For the vaporization of benzene, ΔHvap=30.7 kJ / mol and ΔSvap=87.0 J /(K ·mol) . Calculate ΔSsu

For the vaporization of benzene, ΔHvap=30.7 kJ / mol and...

Key Concepts

Entropy Change of the Surroundings

This concept involves calculating the change in entropy of the surroundings when a system undergoes a process. In phase transitions like vaporization, the surroundings lose or gain entropy based on the heat exchanged at a given temperature. The expression ?S_surr = -?H/T (where ?H is the enthalpy change for the process) is commonly used under the assumption that the surroundings remain at a constant temperature.

Total Entropy Change

Total entropy change (?S_total) is the sum of the entropy changes of both the system and the surroundings. It is a key indicator of a process's spontaneity according to the second law of thermodynamics; if ?S_total is positive, the process is spontaneous, and if it is zero, the process is in equilibrium.

Vaporization Enthalpy and Entropy

Vaporization enthalpy (?H_vap) and vaporization entropy (?S_vap) are fundamental parameters in determining how much energy is required for a substance to change from liquid to vapor and how the disorder of the system increases during this process. The ratio of these values is particularly useful in evaluating the temperature conditions under which phase changes occur.

Normal Boiling Point

The normal boiling point is defined as the temperature at which a liquid's vapor pressure equals the external pressure (typically 1 atm), meaning the liquid is in equilibrium with its vapor. At this temperature, the process is characterized by a balance between the energy absorbed during vaporization and the increase in entropy, and the Gibbs free energy change (?G = ?H - T?S) is zero.

Gibbs Free Energy in Phase Transitions

Gibbs free energy is a thermodynamic potential that combines enthalpy and entropy to determine the spontaneity of a process. In the context of phase transitions like boiling, the condition ?G = 0 identifies the equilibrium state. This balance helps in finding conditions, such as the normal boiling point, where the phase change occurs without net energy gain or loss.

Transcript

00:01 For the vaporization of benzene, and we're going to calculate the surrounding entropy and the total entropy at different temperatures.

00:11 So let's write out our givens here.

00:13 We have our heat of vaporization, which is going to be equal to 30 .7 kilojoules per mole.

00:24 And because we are dealing with entropy, let's go ahead and convert this into joules.

00:28 So we have 1 ,000 joules per one kilojoule.

00:32 So this is going to equal to 30 ,700 joules per mole.

00:39 And we are also given that the entropy of vaporization is going to be equal to 87 .0 joules per mole times kelvin.

00:51 And because this is a vaporization, our entropy of vaporization is also going to equal our our standard entropy for reaction, and same thing for our delta h vaporization is going to equal to our enthalpy of reaction.

01:12 Okay, so now that we have our givens, we can go ahead and start looking at these different temperatures here.

01:17 So for 70 degrees celsius, or 343 kelvin, we have our delta s standard equal to 87, 0 .0 joules per mole kelvin.

01:43 And so now to calculate our delta s surrounding is going to equal to negative delta h over t.

01:53 So we can use our delta h from above, our heat of vaporization, our 30 ,700 joules per mole, divided by our temperature of reaction, which is 343 kelvin.

02:08 So here we get that our surrounding entropy is negative 89 .5 joules per mole kelvin.

02:22 Okay.

02:24 So now that we have our system and surrounding entropy, we can go ahead and calculate our total entropy here.

02:32 So remember our total entropy is equal to our surrounding entropy times our system entropy.

02:39 So we can go ahead and add 87 .0 .8.

02:42 0 joules per mole kelvin, you can add that to our negative 89 .5 joules per more kelvin and get that our total entropy for this reaction is negative 2 .5 joules per mole kelvin.

03:08 Okay, so now let's do the next temperature.

03:11 So our next temperature here is 80 degrees centigrade.

03:16 So that's going to be 3503 kelvin.

03:23 So same process, we have our delta s for our system equal to 87 joules per mole kelvin.

03:35 Our delta s of surroundings is going to be equal to our negative delta h over t, which is going to be equal to our delta h 30 ,700 joules per mole divided by our near temperature here, which is 353.

03:54 And that's going to equal negative 86 .9 joules per mole kelvin.

04:12 Okay, so then our delta s total here, total, is going to equal 87 minus 86 .9.

04:27 So we're going to get 0 .1 joules per mole kelvin has our answer here...

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