Please Help - More Trouble with Phase Changes?

A substance, X, has the following properties.
Change in Hvap = 20.0 kJ/mol (for Hvap, C(solid) = 3.0 J/gCelsius)
Change in Hfus = 5.0 kJ/mol (for Hfus, the C(liquid) = 2.5 J/g
Celsius)
Boiling Point = 75 degrees Celcius (for C(gas), = 1.0 J/g*Celsius)
Melting Point = -15 degrees Celsius

Calculate the energy that must be removed to convert 250. grams of the substance X from a gas at 109°C to a solid -52.4°C. Assume X has a molar mass of 75.0 g/mol.

1 Answer
Jan 29, 2018

Heat capacities are never needed for phase changes... they require the temperature to change.

What should the SIGN of #q# be?


Map it out...

#X(g) stackrel(109.0^@ "C" -> 75.0^@ "C"" ")(->) X(g) stackrel(75.0^@ "C"" ")(->) X(l) stackrel(75.0^@ "C" -> -15.0^@ "C"" ")(->) X(l) stackrel(-15.0^@ "C"" ")(->) X(s) stackrel(-15.0^@ "C" -> -52.4^@ "C"" ")(->) X(s)#

And so we have five steps. But they're always the same steps.

#overbrace("cool")^(q = mC_PDeltaT) -> overbrace("condense")^(q prop nDeltaH_"vap") -> overbrace("cool")^(q = mC_PDeltaT) -> overbrace("freeze")^(q prop nDeltaH_"fus") -> overbrace("cool")^(q = mC_PDeltaT)#

and they're alternating. Oh, and...

You never EVER use heat capacities when condensing, melting, vaporizing, freezing, etc... they are only for heating or cooling.

#1)# Cooling gas

#q_1 = mC_PDeltaT_1#

#= "250. g" cdot "1.0 J/g"^@ "C" cdot (75.0 - 100.0^@ "C")#

#= -"6250 J" = -"6.3 kJ"#

#2)# Condensing gas to liquid

#q_2 = nDeltaH_"cond"#

#= -nDeltaH_"vap"#

#= -"250. g" xx "1 mol"/("75.0 g") cdot "20.0 kJ/mol"#

#= -"66.7 kJ"#

#3)# Cooling liquid

#q_3 = mC_PDeltaT_3#

#= "250. g" cdot "2.5 J/g"^@ "C" cdot (-15.0 - 75.0^@ "C")#

#= -"56250 J" = -"56.3 kJ"#

#4)# Freezing liquid into solid

#q_4 = nDeltaH_"frz"#

#= -nDeltaH_"melt" -= -nDeltaH_"fus"#

#= -"250. g" xx "1 mol"/("75.0 g") cdot "5.0 kJ/mol"#

#= -"16.7 kJ"#

#5)# Cooling solid

#q_5 = mC_PDeltaT_5#

#= "250. g" cdot "3.0 J/g"^@ "C" cdot (-52.4 - (-15.0^@ "C"))#

#= -"28050 J" = -"28.1 kJ"#

And so, the net heat REMOVED is

#color(blue)(|q|) = |q_1+q_2+q_3+q_4+q_5|#

#= |-"6.3 kJ" - "66.7 kJ" - "56.3 kJ" - "16.7 kJ" - "28.1 kJ"|#

#= color(blue)(+"174.1 kJ")#