Heat of VaporizationTable of contentsThe Heat (or Enthalpy) of Vaporization is the quantity of heat that must be absorbed if a certain quantity of liquid is vaporized at a constant temperature. In a solution with both a vaporized and liquidized states, the kinetic energy of the vapor is higher than the kinetic energy of the liquid. Temperature follows kinetic energy, showing a lower temperature in the remaining liquid. IntroductionThe Enthalpy of Vaporization is measured using constant temperature, meaning that something must be added to bring the lower temperature to the temperature of the solution prior to vaporization. To raise the temperature (and as a result, the kinetic energy), heat is added. And the Enthalpy of Vaporization measures the amount of heat added, because it is the quantity that must be absorbed for a liquid to remain at constant temperature during vaporization. The ΔHvap is expressed in kJ/mol.
The EquationΔHvaporization = Hvapor - Hliquid ΔHvaporization corresponds to the change in enthalpy of vaporization. It is usually written as ΔHvap. Hvapor and Hliquid are the absolute enthalpies of the given gas state and liquid state of a compound or element. Because they are not in the process of changing, it is impossible to measure their values. Therefore, Hliquid and Hvapor simply exist to represent the gas and liquid phase enthalpies. However, enthalpy is a function of state meaning that under a specific set of conditions, the change in enthalpy of the gas and liquid has a unique value. Absolute enthalpies cannot be measured, but by subtracting the absolute enthalpy of the vapor state and the absolute enthalpy of the liquid state, the given difference has a unique value that can be measured. Because heat is added to the system to maintain the temperature, vaporization is an endothermic process. This means that the ΔHvap is always positive. The Enthalpy of CondensationCondensation is the opposite of vaporization, and therefore ?Hcondensation is also the opposite of ?Hvaporization. Because ?Hvap is an endothermic process, where heat is lost in a reaction and must be added into the system from the surroundings, ?Hcondensation is an exothermic process, where heat is absorbed in a reaction and must be given off from the system into the surroundings. ΔHcondensation = Hliquid - Hvapor = -ΔHvap Because ΔHcondensation, also written as ΔHcond, is an exothermic process, its value is always negative. Moreover, ?ΔHcond is equal in magnitude to ΔHvap, so the only difference between the two values for one given compound or element is the positive or negative sign.
2.055 liters of steam at 100°C was collected and stored in a cooler container. What was the amount of heat involved in this reaction? (The ΔHvap of water = 44.0 kJ/mol.) Solution 1. First, convert 100°C to Kelvin. °C + 273.15 = K 2. Find the amount involved (in moles). nwater = PV/RT = (1.0 atm)(2.055 L)/(0.08206 L atm mol-1 K-1)(373.15 K) 3. Find the ΔHcond. ΔHcond = -ΔHvap, so 4. Using the ΔHcond of water and the amount in moles, calculate the amount of heat involved in the reaction. To find kJ, multiply the ΔHcond by the amount in moles involved. (ΔHcond)(nwater) Definitions:
Heat of VaporizationIn a system, a liquid is boiled and some of the molecules are converted to gas. The Heat of Vaporization corresponds to the heat that the liquid lost when the molecules phase changed. The Enthalpy of Vaporization, conversely, is the amount of heat applied to the system to boil the liquid. As a result, the temperature of the liquid remained constant, while the given heat was absorbed to convert the molecules. ΔHvap = Heat of Vaporization
If a liquid uses 50 Joules of heat to vaporize one mole of liquid, then what would be the enthalpy of vaporization? Solution:
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