The enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (-33.4°C).Calculate the value of $\Delta$S when 1.00 mole of ammonia is vaporized at -33.4°C and 1.00 atm.

Question

Quizplus · Accepted Answer

The change in entropy ($\Delta S$) during the vaporization of a substance can be calculated using the formula $\Delta S = \frac{\Delta H_{vap}}{T}$, where $\Delta H_{vap}$ is the enthalpy of vaporization and $T$ is the temperature in Kelvin. Given $\Delta H_{vap} = 23.35$ kJ/mol for ammonia and its boiling point at $-33.4^\circ C$ or 239.75 K ($273.15 - 33.4$), the calculation is $\Delta S = \frac{23.35 \times 10^3 \, \text{J/mol}}{239.75 \, \text{K}} = 97.4 \, \text{J/K mol}$, which is approximately $9.74 \, \text{J/K mol}$.

The Enthalpy of Vaporization of Ammonia Is 23 Δ\DeltaΔ S When 100 Mole of Ammonia Is Vaporized at -33

The Enthalpy of Vaporization of Ammonia Is 23 $\Delta$ S When 100 Mole of Ammonia Is Vaporized at -33