What is the change in internal energy of reaction when #"0.721 g"# of titanium is placed into a bomb calorimeter and allowed to combust, if the temperature rises from #25^@ "C"# to #53.8^@ "C"#? #C_"cal" = "9.84 kJ/K"#.

#A)# #"283.4 kJ/mol"#
#B)# #-"283.4 kJ/mol"#
#C)# #1.89 xx 10^4 "kJ/mol"#
#D)# #-1.89 xx 10^4 "kJ/mol"#

1 Answer
Nov 6, 2016

I'm getting #D#.


A bomb calorimeter by definition is a constant-volume calorimeter. These are rigid and closed, generally with a large water bath to insulate the system.

This is in such a way that heat generated or absorbed due to the reaction is equal to the heat absorbed or generated (respectively) by the calorimeter (conservation of energy).

https://www.learner.org/

That means you must recall that #DeltaE = q_V#, where #q_V# is the heat flow #q# at constant volume, since #DeltaE = q + w = q - cancel(PDeltaV)^(0)#.

The following data was given:

  • #m_"Ti" = "0.721 g"#
  • #DeltaT = T_2 - T_1 = 53.8^@ "C" - 25^@ "C" = 28.8^@ "C"#

  • #C_"cal" = "9.84 kJ/K"#, or #"9.84 kJ/"^@ "C"#
    (note: #DeltaT# in #""^@ "C"# is the same as #DeltaT# in #"K"#, but the individual temperature values are NOT.)

Since you were given the heat capacity of the calorimeter, you can calculate #q_V# for the calorimeter itself:

#color(green)(q_(V,"cal")) = m_"cal" c_"cal"DeltaT#

#= C_"cal"DeltaT#

#= ("9.84 kJ/"^@ "C")(28.8^@ "C")#

#=# #color(green)("283.4 kJ")#

where #c# is the specific heat capacity in #"kJ/mol"^@ "C"# and #C# is the heat capacity in #"kJ/"^@ "C"#. Do not confuse them---they are not the same!

So, we can set this equal to #DeltaE_"cal"# as we stated earlier:

#DeltaE_"cal" = q_(V,"cal") = "283.4 kJ"#

Finally, in #"kJ/mol Ti"#, we must calculate the #"mol"#s of #"Ti"# combusted:

#n_"Ti" = m_"Ti"/(M_(m,"Ti"))#

#= "0.721 g" xx "1 mol Ti"/"47.867 g" = "0.0151 mols Ti"#

In the end, calculating the change in internal energy #DeltaE# (some might call it #DeltaU# to not confuse it with the total energy) for the reaction, which is also the heat of reaction in this case, would ask you to realize that:

#q_(V,"rxn") = -q_(V,"cal") = -"283.4 kJ"#

when energy is conserved. Therefore, for the reaction, we have that:

#color(blue)(DeltaE_"rxn") = q_"rxn"/(n_"Ti")#

#= -"283.4 kJ"/("0.0151 mols Ti")#

#= -"18814.3 kJ/mol"#

#= color(blue)(-1.89 xx 10^4)# #color(blue)("kJ/mol")#