# Calorimetry: Measuring Heats of Reactions

- Page ID
- 32666

- Calculate amount of electric energy for heat capacity measurement.
- Perform experiments to measure heats of reactions.
- Calculate the heats of reactions from experimental results.
- Calculate internal energies of reactions from bomb calorimeter experiments.
- Calculate enthalpies of reactions from bomb calorimetry experiments.

## Calorimetry - Measuring Heats of Reactions

**Calorimetry**, derived from the Latin *calor* meaning heat, and the Greek *metry* meaning to measure, is the science of measuring the amount of heat. All calorimetric techniques are therefore based on the measurement of heat that may be generated (exothermic process), consumed (endothermic process) or simply dissipated by a sample. There are numerous methods to measure such heat, and since calorimetry's advent in the late 18th century, a large number of techniques have been developed. Initially techniques were based on simple thermometric (temperature measurement) methods, but more recently, advances in electronics and control have added a new dimension to calorimetry, enabling users to collect data and maintain samples under conditions that were previously not possible.

Any process that results in heat being generated and exchanged with the environment is a candidate for a calorimetric study. Hence it is not surprising to discover that calorimetry has a very broad range of applicability, with examples ranging from drug design in the pharmaceutical industry, to quality control of process streams in the chemical industry, and the study of metabolic rates in biological (people included) systems. Indeed if the full range of applications were to be mentioned, the allocated disk space on this site would soon be used up.

We discuss the basics of two types of calorimetry: measurements based on constant pressure and measurement based on constant volume. The former involves pressure-volume work, whereas the latter does no pressure-volume work.

### Calorimeter

A **calorimeter** is a device used to measure heat of reaction. It can be sophisticated and expensive or simple and cheap. In CHEM120 Labs, a styrofoam cup is used as a calorimeter, because it is a container with good insulated walls to prevent heat exchange with the environment. In order to measure heats of reactions, we often enclose reactants in a calorimeter, initiate the reaction, and measure the temperature difference before and after the reaction. The temperature difference enables us to evaluate the heat released in the reaction. This page gives the basic theory for this technique.

A calorimeter may be operated under constant (atmosphere) pressure, or constant volume. Whichever kind we use, we first need to know its heat capacity. The **heat capacity** is the amount of heat required to raise the temperature of the entire calorimeter by 1 K, and it is usually determined experimentally before or after the actual measurements of heat of reaction.

The heat capacity of the calorimeter is determined by transferring a known amount of heat into it and measuring its temperature increase. Because the temperature differences are very small, extremely sensitive thermometers are required for these measurements. Example 1 shows how it is done.

Example 1

The temperature of a calorimeter increases 0.10 K when 7.52 J of electric energy is used to heat it. What is the heat capacity of the calorimeter?

*Solution:*

Dividing the amount of energy by the temperature increase yields the heat capacity, *C*,

\(C = \dfrac{7.52}{0.10} = \mathrm{75.2\: J/K}\)

**Discussion:**

We often compare the heat capacity of a calorimeter to that of a definite amount of water. The heat capacity of 75.2 J/K for the calorimeter is equivalent to the heat capacity of 1 mole (18 g) of water \(\left (\mathrm{18\: \dfrac{g}{mol}\times 1\: \dfrac{cal}{g\: K}\times 4.184\: \dfrac{J}{cal} = 75.3\: \dfrac{J}{K\: mol}} \right )\).

Do you know that the \(\mathrm{electric\: energy} = q \times dV = i \times dt \times dV\), where *q* is charge; *dV*, voltage; *i*, current; and *t*, time.

#### Measuring the Enthalpy by Calorimetry

By definition, *dH* is the energy (heat) released at constant pressure, whereas *dE* is the energy released at constant volume. These two quantities are related by the equation.

\(\begin{align}

dH &= dE + (d(P V))\\

&= dE + dn R T \hspace{20px}\textrm{(applying the ideal gas law)}

\end{align}\)

where

*dn* = number of moles of gas in the products - number of moles of gas in the reactants.

The P-V work must be taken into consideration for the calculation of depends on the extra amount (*dn* moles) of *dH*, if the calorimetry is performed at constant volume in a bomb calorimeter. A cross-section diagram of the *bomb* is shown here. The wires are for electric ignition, and the sample in the sample holder is in touch with the resistant wire. The bomb's diameter is 10 cm, and its height is 15 cm.

### The Bomb Calorimeter

For combustion reactions, we often enclose all reactants in an explosive-proof steel container, called the **bomb**, whose volume does not change during a reaction. The bomb is then submerged in water or other liquid that absorbs the heat of reaction. The heat capacitor of the bomb plus other things is then measured using the same technique as other calorimeters. Such an instrument is called a **bomb calorimeter**, and its application is called the bomb calorimetry.

The picture shows reading the sensitive thermometer while working with the P6310 Bomb Calorimeter. The *bomb* is inside the tank.

Since volume does not change, a bomb calorimeter measures the heat evolved under constant volume, *q*_{v},

\(q_v = C \times dT\),

where *dT* is the temperature increase. The *q _{v}* so measured is also called the change in internal energy,

*dE*. Note that

\(dE = q_v = C \times dT\)

Example 2

When a 10.0-watt (J/s) heater is used to heat a bomb calorimeter, its temperature increases by 3.0 K in 5.0 min. Calculate the heat capacity of the calorimeter.

formula: \(dE = q_v = C \times dT \) |
---|

*Solution:*

\(\mathrm{Energy\: used = 10 \times 5 \times 60 = 3000\: J}\).

Thus, \(C \mathrm{= \dfrac{3000}{3.0} = 1000\: J/K}\),

This is equivalent to \(\mathrm{\dfrac{1000\: \dfrac{J}{K}}{75.2\:\dfrac{J}{K\: mol}}= 13.3\: moles\: of\: water.}\)

The following **related examples** illustrate the application of bomb calorimeter for the measurement of *dE*, and the derivation of *dH*.

Example 3

A calorimeter with heat capacity equivalent to having 13.3 moles of water is used to measure the heat of combustion from 0.303 g of sugar (\(\ce{C12H22O11}\)). The temperature increase was found to be 5.0 K. Calculate the heat released, the amount of heat released by 1.0 g, and 1.0 mole of sugar.

formula: \(dE = q_v = s \times m \times dT\) s, specific heat; m mass. |
---|

*Solution:*

Heat released, *q _{v}*,

\(\begin{align}

q_v &= \mathrm{13.3\: mol \times 75.2\: \dfrac{J}{K\: mol} \times 5.0\: K}\\

&= \mathrm{5000\: J}

\end{align}\)

The amount of heat released by 1.0 g would be,

\(\mathrm{\dfrac{5000\: J}{0.303\: g} = 16.5\: kJ / g}\)

Since the molecular weight of sugar is 342.3 g/mol, the amount of heat released by 1.0 mole would be

\(\mathrm{16.5\: \dfrac{kJ}{g} \times 342.3\: \dfrac{g}{mol} = 5648\: kJ/mol}\).

Example 4

The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. Calculate the enthalpy of combustion per mole of sugar.

*Solution*

The balanced chemical reaction equation is

\(\ce{C12H22O11_{\large{(s)}} + 12 O_{2\large{(g)}} \rightarrow 12 CO_{2\large{(g)}} + 11 H2O_{\large{(l)}}}\)

Since the total numbers of moles of gases (12) before the reaction is the same as that after the reaction, *dn* = 0.

\(dH = dE = \mathrm{5648\: kJ/mol}\)

**Discussion**

This is the ideal amount of energy released when a mole of sugar is utilized by a living creature such as a person.

Example 5

A table of thermodynamic data gives *dH*_{f} = -285.8 kJ/mol for water. A bomb calorimeter measurement gives the heat of combustion for \(\ce{H2}\) as -282.0 kJ/mol. Estimate the error of the enthalpy measurement.

*Solution*

Reinterpret the problem, we have

\(\mathrm{H_{2\large{(g)}} + 0.5\, O_{2\large{(g)}} \rightarrow H_2O_{\large{(l)}}, \hspace{20px} \mathit{dE} = 282.0\: kJ/mol}\).

Furthermore,

\(\begin{align}

dn &= -1.5\\

dH &= dE + dn R T,\\

&= \mathrm{- 282.0 + (-1.5\: mol \times 8.314\: \dfrac{J}{mol\: K} \times 298\: K)}\\

&= \mathrm{(-282.0 - 3.72)\: kJ}\\

&= \mathrm{-285.7\: kJ}

\end{align}\)

The error is \(\dfrac{285.8-285.7}{285.8} = 0.03\%\)

**Discussion**

More heat is given off if the reaction is carried out at constant pressure, since the *P-V* work (1.5 *R T*) due to the compression of 1.5 moles of gases in the reactants would contribute to *dH*.

If 1.0 mole water is decomposed by electrolysis at constant pressure, we must supply an amount of energy equivalent to enthalpy change, *dH*, a little more than internal energy, *dE*. More energy must be supplied to perform the *P-V* work to be done by the products (\(\ce{H2}\) and \(\ce{O2}\)).

The heat capacity of the calorimeter can also be determined by burning an exactly known amount of a standard substance, whose enthalpy of combustion has been determined. Benzoic acid, \(\ce{C7H6O2}\), is one such standard. The problem below illustrates the calculations.

Example 6

When 0.1025 g of benzoic acid was burnt in a bomb calorimeter the temperature of the calorimeter increased by 2.165° C. For benzoic acid *dH*°_{comb} = -3227 kJ mol^{-1}. Calculate the heat capacity of the calorimeter.

*Solution:*

The equation for the combustion is,

\(\mathrm{C_7H_6O_{2\large{(s)}} + 7.5 O_{2\large{(g)}} \rightarrow 7CO_{2\large{(g)}} + 3H_2O_{\large{(l)}}, \hspace{20px} \mathit{dH}^\circ = 3227\: kJ}\)

Since 7.5 moles of \(\ce{O2}\) gas is needed, and 7 moles of \(\ce{CO2}\) is produced, some pressure-volume work is done to the calorimeter:

\(\begin{align}

P V &= dn R T, \textrm{ where }dn = (7 - 7.5) = \mathrm{- 0.5\: mol}\\

dE &= dH - dn R T\\

&= - 3227 - (-0.5\times8.314298\times298)\\

&= \mathrm{- 3226\: kJ/mol\: (a\: small\: correction)}

\end{align}\)

The amount of heat produced by 0.1025 g benzoic acid is

\(q = \mathrm{\dfrac{0.1025}{122.13}\, mol \times 3226\: \dfrac{kJ}{mol} = 2.680\: kJ}\)

Thus, the heat capacity is

\(C = \dfrac{q}{dT} = \dfrac{2.680}{2.165} = \mathrm{1.238\: kJ / K}\).

After the heat capacity is determined, the calorimeter is ready to be used to measure the enthalpy of combustion of other substances.

Example 7

When 0.7022 g of oxalic acid (\(\ce{C2O4H2}\)) is burnt in the calorimeter under the same conditions as Example 6, the temperature increased by 1.602°C. The heat capacity of the calorimeter is 1.238 kJ/K. Calculate *dH*°_{comb}.

*Solution:*

The balanced equation and various quantities calculated are given in a logical order below:

\(\mathrm{C_2O_4H_{2\large{(s)}} + 0.5\, O_{2\large{(g)}} \rightarrow 2 CO_{2\large{(g)}} + H_2O_{\large{(l)}}}\)

\(dn = 1.5\: q = C\: dT = 1.238\times1.602 = \mathrm{1.984\: kJ}\)

\(n\: \mathrm{of\: oxalic\: acid = \dfrac{0.7022}{90} = 0.00780\: mol}\)

\(\begin{align}

dE &= \mathrm{\dfrac{-1.984}{0.00780} = -354.4\: kJ/mol}\\

dH &= dE + dn R T\\

&= \mathrm{-254.4\: kJ + 1.5 mol \times 0.008314 \dfrac{kJ}{K\: mol}\times 298\: K}\\

&= \mathrm{-250.6\: kJ/mol}

\end{align}\)

**Discussion:**

Can the standard enthalpy of formation *dH*°_{f} for oxalic acid be calculated? What additional data are required?

Example 8

Calculate the enthalpy of formation of oxalic acid, for which the enthalpy of combustion is -251 kJ/mol.

*Solution:*

The following data were looked up from thermodynamic data,

\(\mathrm{

\mathit{dH}^\circ_f (CO_{2\large{(g)}}) = -393.5\: kJ/mol\\

\mathit{dH}^\circ_f (H_2O_{\large{(l)}}) = -285.8\: kJ/mol}\)

Since

\(\mathrm{C_2O_4H_{2\large{(s)}} + 0.5\: O_{2\large{(g)}} \rightarrow 2 CO_{2\large{(g)}} + H_2O_{\large{(l)}}}\)

\(\mathrm{\mathit{dH}^\circ_{comb} = 2 \mathit{dH}^\circ_f(CO_{2\large{(g)}}) + \mathit{dH}^\circ_f(H_2O_{\large{(l)}}) - \mathit{dH}^\circ_f(oxalic\: acid)}\)

\(\mathrm{-250.6 = 2 \times (-393.5) + (-285.8) - \mathit{dH}^\circ_f(oxalic\: acid)}\)

Therefore

\(\begin{align}

\mathrm{\mathit{dH}^\circ_f(oxalic\: acid)} &= 2 \times (-393.5) + (-285.8) - (-250.6)\\

&= \mathrm{-822.2\: kJ/mol}

\end{align}\)

**Discussion:**

It's important to know what additional data is required for problem solving. It's equally important to know where to look for them.

We have given an introduction to calorimetry here, and for a detailed laboratory instruction, consult the CHEM120L manual and the Oxygen Bomb Calorimetry which has several pages giving the details of the measurement. In particular, how to measure the temperature difference is given in Calorimetry Data Analysis.

## Confidence Building Questions

**A microwave oven operates at 500 watt for 2 minutes. How much electric energy (J) is consumed?**Hint: 60000 J

**Skill:**Calculate amount of electric energy for heat capacity measurement. Review Skills.

**A cup of (200 g) water heated in a microwave oven had a temperature increase of 60 K. How much energy (in kJ) was absorbed by the water? Recall that heat capacity of water is 4.184 J/(g K).**Hint: 502.1 kJ

**Skill:**Calculate the amount of energy when the heat capacity and temperature increase is known, a preparation for enthalpy evaluation.

**The temperature of a calorimeter increased 0.25 K when 50.0 J of energy was absorbed. What is its heat capacity (kJ/K)?**Hint: 200 J/K

**Skill:**Measure the heat capacity of a calorimeter.

**When 100 mL of 0.50 M**\(\ce{NH3}\)**is neutralized by 300 mL of a solution containing excess**\(\ce{HCl}\)**, the solution temperature increases 1.6 K. The experiment is performed in a styrofoam cup, and there is no energy loss. Calculate the heat of reaction,***dH*.Hint: 2678 J

**Discussion:**2678 J was generated by 0.05 mol \(\ce{NH3}\) (0.100 L * 0.5 mol/L)

**How much heat (kJ) should be generated if 1.0 mole**\(\ce{NH3}\)**is neutralized by dilute**\(\ce{HCl}\)**solution, if 2675 J was generated when 0.05 mol**\(\ce{NH3}\)**(0.100 L * 0.5 mol/L) was neutralized as described in the previous problem?**Hint: 53.6 kJ

**Discussion:**In fact, 53.6 kJ/mol is called the molar heat of neutralization of \(\ce{NH3}\).

**Combustion of 0.92 g toluene,**\(\ce{C7H8}\)**, in a bomb calorimeter (Cv equivalent to 1003 g of water) raised its temperature by 9.3 K. Estimate***dH*.Hint: -3909 kJ

**Some details***q*= 39040 J;*dE*= 3904 kJ;*dH*= -3909 kJ/mol.**When 0.1234 g of oxalic acid (**\(\ce{C2O4H2}\)**) is burnt in the calorimeter under the same conditions as Example 6, the temperature increased by 2.314°C. Calculate the heat capacity of the calorimeter.**Hint: 1.409 kJ/K

**Skill**Using a standard to determine the heat capacity of a calorimeter.

**When 0.2006 g of oxalic acid (**\(\ce{C2O4H2}\)**) is burnt in the calorimeter whose heat capacity is 1.238 kJ/K, the temperature increased by 0.4577 K. Calculate***dH*°_{comb}.Hint: -250.6 kJ

**Skill:**Measure the enthalpy of combustion by bomb calorimetry.

## Contributors

Chung (Peter) Chieh (Chemistry, University of Waterloo)