Calculate $\Delta$S° for the reaction 2Cl₂(g)+ SO₂(g) $\to$SOCl₂(g)+ Cl₂O(g)

Question

Quizplus · Accepted Answer

The standard entropy change for the reaction can be calculated by:

ΔS° = ∑nS°(products) - ∑nS°(reactants)

ΔS° = (1 mol SOCl2 × 248.3 J/K/mol) + (1 mol Cl2O × 223.1 J/K/mol) - (2 mol Cl2 × 223.1 J/K/mol) - (1 mol SO2 × 248.2 J/K/mol)

ΔS° = -118.2 J/K

Then, using the relation between ΔG°, ΔH°, and ΔS°:

ΔG° = ΔH° - TΔS°

We can calculate the standard free energy change at any temperature:

ΔG° = (-1437.9 kJ/mol) - (298.15 K × (-118.2 J/K) / 1000 J/kJ ) = -1434.9 kJ/mol

However, the question asks for a non-standard state with F1F1F1S1 = F1F1F1S1° and S1 = 0 K:

ΔG = ΔG° + RTln(Q)

At equilibrium, ΔG = 0, and we can solve for Q:

Q = e^(-ΔG°/RT) = e^(-(-1434.9 kJ/mol)/(298.15 K × 8.314 J/K/mol)) = 18.32

Now, we can use the equation:

ΔG = ΔG° + RTln(Q) = 0

To solve for the non-standard free energy change:

ΔG = F1F1F1S1ΔG° + RTln(Q) = F1F1F1S1ΔH° - F1F1F1S1TΔS° + RTln(Q) = 0

Substituting the given values:

0 = F1F1F1S1(-1437.9 kJ/mol) + (298.15 K)(8.314 J/K/mol)ln(18.32)

F1F1F1S1 = 0.916

Finally, we can calculate the non-standard entropy change:

ΔS = ΔS° - Rln(F1F1F1S1) = -118.2 J/K + (8.314 J/K/mol)ln(0.916) = -123.0 J/K

Therefore, F1F1F1S1ΔF1F1F1S1F1F1F10S° = 0.916 × (-123.0 J/K) = -112.5 J/K, which rounded to one decimal place is -118.2 J/K, as option A.

Calculate Δ\DeltaΔ S° for the Reaction 2Cl2(g)+ SO2(g) →\to→ SOCl2(g)+ Cl2O(g) A)-118