## Thermochemistry and Calorimetry

Scientists use calorimetry to measure the change in enthalpy of reactions in a laboratory setting. In calorimetry, we conduct the reaction in an isolated setting and measure the temperature change. We can then use the equation $q = ms\Delta T$ – where $q$ is heat, $m$ is mass, $s$ is specific heat of the substance, and $\Delta T$ is change in temperature – to find the heat of the reaction.

There are two main types of calorimetry: constant pressure, or coffee cup calorimetry, and constant volume, or bomb calorimetry.

Read this text, which describes the two types of calorimetry, and shows worked examples of how to calculate heat of reaction from calorimetry data. Pay attention to the sign conventions here. In calorimetry, we directly measure the temperature change of the surroundings, not the system, and first calculate the heat of the surroundings. Then, we need to convert this to heat of the system.

How are enthalpy changes determined experimentally? First, you must understand that the only thermal quantity that can be observed directly is the heat $q$ that flows into or out of a reaction vessel, and that $q$ is numerically equal to $\Delta H^{o}$ only under the special condition of constant pressure. Moreover, $q$ is equal to the standard enthalpy change only when the reactants and products are both at the same temperature, normally 25°C.

The measurement of $q$ is generally known as calorimetry.

The most common types of calorimeters contain a known quantity of water which absorbs the heat released by the reaction. Because the specific heat capacity of water $(4.184J\: g^{-1}K^{-1})$ is known to high precision, a measurement of its temperature rise due to the reaction enables one to calculate the quantity of heat released.

#### The Calorimeter Constant

In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by the components of the calorimeter itself. It is therefore necessary to calibrate the calorimeter by measuring the temperature change that results from the introduction of a known quantity of heat. The resulting calorimeter constant, expressed in $JK^{-1}$, can be regarded as the heat capacity of the calorimeter. The known source of heat is usually produced by passing a known quantity of electric current through a resistor within the calorimeter, but it can be measured by other means as described in the following problem example.

#### Problem Example 1

In determining the heat capacity of a calorimeter, a student mixes 100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (The specific heat of water is $(4.184J\: g^{-1}K^{-1})$)

After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at 38.7°C. Calculate the heat capacity of the calorimeter in $J/K$.

Solution: The hot water loses heat, the cold water gains heat, and the calorimeter itself gains heat, so this is essentially a thermal balance problem. Conservation of energy requires that

$q_{hot}+ q_{cold} + q_{cal} = 0$

We can evaluate the first two terms from the observed temperature changes:

$q_{hot}=(100g)(38.7-57.0)K(4.184J\: g^{-1}K^{-1})=-7657J$

$q_{cold}=(100g)(38.7-24.2)K(4.184J\: g^{-1}K^{-1})=6067J$

$So\: q_{cal}=(7657-6067)J=1590J$

The calorimeter constant is $\frac{(1590J)}{(38.7-24.2)}K=110J\: K^{-1}$

Note: Strictly speaking, there is a fourth thermal balance term that must be considered in a highly accurate calculation: the water in the calorimeter expands as it is heated, performing work on the atmosphere.

For reactions that can be initiated by combining two solutions, the temperature rise of the solution itself can provide an approximate value of the reaction enthalpy if we assume that the heat capacity of the solution is close to that of the pure water – which will be nearly true if the solutions are dilute.

For example, a very simple calorimetric determination of the standard enthalpy of the reaction $H^{+}(aq) + OH^{-}(aq)\rightarrow H_{2}O(l)$ could be carried out by combining equal volumes of 0.1 M solutions of HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat $q$ will be released into the solution. From the temperature rise and the specific heat of water, we obtain the number of joules of heat released into each gram of the solution, and $q$ can then be calculated from the mass of the solution. Since the entire process is carried out at constant pressure, we have $\Delta H^{o}=q$.

For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly placed within a larger insulated container of water. During the reaction, heat passes between the inner and outer containers until their temperatures become identical. Again, the temperature change of the water is observed, but in this case we need to know the value of the calorimeter constant described above.

#### The Bomb Calorimeter

Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion, since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month.

In order to ensure complete combustion, the experiment is carried out in the presence of oxygen above atmospheric pressure. This requires that the combustion be confined to a fixed volume. Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a bomb, and the technique is known as bomb calorimetry. The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.

Another consequence of the constant-volume condition is that the heat released corresponds to $q_{v}$, and thus to the internal energy change $\Delta U$ rather than to $\Delta H$. The enthalpy change is calculated according to the formula

$\Delta H=q_{v}+\Delta n_{g}RT$

in which $\Delta n_{g}$ is the change in the number of moles of gases in the reaction.

#### Problem Example 3

A sample of biphenyl (C6H5)2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C6H5COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K.

For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1 (that is, $\Delta U^{o}=-3226\: kJ\: mol^{-1}$.) Use this information to determine the standard enthalpy of combustion of biphenyl.

Solution. Begin by working out the calorimeter constant:

Moles of benzoic acid: $\frac{(0.825 g)}{(122.1g\: mol^{-1})}=.00676\: mol$

Heat released to calorimeter: $(.00676\: mol)\times (3226KJ\: mol^{-1}=21.80kJ$

Calorimeter constant: $\frac{(21.80kJ)}{(1.94K)}=11.24kJ\: K^{-1}$

Now determine $\Delta U_{combustion}$ of the biphenyl (
BP):

moles of biphenyl: $\frac{(.526g)}{(154.12g\: mol^{-1})}=.00341\: mol$

heat released to calorimeter: $(1.91K)\times (11.24\: kJ\: K^{-1})=21.46\: kJ$

heat released per mole of biphenyl: $\frac{(21.46\: kJ)}{(.00341\: mol)}=6293\: kJ\: mol^{-1}$

$\Delta U_{combustion}(BP)=-6293\: kJ\: mol^{-1}$

(This is the heat change at constant volume, $q_{v}$ ; the negative sign indicates that the reaction is exothermic, as all combustions are.)

From the balanced reaction equation

$(C_{6}H_{6})_{2}(s)+\frac{29}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(l)$

we have $\Delta n_{g}=12-(\frac{29}{2})=-\frac{5}{2}$. Thus the volume of the system decreases when the reaction takes place.

Converting to $\Delta H$, we substitute into

$\Delta H =q_{v}+\Delta n_{g}RT=\Delta U+\Delta n_{g}RT$

$\Delta H^{o}=\Delta U^{o}-(\frac{5}{2})(8.314\: J\: mol^{-1}K^{-1})(298K)=$

Note that the use of standard temperature, 298 K, defines $\Delta H^{o}$, as opposed to plain $\Delta H$.

$\Delta H^{o} = (–6293 kJ mol–1) – (6194 J mol–1) = (–6293 – 6.2) kJ mol–1 =$ –6299 kJ mol–1

A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in $\Delta H^{o}$ compared to $\Delta U^{o}$ reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation.

#### How do Calorimeters Work?

Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories.

Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms:   The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.

The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C. This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License.