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
I'm getting
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).
That means you must recall that
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
#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" = q_(V,"cal") = "283.4 kJ"#
Finally, in
#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
#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")#