Homework Statement

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was ##20.826^\circ C## and the temperature after the combustion was ##25.000^\circ C##.

This was an adiabatic container.

The heat capacity of the bomb, the water around it, and the contents of the bomb after the combustion was ##10^4\mathrm{JK^{-1}}##.

Calculate ##\Delta_f H^\circ## for ##C_6H_6(l)## at ##298.15\text{K}## from these data.

Relevant Equations

Assume that the water produced in the combustion is in the liquid state and the carbon dioxide produced in the combustion is in the gas state.

I really struggled with this problem and did not make headway on my own. Here is the solution from the solution manual.

I tried to understand the above by drawing the following picture.

Reaction 1 is the balanced equation for the reaction that happened in the constant-volume calorimeter. The actual reaction involved 1g of benzene which is ##1/78.11\ \text{mol}## of benzene.

Since the process was adiabatic and constant-volume then the change in internal energy is zero.

We want the enthalpy of formation of benzene at ##25^\circ C##. We can calculate this from reaction 5 if we are able to determine the reaction enthalpy of 5.

Since we are given the heat capacity of the products of reaction 1 (same products as of reaction 5), then we can compute

$$\Delta U_2=q_{V,2}=\int_{298}^{293.826} 10^4 dT=-41.74\text{kJ}$$

Note that this is the heat required to change the temperature of the products of the combustion of only 1g of benzene (that is, the stoichiometric coefficients involved are not the ones shown above, but something much smaller).

We can, however, determine the change ##\Delta U_2## for 1 mol of benzene.

$$-41.74\mathrm{\frac{kJ}{g}\cdot 78.11\frac{g}{mol}}$$

$$=-3255.7\mathrm{\frac{kJ}{mol}}$$

At this point I got stuck and could not understand what the next steps were.

The solution manual calculations seem to me to be describing the following.

If the given heat capacity given in the problem statement were for the reactants instead of the products, then the calculation above would have given us ##q_{V,6}##.

We could write

$$\Delta U_{V,6}+\Delta U_{V,1}=\Delta U_{V,5}$$

$$\Delta U_{V,5}=\Delta U_{V,6}=-3255.7\mathrm{\frac{kJ}{mol}}$$

Then

$$\Delta H_{V,5}=\Delta U_{V,5}+RT\Delta n_{V,5}$$

$$=-3255.7\mathrm{\frac{kJ}{mol}}+RT(-1.5\ \text{mol})$$

$$=-3259.4\mathrm{kJ\ mol^{-1}}$$

This is the enthalpy of combustion at ##298\text{K}##, ie ##\Delta H_{V,5}=\Delta_cH(298\text{K})##.

At this point we can compute the enthalpy of formation of benzene at ##298\text{K}##.

This latter result relied on my assuming that the problem statement is incorrect, ie the given heat capacity is for the reactants before combustion rather than of the products after combustion.

So, is the problem statement correct, or else what am I getting wrong?

How did you do this estimate?

The approach in the second picture I posted?

Indeed. I was trying to understand the underlying reasons for the calculations, and skipped the calculations.

Here are the calculations as I posed them in second part of the OP (ie the calculations for the 2nd picture)

The solution manual uses ##78\text{g/mol}## for the molar mass of benzene instead of ##78.11\text{g/mol}## which is what I am getting from the ThermophysicalData package from Maple.

Using their figure we have

It is not clear why they got ##41.4\text{kJ}## as their result for the last calculation above. The intermediate results (the first three calculations) all match and the equation in the last expression is the same they are using, so I am confused.

I agree with you about the last calculation.

The issue here is, I think, that you would normally equilibrate your calorimeter at the temperature of interest (e.g. 298K), then do the experiment. Then the book method works.
Let us assume
Initial temperature T0, final temperature Tf; difference ΔT.
Total heat capacity of bomb, contents and water after the experiment Ct
Total heat capacity of bomb, contents and water before the experiment Ct - δ, where δ is the difference in heat capacity between products and reactants.
Referring to your initial picture, following the (3,2) route:
-ΔU(T0) = Ct*ΔT
ΔU(T1) = ΔU(T0) + δ*ΔT
ΔH(T1) = ΔU(T1) + ΔnRT1
The book has omitted the δ*ΔT.
Following the (4,5) route:
-ΔU(T1) = (Ct - δ)*ΔT (you can see that this is equivalent to the previous result)
ΔH(T1) = ΔU(T1) + ΔnRT1
Either way, you need to know δ, which you are not given. I calculated it from heat capacities from Wikipedia (subtracting R from the Cp values for the gases). Here are my results. Note that the difference due to assuming MW (benzene) = 78 is much bigger than the correction due to δ - about 10% of the heat of formation (a small difference between large quantities). Hence the importance of accurate measurements and accurate data input.