# Understanding the calculation of enthalpy of the formation of benzene

• Chemistry
• zenterix
zenterix
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##.

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?

You are correct in your approach. In bomb calorimetry we generally assume that any difference in heat capacity between reactants and products is negligible compared to the total heat capacity. In this case I estimate that the heat capacity of the reactants is lower than that of the products by less than 1 J/K, i.e. less than 1 part in 10,000 of the total heat capacity. The magnitude of ΔU is reduced by ca. 240 J/mol. This is a very small fraction of ΔU, but it means the heat of formation is more like 41.2 than 41.4 kJ/mol - a much larger relative difference. I think that you are meant to assume the heat capacity doesn't change, but that does raise the question how many sig figs are justified in the answer on this assumption.
By comparison, what relative error is introduced by taking the MW of benzene as 78? (BTW, you seem to have copied the book answer for ΔU, rather than doing your own calculation yourself.)

mjc123 said:
I estimate that the heat capacity of the reactants is lower than that of the products by less than 1 J/K, i.e. less than 1 part in 10,000 of the total heat capacity.
How did you do this estimate?
mjc123 said:
You are correct in your approach.
The approach in the second picture I posted?
mjc123 said:
(BTW, you seem to have copied the book answer for ΔU, rather than doing your own calculation yourself.)
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.

 MW Cv,m m n mol C total g/mol J/mol/K g J/K J/K C6H6 78.114​ 134.8​ 1​ 1​ 0.012802​ 1.725683​ O2 31.998​ 21.064​ 7.5​ 0.096014​ 2.022429​ 3.748112​ reactants CO2 44.009​ 28.821​ 6​ 0.076811​ 2.213764​ H2O 18.015​ 75.385​ 3​ 0.038405​ 2.895192​ 5.108956​ products 1.360844​ difference dT 4.174​ K -3259.4​ book CT 10000​ J/K -2361.06​ 6CO2 dU0 -41740​ J -857.49​ 3H2O dU0,m -3260.48​ kJ/mol 40.85​ book result dUf -41734.3​ J dUf,m -3260.03​ kJ/mol dHf,m -3263.75​ kJ/mol dfH 45.20289​ kJ/mol

zenterix

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