Using the following data, estimate the boiling point of carbon disulfide, CS2: CS2 (l) → CS2 (g) ΔH° = 27.2 kJ/mol ΔS° = 86.66 J/K·mol

The Correct Answer and Explanation is:

To estimate the boiling point of carbon disulfide (CS₂), we use the Gibbs free energy equation at equilibrium:

Step 1: Use the Gibbs Free Energy Equation

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ – T\Delta S^\circΔG∘=ΔH∘−TΔS∘

At the boiling point, the liquid and gas phases are in equilibrium, so:ΔG∘=0\Delta G^\circ = 0ΔG∘=0

Thus:0=ΔH∘−TΔS∘⇒T=ΔH∘ΔS∘0 = \Delta H^\circ – T\Delta S^\circ \Rightarrow T = \frac{\Delta H^\circ}{\Delta S^\circ}0=ΔH∘−TΔS∘⇒T=ΔS∘ΔH∘

Step 2: Convert Units

ΔH° = 27.2 kJ/mol = 27,200 J/mol
ΔS° = 86.66 J/mol·K

Step 3: Plug into the Equation

T=27,200 J/mol86.66 J/mol\cdotpK=313.8 KT = \frac{27,200\ \text{J/mol}}{86.66\ \text{J/mol·K}} = 313.8\ \text{K}T=86.66 J/mol\cdotpK27,200 J/mol=313.8 K

Step 4: Convert to Celsius

T(°C)=313.8 K−273.15=40.65∘CT(°C) = 313.8\ \text{K} – 273.15 = \boxed{40.65^\circ \text{C}}T(°C)=313.8 K−273.15=40.65∘C

Final Answer: The estimated normal boiling point of CS₂ is approximately 40.65°C.

Explanation

To determine the normal boiling point of carbon disulfide (CS₂), we apply thermodynamic principles involving phase equilibrium. The boiling point is the temperature at which the liquid and gas phases of a substance are in equilibrium at 1 atm pressure. At this point, the Gibbs free energy change (ΔG°) for vaporization is zero because there is no net change in energy when the substance moves between liquid and vapor states.

We start by using the standard Gibbs free energy equation:ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ – T\Delta S^\circΔG∘=ΔH∘−TΔS∘

At the boiling point, ΔG∘=0\Delta G^\circ = 0ΔG∘=0, so the equation simplifies to:T=ΔH∘ΔS∘T = \frac{\Delta H^\circ}{\Delta S^\circ}T=ΔS∘ΔH∘

Given values are:

Enthalpy of vaporization, ΔH∘=27.2 kJ/mol=27,200 J/mol\Delta H^\circ = 27.2\ \text{kJ/mol} = 27,200\ \text{J/mol}ΔH∘=27.2 kJ/mol=27,200 J/mol
Entropy of vaporization, ΔS∘=86.66 J/mol\cdotpK\Delta S^\circ = 86.66\ \text{J/mol·K}ΔS∘=86.66 J/mol\cdotpK

Substituting into the equation:T=27,20086.66≈313.8 KT = \frac{27,200}{86.66} \approx 313.8\ \text{K}T=86.6627,200≈313.8 K

To convert this to Celsius, subtract 273.15:T(°C)=313.8−273.15=40.65∘CT(°C) = 313.8 – 273.15 = 40.65^\circ \text{C}T(°C)=313.8−273.15=40.65∘C

Thus, the estimated normal boiling point of carbon disulfide is approximately 40.65°C, consistent with its known physical properties. This method provides a useful estimation by applying fundamental thermodynamic relationships.