# Question #3c9e4

##### 1 Answer

#### Explanation:

The first thing to notice here is that you're dealing with a *vapor* to *liquid* **phase change**, so right from the start you should expect the change in entropy, **negative**.

This is the case because you're going from * higher disorder and randomness*, as you get in the gaseous state, to

*, as you get in the liquid state.*

**lower**disorder and randomnessSo the entropy of the system is **decreasing**, i.e. it's going from higher disorder to lower disorder.

Next, notice that the problem tells you that the change in entropy must be expressed in *joules per Kelvin*,

thechange in enthalpyassociated with this phase changethetemperatureat which it takes place, expressed inKelvin

The **molar heat of vaporization** tells you how much heat is **released** when **mole** of water vapor condenses at its boiling point, which is **mole** of liquid water.

In your case, you have

#DeltaH_"vap" = "40.67 kJ mol"^(-1)#

This tells you that when **mole** of water goes from vapor at its boiling point to liquid at its boiling point, **given off**.

Use water's **molar mass** to calculate how many *moles* of water you have in your sample

#39.3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.182 moles H"_2"O"#

Use the molar hear of vaporization to find how much heat is being *given off* here

#2.182 color(red)(cancel(color(black)("moles H"_ 2"O"))) * overbrace("40.76 kJ"/(1color(red)(cancel(color(black)("mole H"_ 2"O")))))^(color(blue)(DeltaH_ "vap")) = "88.94 kJ"#

Now, the trick here is to realize that **heat given off** is associated with a **negative** change in enthalpy,

#DeltaH = - "88.94 kJ"#

Expressed in *joules*, this will be equal to

#-88.94 color(red)(cancel(color(black)("kJ"))) * (10^3"J")/(1color(red)(cancel(color(black)("kJ")))) = -"88,940 J"#

Next, convert the temperature at which the phase change takes places from *Celsius* to *Kelvin* by using the conversion factor

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

You will have

#T = 100.0^@"C" + 273.15 = "373.15 K"#

Now, the equation that connect change in entropy, change in enthalpy, and absolute temperature looks like this

#color(blue)(|bar(ul(color(white)(a/a)DeltaS = (DeltaH)/Tcolor(white)(a/a)|)))#

Plug in your values to find

#DeltaS = (-"88,940 J")/"373.15 K" = color(green)(|bar(ul(color(white)(a/a)color(black)(-"238 J K"^(-1))color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.