delta S^\circ$ and $\Delta G^\circ$ can be built either directly from formation data or, equivalently, by summing the correctly-weighted individual steps — the two routes must agree exactly, which is a built-in check on the arithmetic.

Common pitfall. Forgetting to double reaction (1) before adding it to (2), or forgetting the final ÷2 — stoichiometric coefficients must balance across every intermediate species (SO₂, SO₃) before they cancel.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 6 (Hess's law, standard states, combining reactions).

Question 4 of 7: Thermodynamics of Sulfuric Acid Formation from Elemental Sulfur

Question 4: Thermodynamics of Sulfuric Acid Formation from Elemental Sulfur (20 marks)