Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment is 5.79 kJ/∘C ?

I got #-3.99 xx 10^(3)# #"kJ/mol"#.

Bomb calorimeters are constant-volume calorimeters, which implies that the heat flow #q# is equal to the change in internal energy, #DeltaE#, i.e.

#q_V = DeltaE#

We know that the heat flow is given by:

#q_V = C_VDeltaT#

where you were given that the constant-volume heat capacity is #C_V = "5.79 kJ/"^@ "C"# (rather than the specific heat capacity in #"kJ/g"^@ "C"#).

So, we can first calculate #q_V# to be:

#q_V = "5.79 kJ/"^@ "C" cdot (38.22^@ "C" - 25.74^@ "C")#

#=# #72.2_(592)# #"kJ"#

We then use the definition of #q_V = DeltaE#, dividing by the mols of limiting reagent (hexane) to get:

#q_V/n_(LR) = DeltabarE_(sol n)#,

where #n_(LR)# is the mols of limiting reagent and #DeltabarE_(sol n) -= (DeltaE_(sol n))/n_(LR)# is the change in molar internal energy of the solution.

The heat #q_V# transferred out from the product into the calorimeter is about #"72.26 kJ"#. With respect to the system, #q_V < 0#.

From here, we just need the mols of hexane:

#1.560 cancel"g hexane" xx ("1 mol hexane")/(6cdot12.011 + 14cdot1.0079 cancel"g")#

#= 0.0181_(02)# #"mols hexane combusted"#

Therefore, the change in molar internal energy for the reaction, which is EXOTHERMIC with respect to the system, is:

#color(blue)(DeltabarE_(rxn)) = -DeltabarE_(sol n) = color(red)(-)(q_V)/n_(LR) = color(red)(-)"72.2592 kJ"/"0.018102 mols hexane"#

#=# #color(blue)(-3.99 xx 10^(3))# #color(blue)("kJ/mol hexane")#

to three sig figs.