# 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.

##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 .

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