Given that $\Delta$H_vap is 58.2 kJ/mol,and the boiling point is 83.4°C,1 atm,if one mole of this substance is vaporized at 1 atm,calculate $\Delta$S_surr.

Question

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- The enthalpy change for vaporization (ΔHvap) is given as 58.2 kJ/mol. This means that it takes 58.2 kJ of energy to vaporize one mole of this substance at its boiling point.
- The problem asks us to calculate the entropy change for the surroundings (ΔSsurr). Since we know that the process is occurring at constant temperature and pressure, we can use the equation ΔG° = -TΔS° to relate the changes in free energy and entropy.
- We can rewrite this equation as ΔS° = -ΔG°/T. Since ΔG° = ΔH° - TΔS°, we can rearrange to get ΔS° = -(ΔH°/T) + (ΔG°/T).
- At constant pressure, the change in free energy (ΔG) is equal to the enthalpy change (ΔH) minus the product of the temperature and the change in entropy (ΔS). Therefore, we can rewrite the equation as ΔS° = -(ΔH°/T) + (ΔH°-ΔG°)/T.
- We know the value of ΔH°, but we need to calculate ΔG°. Since the process is occurring at 1 atm, we can assume that the substance is an ideal gas and use the ideal gas law to calculate the change in volume (ΔV) accompanying the vaporization. ΔG° = ΔH° - TΔS° = RTln(P2/P1) = RTln(1 atm/1 atm) = 0. Therefore, ΔS° = -(ΔH°/T) + (ΔH°-ΔG°)/T = -(58.2 kJ/mol)/(356.5 K) = -163 J/K mol. The negative sign indicates that the surroundings are becoming more ordered, which makes sense since a liquid is becoming a gas.

Given That Δ\DeltaΔ Hvap Is 58 Δ\DeltaΔ Ssurr

Given That $\Delta$ H_vap Is 58 $\Delta$ S_surr