Skip to main content
Chemistry LibreTexts

8.4: Standard Enthalpy and Hess’ Law

  • Page ID
    222609
  • Skills to Develop

    • Calculate enthalpy changes for various chemical reactions
    • Explain Hess’s law and use it to compute reaction enthalpies
     

     

    Standard Enthalpy of Formation

    A standard enthalpy of formation \(ΔH^\circ_\ce{f}\) is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.

    The standard enthalpy of formation of CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

    \[\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\hspace{20px}ΔH^\circ_\ce{f}=ΔH^\circ_{298}=−393.5\:\ce{kJ} \label{5.4.9}\]

    starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO2, also at 1 atm and 25 °C. For nitrogen dioxide, \(\ce{NO}_{2(g)}\), \(ΔH^\circ_\ce{f}\) is 33.2 kJ/mol. This is the enthalpy change for the reaction:

    \[\frac{1}{2}\ce{N2}(g)+\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH^\circ_\ce{f}=ΔH^\circ_{298}=+33.2\: \ce{kJ} \label{5.4.10}\]

    A reaction equation with \(\frac{1}{2}\) mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(g).

    You will find a table of standard enthalpies of formation of many common substances in Tables T1 and T2. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

    Example \(\PageIndex{1}\): Evaluating an Enthalpy of Formation

    Ozone, O3(g), forms from oxygen, O2(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, \(ΔH^\circ_\ce{f}\) of ozone from the following information:

    \[\ce{3O2}(g)⟶\ce{2O3}(g)\hspace{20px}ΔH^\circ_{298}=+286\: \ce{kJ} \nonumber\]

    Solution

    \(ΔH^\circ_\ce{f}\) is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, \(ΔH^\circ_\ce{f}\) for O3(g) is the enthalpy change for the reaction:

    \[\dfrac{3}{2}\ce{O2}(g)⟶\ce{O3}(g) \nonumber\]

    For the formation of 2 mol of O3(g), \(ΔH^\circ_{298}=+286\: \ce{kJ}\). This ratio, \(\mathrm{\left(\dfrac{286\:kJ}{2\:mol\:O_3}\right)}\), can be used as a conversion factor to find the heat produced when 1 mole of O3(g) is formed, which is the enthalpy of formation for O3(g):

    \[ΔH^\circ \ce{\:for\:1\:mole\: of\:O_3}(g)=\mathrm{1\:\cancel{mol\:O_3}×\dfrac{286\:kJ}{2\:\cancel{mol\:O_3}}=143\:kJ} \nonumber\]

    Therefore, \(ΔH^\circ_\ce{f}[\ce{O3}(g)]=\ce{+143\: kJ/mol}\).

    Exercise \(\PageIndex{1}\)

    Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

    Answer

    For the reaction

    \[\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−184.6\:kJ}\nonumber\]

    Example \(\PageIndex{2}\): Writing Reaction Equations for \(ΔH^\circ_\ce{f}\)

    Write the heat of formation reaction equations for:

    1. \(\ce{C2H_5OH}_{(l)}\)
    2. \(\ce{Ca_3(PO_4)}_{2(s)}\)

    Solution

    Remembering that \(ΔH^\circ_\ce{f}\) reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

    1. \(\ce{2C}(s,\:\ce{graphite})+\ce{3H2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{C2H5OH}(l)\)
    2. \(\ce{3Ca}(s)+\frac{1}{2}\ce{P4}(s)+\ce{4O2}(g)⟶\ce{Ca3(PO4)2}(s)\)

    Note: The standard state of carbon is graphite, and phosphorus exists as \(P_4\).

    Exercise \(\PageIndex{2}\)

    Write the heat of formation reaction equations for:

    1. \(\ce{C_2H_5OC_2H}_{5(l)}\)
    2. \(\ce{Na_2CO}_{3(s)}\)
    Answer a

    \(\ce{4C}(s,\:\ce{graphite})+\ce{5H2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{C2H5OC2H5}(l)\);

    Answer b

    \(\ce{2Na}(s)+\ce{C}(s,\:\ce{graphite})+\dfrac{3}{2}\ce{O2}(g)⟶\ce{Na2CO3}(s)\)

    ​​​​Hess’s Law

    There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

    This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

    \[\ce{C}_{(s)}+\ce{O}_{2(g)}⟶\ce{CO}_{2(g)}\;\;\;ΔH^∘_{298}=\mathrm{−394\;kJ} \label{ 5.4.11}\]

    In the two-step process, first carbon monoxide is formed:

    \[\ce{C}_{(s)}+\dfrac{1}{2}\ce{O}_{2(g)}⟶\ce{CO}_{(g)}\;\;\;ΔH^∘_{298}=\mathrm{−111\;kJ} \label{ 5.4.12}\]

    Then, carbon monoxide reacts further to form carbon dioxide:

    \[\ce{CO} {(g)}+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CO}_2 {(g)}\;\;\;ΔH^∘_{298}=\mathrm{−283\;kJ} \label{ 5.4.13}\]

    The equation describing the overall reaction is the sum of these two chemical changes:

    \[\begin {align*} &\textrm{Step 1:} \:\ce{C}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)\\ &\underline{\textrm{Step 2:} \:\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO2}(g)}\\ &\textrm{Sum:} \:\ce{C}(s)+\frac{1}{2}\ce{O2}(g)+\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)+\ce{CO2}(g) \end {align*} \label{5.4.14}\]

    Because the CO produced in Step 1 is consumed in Step 2, the net change is:

    \[\ce{C}_{(s)}+\ce{O}_{2(g)}⟶\ce{CO}_{2(g)} \label{5.4.15}\]

    According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in Table \(\PageIndex{1}\) to find the enthalpy change of the entire reaction from its two steps:

    \[\begin {align*}
    &\ce{C}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)&&ΔH^\circ_{298}=\mathrm{−111\:kJ}\\
    &\ce{CO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO2}(g)&&ΔH^\circ_{298}=\mathrm{−283\:kJ}\\
    &\overline{\ce{C}(s)+\ce{O2}(g)⟶\ce{CO2}(g)\hspace{25px}}&&\overline{ΔH^\circ_{298}=\mathrm{−394\:kJ}}
    \end {align*} \label{5.4.16}\]

    The result is shown in Figure \(\PageIndex{6}\). We see that ΔH of the overall reaction is the same whether it occurs in one step or two. This finding (overall ΔH for the reaction = sum of ΔH values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

    CNX_Chem_05_03_HessCO2.jpg

    Figure \(\PageIndex{2}\): The formation of CO2(g) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.

    Before we further practice using Hess’s law, let us recall two important features of ΔH.

    1. ΔH is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO2(g) is +33.2 kJ: \[\frac{1}{2}\ce{N2}(g)+\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH=\mathrm{+33.2\: kJ} \label{5.4.17}\]

      When 2 moles of NO2 (twice as much) are formed, the ΔH will be twice as large: \[\ce{N2}(g)+\ce{2O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{+66.4\: kJ} \label{5.4.18}\]

      In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.

    2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the reverse direction. For example, given that: \[\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH=\mathrm{−184.6\:kJ} \label{5.4.19}\]

      Then, for the “reverse” reaction, the enthalpy change is also “reversed”: \[\ce{2HCl}(g)⟶\ce{H2}(g)+\ce{Cl2}(g)\hspace{20px}ΔH=\mathrm{+184.6\: kJ} \label{5.4.20}\]

    Example \(\PageIndex{3}\): Stepwise Calculation of \(ΔH^\circ_\ce{f}\)

    Using Hess’s Law Determine the enthalpy of formation, \(ΔH^\circ_\ce{f}\), of FeCl3(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

    \[\ce{Fe}(s)+\ce{Cl2}(g)⟶\ce{FeCl2}(s)\hspace{20px}ΔH°=\mathrm{−341.8\:kJ} \nonumber\]

    \[\ce{FeCl2}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH°=\mathrm \nonumber{−57.7\:kJ} \]

    Solution

    We are trying to find the standard enthalpy of formation of FeCl3(s), which is equal to ΔH° for the reaction:

    \[\ce{Fe}(s)+\frac{3}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH^\circ_\ce{f}=\:? \nonumber\]

    Looking at the reactions, we see that the reaction for which we want to find ΔH° is the sum of the two reactions with known ΔH values, so we must sum their ΔHs:

    \[\ce{Fe}(s)+\ce{Cl2}(g)⟶\ce{FeCl2}(s)\hspace{59px}ΔH°=\mathrm{−341.8\:kJ}\\ \underline{\ce{FeCl2}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{20px}ΔH°=\mathrm{−57.7\:kJ}}\\ \ce{Fe}(s)+\frac{1}{2}\ce{Cl2}(g)⟶\ce{FeCl3}(s)\hspace{43px}ΔH°=\mathrm{−399.5\:kJ} \nonumber\]

    The enthalpy of formation, \(ΔH^\circ_\ce{f}\), of FeCl3(s) is −399.5 kJ/mol.

    Exercise \(\PageIndex{3}\)

    Calculate ΔH for the process:

    \[\ce{N2}(g)+\ce{2O2}(g)⟶\ce{2NO2}(g) \nonumber\]

    from the following information:

    \[\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)\hspace{20px}ΔH=\mathrm{180.5\:kJ} \nonumber\]

    \[\ce{NO}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{NO2}(g)\hspace{20px}ΔH=\mathrm{−57.06\:kJ} \nonumber\]

    Answer

    66.4 kJ

    Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.

    Example \(\PageIndex{4}\): A More Challenging Problem

    Using Hess’s Law Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

    (i) \(\ce{ClF}(g)+\ce{F2}(g)⟶\ce{ClF3}(g)\hspace{20px}ΔH°=\:?\)

    Use the reactions here to determine the ΔH° for reaction (i):

    (ii) \(\ce{2OF2}(g)⟶\ce{O2}(g)+\ce{2F2}(g)\hspace{20px}ΔH^\circ_{(ii)}=\mathrm{−49.4\:kJ}\)

    (iii) \(\ce{2ClF}(g)+\ce{O2}(g)⟶\ce{Cl2O}(g)+\ce{OF2}(g)\hspace{20px}ΔH^\circ_{(iii)}=\mathrm{+205.6\: kJ}\)

    (iv) \(\ce{ClF3}(g)+\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)\hspace{20px}ΔH^\circ_{(iv)}=\mathrm{+266.7\: kJ}\)

    Solution

    Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that \(\ce{ClF}_{(g)}\) is needed as a reactant. This can be obtained by multiplying reaction (iii) by \(\frac{1}{2}\), which means that the ΔH° change is also multiplied by \(\frac{1}{2}\):

    \[\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)\hspace{20px} ΔH°=\frac{1}{2}(205.6)=+102.8\: \ce{kJ} \nonumber\]

    Next, we see that \(\ce{F_2}\) is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the ΔH° changes sign and is halved:

    \[\frac{1}{2}\ce{O2}(g)+\ce{F2}(g)⟶\ce{OF2}(g)\hspace{20px}ΔH°=+24.7\: \ce{kJ} \nonumber\]

    To get ClF3 as a product, reverse (iv), changing the sign of ΔH°:

    \[\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)⟶\ce{ClF3}(g)+\ce{O2}(g)\hspace{20px}ΔH°=\mathrm{−266.7\: kJ} \nonumber\]

    Now check to make sure that these reactions add up to the reaction we want:

    \[\begin {align*}
    &\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)&&ΔH°=\mathrm{+102.8\: kJ}\\
    &\frac{1}{2}\ce{O2}(g)+\ce{F2}(g)⟶\ce{OF2}(g)&&ΔH°=\mathrm{+24.7\: kJ}\\
    &\frac{1}{2}\ce{Cl2O}(g)+\dfrac{3}{2}\ce{OF2}(g)⟶\ce{ClF3}(g)+\ce{O2}(g)&&ΔH°=\mathrm{−266.7\:kJ}\\
    &\overline{\ce{ClF}(g)+\ce{F2}⟶\ce{ClF3}(g)\hspace{130px}}&&\overline{ΔH°=\mathrm{−139.2\:kJ}}
    \end {align*}\]

    Reactants \(\frac{1}{2}\ce{O2}\) and \(\frac{1}{2}\ce{O2}\) cancel out product O2; product \(\frac{1}{2}\ce{Cl2O}\) cancels reactant \(\frac{1}{2}\ce{Cl2O}\); and reactant \(\dfrac{3}{2}\ce{OF2}\) is cancelled by products \(\frac{1}{2}\ce{OF2}\) and OF2. This leaves only reactants ClF(g) and F2(g) and product ClF3(g), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified ΔH° values will give the desired ΔH°:

    \(ΔH°=\mathrm{(+102.8\:kJ)+(24.7\:kJ)+(−266.7\:kJ)=−139.2\:kJ}\)

    Exercise \(\PageIndex{4}\)

    Aluminum chloride can be formed from its elements:

    (i) \(\ce{2Al}(s)+\ce{3Cl2}(g)⟶\ce{2AlCl3}(s)\hspace{20px}ΔH°=\:?\)

    Use the reactions here to determine the ΔH° for reaction (i):

    (ii) \(\ce{HCl}(g)⟶\ce{HCl}(aq)\hspace{20px}ΔH^\circ_{(ii)}=\mathrm{−74.8\:kJ}\)

    (iii) \(\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH^\circ_{(iii)}=\mathrm{−185\:kJ}\)

    (iv) \(\ce{AlCl3}(aq)⟶\ce{AlCl3}(s)\hspace{20px}ΔH^\circ_{(iv)}=\mathrm{+323\:kJ/mol}\)

    (v) \(\ce{2Al}(s)+\ce{6HCl}(aq)⟶\ce{2AlCl3}(aq)+\ce{3H2}(g)\hspace{20px}ΔH^\circ_{(v)}=\mathrm{−1049\:kJ}\)

    Answer

    −1407 kJ

    We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with \(\sum\) representing “the sum of” and n standing for the stoichiometric coefficients:

    \[ΔH^\circ_\ce{reaction}=\sum n×ΔH^\circ_\ce{f}\ce{(products)}−\sum n×ΔH^\circ_\ce{f}\ce{(reactants)} \label{5.4.20B}\]

    The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.

    Example \(\PageIndex{5}\): Using Hess’s Law

    What is the standard enthalpy change for the reaction:

    \[\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g)\hspace{20px}ΔH°=\:? \nonumber\]

    Solution 1: Using the Equation

    Alternatively, we could use the special form of Hess’s law given previously:

    \[ΔH^\circ_\ce{reaction}=∑n×ΔH^\circ_\ce{f}\ce{(products)}−∑n×ΔH^\circ_\ce{f}\ce{(reactants)} \nonumber\]

    \[\begin {align*} &=\mathrm{\left[2\:\cancel{mol\:HNO_3}×\dfrac{−207.4\:kJ}{\cancel{mol\:HNO_3\:(\mathit{aq})}}+1\:\cancel{mol\: NO\:(\mathit{g})}×\dfrac{+90.2\: kJ}{\cancel{mol\: NO\:(\mathit{g})}}\right]}\\ &\mathrm{\:−\,\left[3\:\cancel{mol\:NO_2(\mathit{g})}×\dfrac{+33.2\: kJ}{\cancel{mol\:NO_2\:(\mathit{g})}}+1\:\cancel{mol\:H_2O\:(\mathit{l})}×\dfrac{−285.8\:kJ}{\cancel{mol\:H_2O\:(\mathit{l})}}\right]}\\ &=\mathrm{2(−207.4\:kJ)+1(+90.2\: kJ)−3(+33.2\: kJ)−1(−285.8\:kJ)}\\ &=\mathrm{−138.4\:kJ}\end {align*}\]

    Solution 2: Supporting Why the General Equation Is Valid

    We can write this reaction as the sum of the decompositions of 3NO2(g) and 1H2O(l) into their constituent elements, and the formation of 2 HNO3(aq) and 1 NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the \(ΔH^\circ_\ce{f}\) values for these compounds (from Tables T1 and T2 ), we have:

    \[\ce{3NO2}(g)⟶ \dfrac{3}{2} \ce{N2}(g)+ 3 \ce{O2}(g)\hspace{20px}ΔH^\circ_{1}=\mathrm{−99.6\:kJ} \nonumber\]

    \[\ce{H2O}(l)⟶\ce{H2}(g)+\frac{1}{2}\ce{O2}(g)\hspace{20px}ΔH^\circ_{2}=+285.8\: \ce{kJ}\:[−1×ΔH^\circ_\ce{f}(\ce{H2O})]\nonumber\]

    \[\ce{H2}(g)+\ce{N2}(g)+ 3 \ce{O2}(g)⟶\ce{2HNO3}(aq)\hspace{20px}ΔH^\circ_{3}=−414.8\:kJ\:[2×ΔH^\circ_\ce{f}(\ce{HNO3\nonumber})]\]

    \[\frac{1}{2}\ce{N2}(g)+\frac{1}{2}\ce{O2}(g)⟶\ce{NO}(g)\hspace{20px}ΔH^\circ\nonumber_{4}=+90.2\: \ce{kJ}\:[1×(\ce{NO})]\]

    Summing these reaction equations gives the reaction we are interested in:

    \[\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g)\nonumber\]

    Summing their enthalpy changes gives the value we want to determine:

    \[\begin {align*} ΔH^\circ_\ce{rxn}&=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3+ΔH^\circ_4=\mathrm{(−99.6\:kJ)+(+285.8\: kJ)+(−414.8\:kJ)+(+90.2\: kJ)}\\ &=\mathrm{−138.4\:kJ} \end {align*}\]

    So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ.

    Note that this result was obtained by:

    1. multiplying the \(ΔH^\circ_\ce{f}\) of each product by its stoichiometric coefficient and summing those values,
    2. multiplying the \(ΔH^\circ_\ce{f}\) of each reactant by its stoichiometric coefficient and summing those values, and then
    3. subtracting the result found in step 2 from the result found in step 1.

    This is also the procedure in using the general equation, as shown.

    Exercise \(\PageIndex{5}\)

    Calculate the heat of combustion of 1 mole of ethanol, C2H5OH(l), when H2O(l) and CO2(g) are formed. Use the following enthalpies of formation: C2H5OH(l), −278 kJ/mol; H2O(l), −286 kJ/mol; and CO2(g), −394 kJ/mol.

    Answer

    −1368 kJ/mol

    Summary

    Video \(\PageIndex{1}\): A summary of Enthalpy and Hess' Law from Crash Course Chemistry.

    If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the change is called the enthalpy change with the symbol ΔH, or \(ΔH^\circ_{298}\) for reactions occurring under standard state conditions. The value of ΔH for a reaction in one direction is equal in magnitude, but opposite in sign, to ΔH for the reaction in the opposite direction, and ΔH is directly proportional to the quantity of reactants and products. Examples of enthalpy changes include enthalpy of combustion, enthalpy of fusion, enthalpy of vaporization, and standard enthalpy of formation. The standard enthalpy of formation, \(ΔH^\circ_\ce{f}\), is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar (standard state). Many of the processes are carried out at 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.

    Key Equations

    • \(ΔU=q+w\)
    • \(ΔH^\circ_\ce{reaction}=∑n×ΔH^\circ_\ce{f}\ce{(products)}−∑n×ΔH^\circ_\ce{f}\ce{(reactants)}\)

    Footnotes

    1. For more on algal fuel, see http://www.theguardian.com/environme...n-fuel-problem.

    Glossary

    chemical thermodynamics
    area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes
    enthalpy (H)
    sum of a system’s internal energy and the mathematical product of its pressure and volume
    enthalpy change (ΔH)
    heat released or absorbed by a system under constant pressure during a chemical or physical process
    expansion work (pressure-volume work)
    work done as a system expands or contracts against external pressure
    first law of thermodynamics
    internal energy of a system changes due to heat flow in or out of the system or work done on or by the system
    Hess’s law
    if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps
    hydrocarbon
    compound composed only of hydrogen and carbon; the major component of fossil fuels
    internal energy (U)
    total of all possible kinds of energy present in a substance or substances
    standard enthalpy of combustion (\(ΔH^\circ_\ce{c}\))
    heat released when one mole of a compound undergoes complete combustion under standard conditions
    standard enthalpy of formation (\(ΔH^\circ_\ce{f}\))
    enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions
    standard state
    set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations, usually at a temperature of 298.15 K
    state function
    property depending only on the state of a system, and not the path taken to reach that state

    Contributors

    • Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

    • Adelaide E. Clark, Oregon Institute of Technology
    • Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes.

     

    • Was this article helpful?