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# The normal boiling point of water is 100.0 °C and its molar enthalpy of vaporization is 40.67 kJ/mol. What is the change in entropy in the system in J/K when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point?

##DeltaS = -"238 J K"^(-1)##

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

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

So 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, ##"J K"^(-)##. This lets you know that you're essentially looking for two things here

- the
**change in enthalpy**associated with this phase change - the
**temperature**at which it takes place, expressed in**Kelvin**

The **molar heat of vaporization** tells you how much heat is **released** when ##1## **mole** of water vapor condenses at its boiling point, which is ##100.0^@"C"## at normal pressure, to form ##1## **mole** of liquid water.

In your case, you have

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

This tells you that when ##1## **mole** of water goes from vapor at its boiling point to liquid at its boiling point, ##"40.76 kJ"## of heat are being **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 , ##DeltaH##. You can thus say that when ##"88.94 kJ"## of heat are being given off, the change in enthalpy is

##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 .