# 1.7: Heats of Reactions - ΔU and ΔH


##### Learning Objectives
• To understand how enthalpy pertains to chemical reactions

When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction ($$ΔH_{rxn}$$), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so $$ΔH_{rxn}$$ is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so $$ΔH_{rxn}$$ is positive. Thus $$ΔH_{rxn} < 0$$ for an exothermic reaction, and $$ΔH_{rxn} > 0$$ for an endothermic reaction. In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:

Reaction Type q $$ΔH_{rxn}$$
exothermic < 0 < 0 (heat flows from a system to its surroundings)
endothermic > 0 > 0 (heat flows from the surroundings to a system)

If $$ΔH_{rxn}$$ is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Figure $$\PageIndex{1a}$$). Conversely, if $$ΔH_{rxn}$$ is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure $$\PageIndex{1b}$$).

Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.

Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.

• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):

\begin{align} \text{heat} + \ce{H_{2}O(s)} & \ce{ -> H_{2}O(l)} & \Delta H > 0 \label{7.6.7} \\[4pt] \ce{H2O (l)} & \ce{-> H2O(s) + heat} & \Delta H < 0 \end{align} \label{7.6.8}

In both cases, the magnitude of the enthalpy change is the same; only the sign is different.

• Enthalpy is an extensive property (like mass). The magnitude of $$ΔH$$ for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation $$\ref{7.6.9}$$, which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:

$\ce{ 2Al (s ) + Fe2O3 (s ) -> 2Fe (s) + Al2O3 (s )} + 851.5 \; kJ \label{7.6.9}$

Thus $$ΔH = −851.5 \,kJ/mol$$ of $$\ce{Fe2O3}$$. We can also describe $$ΔH$$ for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for $$ΔH$$, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation \ref{7.6.9}, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:

$\ce{ 2Al (s) + Fe2O3 (s) -> 2Fe (s) + Al2O3 (s)} \quad\quad \Delta H_{rxn}= - 851.5 \; kJ \label{7.6.10}$

If 4 mol of Al and 2 mol of Fe2O3 react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:

$- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{7.6.6}$

The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example $$\PageIndex{1}$$.

##### Example $$\PageIndex{1}$$

Certain parts of the world, such as southern California and Saudi Arabia, are short of fresh water for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If $$ΔH$$ is 6.01 kJ/mol for the reaction $$\ce{H2O(s) → H2O(l)}$$ at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.)

Given: energy per mole of ice and mass of iceberg

Asked for: energy required to melt iceberg

Strategy:

1. Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
2. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice.

Solution:

A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):

\begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{metric \; tons}\, H_{2}O \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{metric \; ton}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right )\\[4pt] & = 5.55\times 10^{10} \; mol H_{2}O \end{align*}

B The energy needed to melt the iceberg is thus

$\left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber$

Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below.

Possible sources of the approximately 3.34 × 1011 kJ needed to melt a 1.00 × 106 metric ton iceberg

• Combustion of 3.8 × 103 ft3 of natural gas
• Combustion of 68,000 barrels of oil
• Combustion of 15,000 tons of coal
• 1.1 × 108 kilowatt-hours of electricity
##### Exercise $$\PageIndex{1}$$

If 17.3 g of powdered aluminum are allowed to react with excess $$\ce{Fe2O3}$$, how much heat is produced?

• 273 kJ

## Types of Enthalpies of Reactions

One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:

• Enthalpy of combustion ($$ΔH_{comb}$$) is the change in enthalpy that occurs during a combustion reaction. Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance.
• Enthalpy of fusion ($$ΔH_{fus}$$) is the enthalpy change that accompanies the melting (fusion) of 1 mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds.
• Enthalpy of vaporization ($$ΔH_{vap}$$) is the enthalpy change that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds.
• Enthalpy of solution ($$ΔH_{soln}$$) is the change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent.
Table $$\PageIndex{1}$$: Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points
Substance ΔHvap (kJ/mol) ΔHfus (kJ/mol)
argon (Ar) 6.3 1.3
methane (CH4) 9.2 0.84
ethanol (CH3CH2OH) 39.3 7.6
benzene (C6H6) 31.0 10.9
water (H2O) 40.7 6.0
mercury (Hg) 59.0 2.29
iron (Fe) 340 14

The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.

Enthalpy of Reaction: https://youtu.be/z2KUaIEF9qI

## Summary

Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy ($$ΔH$$) can be measured. A negative $$ΔH$$ means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the enthalpy of reaction ($$ΔH_{rxn}$$) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of $$ΔH_{rxn}$$.

The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion ($$ΔH_{fus}$$) and the enthalpy of vaporization ($$ΔH_{vap}$$), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion ($$ΔH_{comb}$$) is the enthalpy change that occurs when a substance is burned in excess oxygen.

## Contributors and Attributions

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