# Constant Volume Calorimetry

Skills to Develop

• Write a balanced thermochemical equation that expresses the enthalpy change  for a  given  chemical reaction. Be sure to correctly specify the physical state (and, if necessary, the  concentration) of each component.
• Write an equation that defines the standard enthalpy of formation of a given chemical  species.
• Define the standard enthalpy change of a chemical reaction. Use a table of standard  enthalpies of formation to evaluate $$\Delta H_f^o$$.
• State the principle on which Hess' law  depends, and explain why this  law is so useful.
• Describe  a  simple calorimeter and explain how it is employed and how its heat capacity  is  determined

Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.

### Introduction

Calorimetry is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a coffee-cup calorimeter is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions.

### The Bomb Calorimeter

Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion $$\Delta H_{cumbustion}$$, 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 a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done.  A bomb calorimeter structure consists of the following:

• Steel bomb which contains the reactants
• Water bath in which the bomb is submerged
• Thermometer
• A motorized stirrer
• Wire for ignition

All of these components are contained within the double-walled outer part of the calorimeter (Figure 1). After the initial temperature of the water is measured, the heated wire inside the bomb starts the reaction. After combustion the final temperature of the water is measured, and then the change in temperature of the reactants can be calculated. Through the combustion reaction, the temperature rises due to the conversion from chemical energy to thermal energy that occurs through the reaction. 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.

Figure 1: A simplified bomb calorimeter

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 ΔU rather than to ΔH. The enthalpy change is calculated  according to the formula

$ΔH = q_v + Δn_gRT \tag{1}$

in which Δng  is the change in the number of moles of gases in the reaction.

Example 1

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, ΔU = –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:

$\dfrac{(0.825 g}{122.1 \;g/mol} = 0.00676\; mol$

• Heat released to  calorimeter:

$(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ$

• Calorimeter constant:

$\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K$

Now determine $$ΔU_{combustion}$$ of the  biphenyl ("BP"):

• moles of biphenyl:

$\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol$

• heat released to  calorimeter:

$(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ$

• heat  released per mole  of biphenyl:

$\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol$

$ΔU_{combustion} (BP) = –6,293\; kJ/mol$

(This is the heat change at constant volume, $$q_v$$; the negative  sign indicates that the reaction is exothermic, as all combustion reactions are.)

From the reaction equation

$(C_6H_5)_{2(s)} + \frac{19}{2} O_{2(g)} \rightarrow 12 CO_{2(g)} + 5 H_2O_{(l)}$

we have

$Δn_g = 12 - \frac{19}{2} = \frac{-5}{2}$

Thus the volume of the system decreases when  the reaction takes place. Converting to ΔH, we can write the following equation.  Additionally, recall that at constant volume, $$ΔU = q_V$$.

$ΔH = q_V + Δn_gRT$

$ΔH = ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K)=$

$ΔH=(-6,293 \; kJ/mol)–(6,194\; J/mol)=(-6,293-6.2)\;kJ/mol= -6299 \; kJ/mol$

A common mistake here is to forget  that the subtracted  term is in J, not  kJ. Note  that the additional 6.2 kJ  in $$ΔH$$ compared to  $$ΔU$$ reflects the work that the surroundings do on the system as the volume of  gases decreases according  to the reaction  equation.

### Determining the Heat of Reaction

The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the heat of reaction. The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents ($$q_{calorimeter}$$) through the combustion reaction.

$q_{rxn} = -q_{calorimeter} \tag{2}$

where

$q_{calorimeter} = q_{bomb} + q_{water} \tag{3}$

If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula:

$q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \tag{4}$

Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate $$q_{rxn}$$ from $$q_{calorimeter}$$ calculated by Equation 2. The heat capacity of the calorimeter can be determined by conducting an experiment.

Example 2

1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, $$C_{12}H_{22}O_{11}$$, in kJ per mole of $$C_{12}H_{22}O_{11}$$.

Given:

• mass of $$C_{12}H_{22}O_{11}$$: 1.150 g
• $$T_{initial}$$: 23.42°C
• $$T_{final}$$:27.64°C
• Heat Capacity of Calorimeter: 4.90 kJ/°C

SOLUTION

Using Equation 4 to calculate $$q_{calorimeter}$$:

$q_{calorimeter} = (4.90\; kJ/°C) \times (27.64 - 23.42)°C = (4.90 \times 4.22) \;kJ = 20.7\; kJ$

Plug into Equation 2:

$$q_{rxn} = -q_{calorimeter} = -20.7 \; kJ \;$$

But the question asks for kJ/mol $$C_{12}H_{22}O_{11}$$, so this needs to be converted:

$$q_{rxn} = \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} = \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}}$$

Per Mole $$C_{12}H_{22}O_{11}$$:

$$q_{rxn} = \dfrac{-18.0 \; kJ}{g \; C_{12}H_{22}O_{11}} \times \dfrac{342.3 \; g \; C_{12}H_{22}O_{11}}{1 \; mol \; C_{12}H_{22}O_{11}} = \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; C_{12}H_{22}O_{11}}$$

### Problems

1. After going through combustion in a bomb calorimeter a sample gives off 5,435 cal. The calorimeter experiences an increase of 4.27°C  in its temperature. Using this information, determine the heat capacity of the calorimeter in kJ/°C.
2. Referring to the example given above about the heat of combustion, calculate the temperature change that would occur in the combustion of 1.732 g $$C_{12}H_{22}O_{12}$$ in a bomb calorimeter that had the heat capacity of 3.87 kJ/°C.
3. Given the following data calculate the heat of combustion in kJ/mol of xylose,$$C_{5}H_{10}O_{5}$$(s), used in a bomb calorimetry experiment: mass of $$C_{5}H_{10}O_{5}$$(s) = 1.250 g, heat capacity of calorimeter = 4.728 kJ/°C, Initial Temperature of the calorimeter = 24.37°C, Final Temperature of calorimeter = 28.29°C.
4. Determine the heat capacity of the bomb calorimeter if 1.714 g of naphthalene, $$C_{10}H_{8}$$(s), experiences an 8.44°C increase in temperature after going through combustion. The heat of combustion of naphthalene is -5156 kJ/mol $$C_{10}H_{8}$$.
5. What is the heat capacity of the bomb calorimeter if a 1.232 g sample of benzoic acid causes the temperature to increase by 5.14°C? The heat of combustion of benzoic acid is -26.42 kJ/g.

1. Use equation 4 to calculate the heat of capacity:

$$q_{calorimeter} = \; heat \; capicity \; of \; calorimeter \; x \; \Delta{T}$$

5435 cal = heat capacity of calorimeter x 4.27°C

Heat capacity of calorimeter = (5435 cal/ 4.27°C) x (4.184 J/1 cal) x (1kJ/1000J) = 5.32 kJ/°C

2. The temperature should increase since bomb calorimetry releases heat in an exothermic combustion reaction.

Change in Temp = (1.732 g $$C_{12}H_{22}O_{11}$$) x (1 mol $$C_{12}H_{22}O_{11}$$/342.3 g $$C_{12}H_{22}O_{11}$$) x (6.61 x 10³ kJ/ 1 mol $$C_{12}H_{22}O_{11}$$) x (1°C/3.87kJ) = 8.64°C

3. [(Heat Capacity x Change in Temperature)/mass] =[ ((4.728 kJ/°C) x(28.29 °C – 24.37 °C))/1.250 g] = 14.8 kJ/g xylose

$$q_{rxn}$$ = (-14.8 kJ/g xylose) x (150.13 g xylose/ 1 mol xylose) = -2.22x10³ kJ/mol xylose

4. Heat Capacity = [(1.714 g $$C_{10}H_{8}$$)  x (1 mol $$C_{10}H_{8}$$/128.2 g $$C_{10}H_{8}$$) x (5.156x10³ kJ/1 mol $$C_{10}H_{8}$$)]/8.44°C = 8.17 kJ/ °C

5. Heat Capacity = [(1.232 g benzoic acid)  x  (26.42 kJ/1 g benzoic acid)]/5.14°C = 6.31 kJ/ °C

### References

1. Petrucci, Ralph H. Herring, F Geoffrey. Madura, Jeffery D. Bissonnette, Carey. GENERAL CHEMISTRY Principles and Modern Applications. Custom Edition for Chem 2 UCD. Upper Saddle River, New Jersey; Pearson Canada, 2011.
2. Oxtoby, David W. Gillis, H.P. Campion, Alan. Principles of Modern Chemistry. Sixth Edition. Belmont, Ca; Thompson Learning, Inc., 2008.

### Contributors

Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook