Trouton's rule states that for many liquids at their normal...

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about $88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$. (a) Estimate the normal boiling point of bromine, $\mathrm{Br}_{2}$, by determining $\Delta H_{\text {vap }}^{\circ}$ for $\mathrm{Br}_{2}$ using data from Appendix C. Assume that $\Delta H_{\text {vap }}^{\circ}$ remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of $\mathrm{Br}_{2}$ in a chemistry handbook or at the WebElements Web site (www.webelements.com).

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about $88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$.
(a) Estimate the normal boiling point of bromine, $\mathrm{Br}_{2}$, by determining $\Delta H_{\text {vap }}^{\circ}$ for $\mathrm{Br}_{2}$ using data from Appendix C. Assume that $\Delta H_{\text {vap }}^{\circ}$ remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of $\mathrm{Br}_{2}$ in a chemistry handbook or at the WebElements Web site (www.webelements.com).

Key Concepts

Trouton's Rule

Trouton's rule is an empirical observation that many liquids have a roughly constant standard molar entropy of vaporization (about 88 J/mol·K) at their normal boiling points. This rule allows for simple estimations of boiling points when the enthalpy of vaporization is known, as the entropy change during vaporization is assumed to be nearly constant for many substances.

Standard Molar Entropy of Vaporization

The standard molar entropy of vaporization is the change in entropy when one mole of a liquid converts to its gas phase at its boiling point and under standard conditions. It quantifies the increase in disorder as molecules transition from the ordered liquid phase to a more disordered gas phase.

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of heat required to convert one mole of a liquid into its gaseous form at constant pressure. It is a key thermodynamic parameter that, when combined with the entropy of vaporization, can be used to determine the boiling point using relations derived from thermodynamic equilibrium conditions.

Normal Boiling Point

The normal boiling point is defined as the temperature at which a liquid's vapor pressure equals standard atmospheric pressure (1 atm). At this temperature, the liquid and vapor phases are in equilibrium, and the relationship between enthalpy and entropy of vaporization is critical in determining this equilibrium state.

Transcript

00:01 Let's take a look at the vaporization of bromine, so going from the liquid to the gas state.

00:08 And let's say we wanted to estimate the temperature at which this is going to happen, the vaporization temperature or the boiling temperature.

00:19 Well, one way that we can do this estimation is to use trouton's rule.

00:24 And trouton's rule says that liquids, generally have a molar entropy of vaporization of about 88 joules per mole kelvin.

00:44 And so the question is, if we use this estimation, what do we wind up with for the vaporization or boiling point of bromine? and then let's compare that to the actual value and see how close we get.

01:02 So we're dealing with an equilibrium when we're dealing with a phase change.

01:06 And when we're dealing with an equilibrium, we know that the gibbs free energy of that process is going to be zero.

01:12 We also know that the gibbs free energy is going to be given as delta h minus t delta s.

01:26 And so we're dealing with vaporization here.

01:29 And we know that's equal to zero at equilibrium.

01:37 So what that means for us is that delta h of vaporization is equal to t delta s of vaporization.

01:45 And so if we want to know that t, which is our t of vaporization, that should be equal to the delta h of vaporization over the delta s of vaporization.

02:01 Okay, well, we're going to look at our change in vaporization here in terms of the delta h of vaporization.

02:08 Of our delta s.

02:10 So we know that we're going to plug in 88 for our entropy, but what are we going to do for our enthalpy? well, for our enthalpy, we can take a look at the delta h of our vaporization that's going from liquid to gas.

02:28 And so that's going to wind up being the sum of the number of moles times the enthalpy of formation our products minus the sum of the delta of the h of enthalpy of our reactants so we do have to look up the entope of the entopees of liquid bromine and gaseous bromine and do that math so product first the enthalpy of formation for gaseous bromine is 30 .71 kilojoules.

03:20 There's only one mole there, so times one...

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