- #1
zenterix
- 689
- 83
- 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?
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?