#"1.659 g"# of sample benzoic acid was placed in a bomb calorimeter. The complete combustion raised the temperature of the surrounding water from #27.72^@ "C"# to #34.04^@ "C"#? What is the enthalpy of combustion?

Basically, the exothermic #DeltaH_C# is the negatively-signed result of the following:

The heat capacity of the calorimeter in #"J/K"# multiplied by the change in temperature in #"K"#, the quantity plus #RT_"room"Deltan_"gas"#, with #Deltan_"gas"# gotten from multiplying #(1.659g)/(122.12g)# by the change in moles of gas of:

#2C_6H_5COOH(s) + 13O_2(g) -> 12CO_2(g) + 6H_2O(g)#

which is #5 "mol" *(1.659g)/(2 cdot 122.12g)#.

The answer turns out to be #color(blue)(-64.36 "kJ")# for #1.659 "g"#, or #color(blue)(-38.79 "kJ/g")#. I intentionally did not do this on a per-mol basis because sometimes you don't know what your analyte fuel is, or perhaps you don't know the exact composition (like of random vegetable oil).

(I've color-coded the important parts of the equations below, and you don't need to know Calculus to use the equations)

More detailed explanation:
We can assume that the bomb calorimeter is a system closed off to the surroundings, because one can do the experiment with a large amount of water surrounding the bomb (near #2"L"# is possible, for example); thus, the heat transfer out into the surroundings is minimized.

Knowing that this is a constant-volume system, the enthalpy of combustion can be acquired by working backwards:

#DeltaH_C = DeltaU_"cal" + RT_"room"Deltan_"gas"#

where #H# is enthalpy, #U# is internal energy, #R# is #8.314472 "J/mol" cdot "K"#, #T_"room"# is the temperature of the room in #"K"#, and #Deltan_"gas"# is the change in the moles of gas in the reaction. (#C# stands for combustion, and #"cal"# means calorimeter.)

To acquire the internal energy, note that at a constant volume:

#((delU)/(delT))_V = C_V#

where #C_V# is the heat capacity of something at a constant volume in #"J/K"#, and #((delU)/(delT))_V# means "the change in internal energy due to a change in temperature while in a constant volume situation".

Rearranging the equation and integrating it gives:

#dU = C_VdT#
#DeltaU_C = color(green)(DeltaU_"cal") = int_(T_1)^(T_2) C_VdT color(green)(= C_VDeltaT_"cal")#

• Using the heat capacity of the entire system #(C_V)# and the change in temperature #(DeltaT_"cal")#, we have the change in internal energy for the combustion #(DeltaU_C)#, which is equal to the internal energy for the calorimeter #(DeltaU_"cal")#.
• Using #DeltaU_"cal"# in #"J"#, #R = 8.314472 "J/mol" cdot "K"#, #T_"room" = 298.15 "K"#, and #Deltan_"gas"#, we'll have the enthalpy of combustion.

We now need #Deltan_"gas"#. The reaction for complete combustion of a solid benzoic acid (BA) pellet is:

#2C_6H_5COOH(s) + 13O_2(g) -> 12CO_2(g) + 6H_2O(g)#

Therefore, for a reaction using #2 "mol"# of BA, #Deltan_"gas"# is #6+12 - 13 = 5#. Since BA is clearly the limiting reagent in the presence of excess oxygen, normalizing down to the scale of the experiment itself, we get:

#Deltan_"gas" = ((1.659 "g")/(2 cdot 122.12"g") "BA") * (5 "mol gas") ~~ 0.03396 "mol gas"# on the scale of the reaction.

Finally, we can get the enthalpy for the combustion of BA like so, using #DeltaH_C = DeltaH_"cal"# and #DeltaU_C = DeltaU_"cal" = C_VDeltaT_"cal"#:

#color(blue)(DeltaH_C) = color(green)(DeltaU_"cal") + RTDeltan_"gas"#

#= color(green)(C_VDeltaT_"cal") + RTDeltan_"gas"#

#= (1.017xx10^4 "J/K")*(34.04"K"-27.72 "K") + (8.314472 "J/mol" cdot "K")(298.15 "K")(0.03396 "mol")#

#~~ 6.4359xx10^4 "J"#

Then since combustion is exothermic, the sign is negative:

#color(blue)(~~ -64.36 "kJ")#

Or:

#(64.36 kJ)/(1.659 g "BA") color(blue)(~~ -38.79 "kJ/g")#

#~~ color(blue)(-9.272 "kcal/g")#

The temperature change #DeltaT=27.72-34.04=-6.32"K"#

To get the amount of heat produced we multiply this by the heat capacity of the bomb calorimeter.

This means we don't have to bother with any complicated thermodynamic equations as that is what they are designed for.
#rArr#

#H= 10.17xx(-6.32)=-64.27"kJ"#

This is the heat produced from burning 1.659g. Now we just need to work out what burning 1 mole would produce.

#M_r=122.12#

#1.659"g"rarr-64.27"kJ"#

#1"g"rarr-64.27/1.659"kJ"#

#122.12"g"rarr-(64.27)/(1.659)xx122.12=-4730"J/mol"#

#DeltaH_c=-4.73"kJ/mol"#