The energy required to completely separate the molecules(in case of vaporization), i.e to convert liquid into gas, is much greater that if you were just to reduce their separation, solid to liquid. Why is entropy of vaporization greater than entropy of fusion? If each configuration is equally probable, then the entropy is the natural logarithm of the number of configurations, multiplied by Boltzmann’s constant: S = k B ln W.Entropy is a measure of probability and the molecular disorder of a macroscopic system. ![]() Since G<0, the vaporization of water is predicted to occur spontaneously and irreversibly at 110 C. Thus G=0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions. The higher than expected value for water (data book value +116.9 J K –1 mol –1 ) arises due to its more ordered liquid state, as a result of hydrogen bonding, so vapoursation leads to a much greater increase of disorder.The energy required for vaporization offsets the increase in entropy of the system. ![]() ∆S vapourisation= ∆H/T=43.6 x 1000/373 = +116.9 J K –1 mol –1Ī typical liquid, that obeys Trouton’s rule, has a ΔS ⦵ vap of around +85 J K –1 mol –1 as the ratio ΔH ⦵ vap / T B is a constant. Moles of liquid water vapourised per second = 1/18 = 0.055 mol/s Typically a 2.4kW kettle, delivering 2400 J/s of energy, will boil away 1g of water per second. 100s) is used to determine: mass of water vapourised and the energy needed to achieved this vapourisation. The change in mass (or volume) of water at its boiling point (373K) for a known time period (e.g. In this experiment, a kettle, of known electrical power and containing an exact mass (or volume) of water is allowed to reach boiling point. ∆S= ∆H/T=44.3 x 1000/373 = +118.8 J K –1 mol –1Įxperimental determination of ∆S for vaporisation of water using a kettle Moles of liquid water = 1.53/18 So, ΔH ⦵= 3.766 ÷ 1.53/18 = +44.3 kJ mol –1Īssuming liquid to gas phase change occurs at constant temperature at 373K, then ∆G =0, so ∆G=∆H – T∆S=0 If 3.766 kJ heat energy converts 1.53g of liquid water into steam at 373K and 100kPa, calculate ΔH ⦵ (kJ mol –1) and ΔS ⦵ ( J K –1 mol –1) enthalpy change for H 2O(l) → H 2O(g) Thus, T= ∆H/ ∆S = 273 K So, if T>273K then ∆G ΔH, enabling ΔG<0.Įxamples of energy calculations relating to water Melting ice is an endothermic process: energy must be supplied in order to break hydrogen bonds between water molecules arranged in an ice crystalline structure.įrom ∆G = ∆H – T∆S, the temperature at which ∆ G ϴ = 0 for melting ice occurs when ∆H = T∆S
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