One of the most important applications of the First Law of Thermodynamics is the experimental determination of energy changes associated with chemical reactions. The technique most commonly used to measure the energy released during combustion reactions is bomb calorimetry.
A bomb calorimeter consists of a heavy steel container, called the bomb, that is designed to withstand high pressures. A sample of the substance to be studied is placed inside the bomb along with excess oxygen. The bomb is then submerged in a known quantity of water, and the entire assembly is thermally insulated from the surroundings.
The reaction is initiated electrically using a short length of fuse wire. As the reaction proceeds, heat released by the reaction warms the bomb and the surrounding water. By measuring the resulting temperature change, the amount of energy released by the reaction can be determined.
The walls of the bomb prevent the reaction mixture from expanding or contracting significantly. As a result, the reaction occurs at approximately constant volume:
\[ \Delta V = 0 \]
Since pressure-volume work is given by
\[ w=-P_{\mathrm{ext}}\Delta V \]
it follows that
\[ w=0 \]
The First Law then becomes
\[ \Delta U = q_V \]
Therefore, the heat measured by a bomb calorimeter is equal to the change in internal energy of the reaction.
Before a calorimeter can be used to measure the energy of an unknown reaction, it must be calibrated. This is accomplished by carrying out a reaction whose energy change is known accurately, most commonly the combustion of benzoic acid.
The calibration determines the water equivalent (sometimes called the calorimeter constant), which relates a measured temperature change to the amount of energy absorbed by the calorimeter:
\[ \varepsilon = \frac{q} {\Delta T} \]
Once the water equivalent is known, the energy released by an unknown reaction can be determined directly from the observed temperature change.
During a bomb calorimetry experiment, energy released by the reaction is absorbed by the calorimeter:
\[ q_{\mathrm{rxn}} + q_{\mathrm{cal}} = 0 \]
Therefore,
\[ q_{\mathrm{rxn}} = -q_{\mathrm{cal}} \]
and
\[ q_{\mathrm{cal}} = \varepsilon\Delta T \]
A positive temperature change indicates that the reaction released energy, while a negative temperature change indicates that the reaction absorbed energy from the calorimeter.
Chemists are often more interested in the enthalpy change of a reaction than the internal-energy change because most reactions are carried out at constant pressure.
The relationship between these quantities is
\[ \Delta H = \Delta U + \Delta(pV) \]
For reactions involving ideal gases at constant temperature,
\[ \Delta H = \Delta U + RT\Delta n_{\mathrm{gas}} \]
where \(\Delta n_{\mathrm{gas}}\) is the change in the number of moles of gas during the reaction.
This correction allows bomb calorimetry data to be converted into standard reaction enthalpies.
Big picture: Bomb calorimetry is a constant-volume technique used to determine the internal-energy changes of chemical reactions. By measuring temperature changes and applying the First Law of Thermodynamics, reaction energies can be determined experimentally and converted into enthalpy changes when needed.
A bomb calorimeter is calibrated by burning \(0.7842\ \mathrm{g}\) of benzoic acid, producing a temperature increase of \(2.02^\circ\mathrm{C}\). The internal energy of combustion of benzoic acid is \(-3225.7\ \mathrm{kJ\,mol^{-1}}\).
A \(0.5348\ \mathrm{g}\) sample of naphthalene, \(C_{10}H_8\), is then burned in the same calorimeter, producing a temperature increase of \(2.24^\circ\mathrm{C}\). Calculate \(\Delta H_c\) for naphthalene.
\[ n_{\mathrm{BA}} = \frac{0.7842\ \mathrm{g}} {122.12\ \mathrm{g\,mol^{-1}}} = 0.006422\ \mathrm{mol} \]
\[ q_{\mathrm{cal}} = -q_{\mathrm{rxn}} = (0.006422\ \mathrm{mol}) (3225.7\ \mathrm{kJ\,mol^{-1}}) = 20.71\ \mathrm{kJ} \]
\[ \varepsilon = \frac{q_{\mathrm{cal}}}{\Delta T} = \frac{20.71\ \mathrm{kJ}} {2.02^\circ\mathrm{C}} = 10.25\ \mathrm{kJ\,^\circ C^{-1}} \]
\[ q_{\mathrm{cal}} = \varepsilon\Delta T = (10.25\ \mathrm{kJ\,^\circ C^{-1}}) (2.24^\circ\mathrm{C}) = 22.97\ \mathrm{kJ} \]
Since the calorimeter gains heat, the reaction releases heat:
\[ q_{\mathrm{rxn}} = -22.97\ \mathrm{kJ} \]
In a bomb calorimeter, the reaction occurs at constant volume, so
\[ q_V = \Delta U \]
\[ n_{\mathrm{naph}} = \frac{0.5348\ \mathrm{g}} {128.17\ \mathrm{g\,mol^{-1}}} = 0.004173\ \mathrm{mol} \]
\[ \Delta U_c = \frac{-22.97\ \mathrm{kJ}} {0.004173\ \mathrm{mol}} = -5.50\times10^3\ \mathrm{kJ\,mol^{-1}} \]
The combustion reaction is
\[ C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l) \]
The change in moles of gas is
\[ \Delta n_{\mathrm{gas}} = 10 - 12 = -2 \]
\[ \Delta H = \Delta U + RT\Delta n_{\mathrm{gas}} \]
\[ \Delta H_c = -5505\ \mathrm{kJ\,mol^{-1}} + (8.314\times10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}) (298.15\ \mathrm{K}) (-2) \]
\[ \Delta H_c = -5510\ \mathrm{kJ\,mol^{-1}} \]
Physical interpretation: The reaction is strongly exothermic, so the calorimeter temperature rises. Bomb calorimetry directly gives \(\Delta U_c\), and the small \(RT\Delta n_{\mathrm{gas}}\) correction converts this value to the constant-pressure enthalpy of combustion, \(\Delta H_c\).
Use the calibration data to determine the calorimeter constant, then use the unknown sample data to calculate \(\Delta H_c\). Choose the correct answer.
Many chemical reactions are carried out in containers that are open to the atmosphere. Under these conditions, the pressure remains approximately constant, so the heat measured by the calorimeter is related to the enthalpy change of the process.
\[ q_p = \Delta H \]
This makes constant-pressure calorimetry especially useful in chemistry, because reaction enthalpies are usually reported as \(\Delta H\) values.
A common constant-pressure calorimeter is the coffee-cup calorimeter. In this setup, the reaction occurs in an insulated cup, usually in aqueous solution. The solution acts as the surroundings, and its temperature change is used to determine how much heat was released or absorbed by the reaction.
The heat absorbed by the solution is usually calculated using
\[ q_{\mathrm{solution}} = mC\Delta T \]
where \(m\) is the mass of the solution, \(C\) is the specific heat capacity, and \(\Delta T\) is the temperature change of the solution.
If the calorimeter itself absorbs a significant amount of heat, a calorimeter correction can also be included:
\[ q_{\mathrm{cal}} = C_{\mathrm{cal}}\Delta T \]
Energy is conserved during a calorimetry experiment. Heat released by the reaction is absorbed by the solution and calorimeter, while heat absorbed by the reaction is supplied by the solution and calorimeter.
\[ q_{\mathrm{rxn}} + q_{\mathrm{solution}} + q_{\mathrm{cal}} = 0 \]
If the heat absorbed by the calorimeter is negligible, this simplifies to
\[ q_{\mathrm{rxn}} = -q_{\mathrm{solution}} \]
Since the experiment occurs at constant pressure,
\[ \Delta H_{\mathrm{rxn}} = q_p = q_{\mathrm{rxn}} \]
The sign of the temperature change tells us whether the reaction released or absorbed heat.
| Observation | Meaning | Sign of \(\Delta H_{\mathrm{rxn}}\) |
|---|---|---|
| Solution temperature increases | Reaction releases heat | \(\Delta H_{\mathrm{rxn}} < 0\) |
| Solution temperature decreases | Reaction absorbs heat | \(\Delta H_{\mathrm{rxn}} > 0\) |
Big picture: Constant-pressure calorimetry measures heat flow under conditions where \(q_p=\Delta H\). By measuring the temperature change of the surroundings, we can determine the enthalpy change of the reaction.
A student mixes \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{M}\ HCl\) with \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{M}\ NaOH\). Both solutions are initially at \(19.0^\circ\mathrm{C}\), and the final temperature is \(19.7^\circ\mathrm{C}\). Calculate the enthalpy of reaction.
\[ HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l) \]
First calculate the moles reacted:
\[ n = (0.0250\ \mathrm{L}) \left( 0.100\ \mathrm{mol\,L^{-1}} \right) = 0.00250\ \mathrm{mol} \]
The temperature change of the solution is
\[ \Delta T = 19.7^\circ\mathrm{C} - 19.0^\circ\mathrm{C} = 0.7^\circ\mathrm{C} \]
Assuming the final solution has a density of \(1.00\ \mathrm{g\,mL^{-1}}\), the mass of solution is approximately
\[ m = 50.0\ \mathrm{g} \]
The heat absorbed by the solution is
\[ q_{\mathrm{solution}} = mC\Delta T \]
\[ q_{\mathrm{solution}} = (50.0\ \mathrm{g}) \left( 4.184\ \mathrm{J\,g^{-1}\,^\circ C^{-1}} \right) (0.7^\circ\mathrm{C}) = 146\ \mathrm{J} \]
Since the solution warms up, the reaction released heat:
\[ q_{\mathrm{rxn}} = -q_{\mathrm{solution}} = -146\ \mathrm{J} \]
At constant pressure,
\[ q_p = \Delta H \]
Therefore, the molar enthalpy of reaction is
\[ \Delta H_{\mathrm{rxn}} = \frac{-146\ \mathrm{J}} {0.00250\ \mathrm{mol}} = -5.86\times10^4\ \mathrm{J\,mol^{-1}} \]
\[ \Delta H_{\mathrm{rxn}} = -58.6\ \mathrm{kJ\,mol^{-1}} \]
Physical interpretation: The temperature of the solution increases, so the neutralization reaction is exothermic. Because the experiment occurs at constant pressure, the heat released by the reaction is interpreted as the reaction enthalpy.
A generic reaction \(A + B \rightarrow \text{products}\) is carried out in a coffee-cup calorimeter. Choose the correct molar enthalpy of reaction.
Big picture: Calorimetry connects measurable temperature changes to thermodynamic quantities such as \(\Delta U\) and \(\Delta H\). Bomb calorimetry measures internal-energy changes under constant-volume conditions, while coffee-cup calorimetry measures enthalpy changes under constant-pressure conditions.