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Chemistry LibreTexts

Chemical Energy

  • Page ID
  • Skills to Develop

    • Discuss the energy aspect of chemical reactions.
    • Identify chemical reactions that provide energy.
    • Calculate the amount of energy accompanying a chemical reaction.
    • Identify exothermic and endothermic changes.
    • Estimate work performed due to volume change under constant pressure, including the application of ideal gas law.
    • Explain enthalpy and standard enthalpy of a reaction.
    • Calculate the energy change using standard enthalpy of reaction and standard enthalpy of formation.

    Chemical Energy

    firework.jpgPropane (\(\ce{C3H8}\)), natural gas (\(\ce{CH4}\)), and phosphorous (\(\ce{P4}\)) react with oxygen (\(\ce{O2}\)), and these reactions release energy in the form of heat and light. No doubt, according to the principle of conservation of energy, energy is required to reverse the reactions. Thus, energy stored in chemicals (compounds) and energy released or absorbed in chemical reactions are called chemical energy, which also covers topics such as bond energy, ionization potential, electron affinity, electronegativity, lattice energy, etc.

    For example, at standard conditions, the combustion of 1.0 mole hydrogen with oxygen releases 285.8 kJ of energy. We represent the reaction:

    \(\mathrm{H_{2\large{(g)}} + \dfrac{1}{2} O_2 \rightarrow H_2O_{\large{(l)}}}, \hspace{20px} dH = \mathrm{-285.8\: kJ/mol}\)

    where dH represent the heat (or enthalpy) of reaction, and a negative value means that the heat is released. Usually, dH is represented by DH in textbooks, but using dH notation is much less work on Internet documents.

    For the reverse reaction, 285.8 kJ/mol is required, and the sign for dH value changes.

    \(\mathrm{H_2O_{\large{(l)}} \rightarrow H_{2\large{(g)}} + \dfrac{1}{2} O_2}, \hspace{20px} dH = \mathrm{+285.8\: kJ/mol}\)

    A Chemical Energy Level Diagram
      ------------H2(g) + 1/2O2
            ­ |
            | |
     286 kJ | |  -286 kJ
            | |
            | ¯
      ------------   H2O

    We can also use an energy level diagram to show the relative content of energy. The energy content of \(\mathrm{H_{2\large{(g)}} + 0.5\, O_2}\) is 285.8 kJ higher than a mole of water, \(\ce{H2O}\).

    Oil, gas, and food are often called energy by the news media, but more precisely they are sources of (chemical) energy -- energy stored in chemicals with a potential to be released in a chemical reaction. The released energy performs work or causes physical and chemical changes.

    It is obvious that the amount of energy released in a chemical reaction is related to the amount of reactants. For example, when the amount is doubled, so is the amount of energy released.

    \(\mathrm{2 H_{2\large{(g)}} + O_2 \rightarrow 2 H_2O_{\large{(l)}}}, \hspace{20px} dH = \mathrm{-571.6\: kJ/mol}\)

    Example 1 shows the calculation when the amount of reactants is only a fraction of a mole.

    Example 1

    How much energy is released when a balloon containing 0.15 mole of hydrogen is ignited in the air?

    The amount released is \(\mathrm{0.15\: mol \times 285.8\: kJ/mol = 42.9\: kJ}\)

    The sudden release of energy causes an explosion.

    Endothermic and Exothermic Reactions

    A reaction that releases energy is called exothermic reaction. Energy is released in the form of heat, light and (pressure-volume) work. For example, when methane or propane is oxidized by \(\ce{O2}\), the heat released causes the gas to expand (explosion in some cases), releasing heat & light and doing work at the same time. In this case, the energy source came from chemical reactions instead of at the expense of internal energy described in the previous module.

    Endothermic reactions absorb energy, and in all cases, the energy is supplied from another source, in the form of electrical energy, heat or light.

    Pressure-Volume Work in Chemical Reactions

    Many chemical reactions involve gases, and when a gas is formed, it displaces other gases by pushing them out against a pressure. Work, defined in Newtonian physics as a force times the distance along the force direction, is performed in such an action. The work is called the pressure-volume (P-V) work, which is a form of energy and it must be analyzed and its quantity included in chemical energy calculations.

    The SI units for pressure are N m-2 and that of the volume is m3. Pressure times volume gives the unit of N m, which is the definition of joule,

    \mathrm{1\: Pa \times 1\: m^3}
    &= \mathrm{1\: N\: m^{-2}\: m}\\
    &= \textrm{1 N m}\\
    &= \textrm{1 J}

    Since 1 atm = 101300 Pa, and 1 L = 0.001 m3, thus

    \(\mathrm{1\: atm\: L = 101.3\: J}\)

    The P-V work under constant pressure (P) is simply the pressure times the change in volume dV.

    \(w = - P\, dV\)

    This method applies to reactions that produce gases, which are released into the atmosphere. When work is done by the system, the work is negative, as the formula indicates. In reactions where gases are consumed to produce a liquid or solid, work is done on the system by its environment. The work is positive.

    In cases where the pressure varies, an integral approach is required to evaluate the pressure volume work.

    w &= - \int \mathrm d (P V)\\
    &= - \int P \mathrm dV - \int V \mathrm dP

    The negative sign is retained, because the work done by the system is negative. However, the integral of P V work depends on the path, and we will not get into a detailed discussion at this stage.

    Example 2

    In the reaction to produce oxygen,

    \(\mathrm{KClO_{3\large{(s)}} \rightarrow KCl_{\large{(s)}} + \dfrac{3}{2} O_{2\large{(g)}}}\)

    calculate the pressure-volume work done by 8.2 g of \(\ce{KClO3}\).

    The molar mass of \(\ce{KClO3}\) is 123.5 g/mol and 8.2 g is 0.067 mol. Thus, the amount of oxygen produced is 0.10 (= 0.067*2/3) mol. Apply the ideal gas law to the pressure volume work (P V), and we have

    \(P V = n R T\)

    w &= - \mathrm D P V\\
    &= - \mathrm D n R T\\
    &= \mathrm{- 0.10\: mol\times8.312\: \dfrac{J}{mol\: K}\times 298\: K}\\
    &= \mathrm{- 248\: J}


    The work done is due to the formation of gas \(\ce{O2}\) which expands against the atmosphere of 1.0 atm or 101.3 kPa. The volume changes of the solids are insignificant compared to that of the gas.

    In cases where both pressure and volume change, the work is the difference of the pressure-volume product, DP V.


    The enthalpy, usually represented by H, is the energy released in a chemical reaction under constant pressure, H = qP. The enthalpy is a convenient property to evaluate for reactions taking place at constant pressure. Enthalpy differs from internal energy, E, defined in Energy as the energy input to a system at constant volume. The energy released in a chemical reaction raises the internal energy, E, and does work under constant pressure at the expense of energy stored in compounds. Thus,

    \(H = q_{\mathrm P} = E + P\, dV\)

    Of course, the enthalpy change (dH) of a chemical reaction depends on the amount of reactants, the temperature, and pressure. Under normal conditions, the ideal gas law can be applied to give reasonable results.

    Like the internal energy, enthalpy is also a thermodynamic state function, depending only on the initial and final states of the system, but not on the rate of reaction.

    Standard Enthalpy of Reaction

    In order to make the data useful for scientific and engineering applications, there is a general agreement to report and tabulate enthalpy changes for a mole of reaction at a standard temperature and pressure. Such quantities are called the standard enthalpy of reaction.

    In handbooks and textbooks, the standard enthalpy change is represented by \(\Delta\)Ho. For simplicity, we use dHo to represent the changes of standard enthalpy in our discussion to avoid (very) slow loading of the delta onto your computer.

    Example 3

    The standard enthalpy for the combustion of methane is 890.4 kJ per mole,

    \(\mathrm{CH_{4\large{(g)}} + 2 O_{2\large{(g)}} \rightarrow CO_{2\large{(g)}} + 2 H_2O_{\large{(g)}}}, \hspace{20px} dH^\circ = \mathrm{-890.4\: kJ/mol}\)

    Calculate the standard enthalpy change when 1.0 cubic meter of natural gas is burned converting to gaseous products.

    When 1.0 mol or 22.4 L of \(\ce{CH4}\), at 273K and 1 atm, is oxidized completely, the standard enthalpy change is 890.4 kJ. One cubic meter is 1000 L (/22.4 = 44.6 mol). Thus, the standard enthalpy of change is

    \(dH = \mathrm{44.6\: mol \times 890.4\: kJ/mol = 39712\: kJ\: or\: 39\: million\: joules}\).

    A problem can be made up using any of the following standard enthalpy of reactions. These are given here to illustrate the type of reactions and the representation of enthalpy of reactions.

    &\mathrm{2 H_{\large{(g)}} \rightarrow H_{2\large{(g)}}} &&dH^\circ = \mathrm{-436\: kJ/mol}\\
    &\mathrm{2 O_{\large{(g)}} \rightarrow O_{2\large{(g)}}} &&dH^\circ = \mathrm{-498\: kJ/mol}\\
    &\mathrm{H_2O_{\large{(l)}} \rightarrow H_2O_{\large{(g)}}} &&dH^\circ = \mathrm{44\: kJ/mol\: at\: 298\: K}\\
    &\mathrm{H_2O_{\large{(l)}} \rightarrow H_2O_{\large{(g)}}} &&dH = \mathrm{41\: kJ/mol\: at\: 373\: K,\: \textrm{non-standard}\: condition}\\
    &\mathrm{Mg_{\large{(s)}} + S_{\large{(s)}} \rightarrow MgS_{\large{(s)}}} &&dH^\circ = \mathrm{-598\: kJ/mol}\\
    &\mathrm{2 H_{\large{(g)}} + O_{\large{(g)}} \rightarrow H_2O_{\large{(g)}}} &&dH^\circ =\mathrm{-847\: kJ/mol}\\
    &\mathrm{Cu_{\large{(s)}} + \dfrac{1}{2}O_{2\large{(g)}} \rightarrow CuO_{\large{(s)}}} &&dH^\circ = \mathrm{-157\: kJ/mol}\\
    &\mathrm{\dfrac{1}{2}N_{2\large{(g)}} + O_{2\large{(g)}} \rightarrow NO_{2\large{(g)}}} &&dH^\circ = \mathrm{34\: kJ/mol}\\
    &\mathrm{Mg_{\large{(s)}} + \dfrac{1}{2}O_{2\large{(g)}} \rightarrow MgO_{\large{(s)}}} &&dH^\circ = \mathrm{-602\: kJ/mol}\\
    &\mathrm{2 P_{\large{(s)}} + 3 Cl_{2\large{(g)}} \rightarrow 2 PCl_{3\large{(s)}}} &&dH^\circ = \mathrm{-640\: kJ/mol}\\
    &\mathrm{2 P_{\large{(s)}} + 5 Cl_{2\large{(g)}} \rightarrow 2 PCl_{5\large{(s)}}} &&dH^\circ = \mathrm{-880\: kJ/mol}\\
    &\mathrm{C_{\large{(graphite)}} + 2 O_{\large{(g)}} \rightarrow CO_{2\large{(g)}}} &&dH^\circ = \mathrm{-643\: kJ/mol}\\
    &\mathrm{C_{\large{(graphite)}} + O_{2\large{(g)}} \rightarrow CO_{2\large{(g)}}} &&dH^\circ = \mathrm{-394\: kJ/mol}\\
    &\mathrm{C_{\large{(graphite)}} + 2 H_{2\large{(g)}} \rightarrow CH_{4\large{(g)}}} &&dH^\circ = \mathrm{-75\: kJ/mol}\\
    &\mathrm{2 Al_{\large{(s)}} + Fe_2O_{3\large{(s)}} \rightarrow Al_2O_{3\large{(s)}} + 2Fe_{\large{(s)}}} &&dH^\circ = \mathrm{-850\: kJ/mol}

    As we shall see, the application of Hess's Law will make these data very useful. For example, applying Hess's Law using a few of these reactions enables us to calculate the heat of combustion of methane to form liquid water (as opposed to gaseous water) and carbon dioxide,

    \(\mathrm{CH_4 + 2 O_2 \rightarrow 2 H_2O_{\large{(l)}} + CO_{2\large{(g)}}} \hspace{20px} dH = \mathrm{-980\: kJ/mol}\).

    Enthalpy is an important topic in thermodynamics. Various methods have been devised for the accurate measurement of heat of reaction under constant pressure or under constant volume. This link gives a more advanced treatment on enthalpy.

    Standard Enthalpy of Formation, dHf

    When the standard enthalpy is for a reaction that forms a compound from its basic elements also at the standard state, the standard enthalpy of reaction is called the standard enthalpy of formation, represented by dHof. Unless specified, the temperature is 298 K.

    Table of dHof
    Compound dHof
    \(\ce{MgS}\) -598 kJ/mol
    \(\ce{CuO}\) -157
    \(\ce{PCl3}\) -320
    \(\ce{PCl5}\) -440
    \(\ce{H2O}\) -286
    \(\ce{NO2}\) + 34
    \(\ce{MgO}\) -602
    \(\ce{CO2}\) -394
    \(\ce{CH4}\) -75

    In the above list, some of the equations lead to the formation of a compound from its elements at their standard state. These equations and their enthalpy of formation are given below:

    &\mathrm{Mg_{\large{(s)}} + S_{\large{(s)}} \rightarrow MgS_{\large{(s)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-598\: kJ/mol}\\
    &\mathrm{P_{\large{(s)}} + \dfrac{3}{2} Cl_{2\large{(g)}} \rightarrow PCl_{3\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-320\: kJ/mol}\\
    &\mathrm{P_{\large{(s)}} + \dfrac{5}{2} Cl_{2\large{(g)}} \rightarrow PCl_{5\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{- 440\: kJ/mol}\\
    &\mathrm{H_{2\large{(g)}} + \dfrac{1}{2} O_{2\large{(g)}} \rightarrow H_2O_{\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-286\: kJ/mol}\\
    &\mathrm{\dfrac{1}{2}N_{2\large{(g)}} + O_{2\large{(g)}} \rightarrow NO_{2\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{+ 34\: kJ/mol}\\
    &\mathrm{Cu_{\large{(s)}} + \dfrac{1}{2} O_{2\large{(g)}} \rightarrow CuO_{\large{(s)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-157\: kJ/mol}\\
    &\mathrm{Mg_{\large{(s)}} + \dfrac{1}{2} O_{2\large{(g)}} \rightarrow MgO_{\large{(s)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-602\: kJ/mol}\\
    &\mathrm{C_{\large{(graphite)}} + O_{2\large{(g)}} \rightarrow CO_{2\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-394\: kJ/mol}\\
    &\mathrm{C_{\large{(graphite)}} + 4 H_{2\large{(g)}} \rightarrow CH_{4\large{(g)}}}
    &&dH^\circ_{\mathrm f} = \mathrm{-75\: kJ/mol}

    In all the above equations of reaction, the right hand side has only one product and its coefficient is 1. A general rule is to consider standard enthalpy of formation of all elements at the standard condition to be zero. Then, there is no need to write the complete equation for the tabulation of the standard enthalpy of formation. The above list can be simplified to give the table shown here.

    A simple application of the standard enthalpy of formation is illustrated by Example 4.

    Example 4

    For \(\ce{NH3}\), dHf = -46.1 kJ/mol. Estimate energy released when 10 g of \(\ce{N2}\) react with excess of \(\ce{H2}\) to form ammonia.

    Ten grams of nitrogen is less than 1 mol, and we carry out the calculation in the following manner:

    \(\mathrm{10\: g\: N_2\times\dfrac{1\: mol\: N_2}{28.1\: g\: N_2}\times\dfrac{- 46.1\: kJ}{0.5\: mol\: N_2}= - 32.9\: kJ}\)

    Thus, 32.9 kJ is released when 10 g of \(\ce{N2}\) is consumed.

    Standard enthalpies of formation and standard entropies are important thermodynamic data, and this link gives an extensive table of values for some key compounds.

    The principle of conservation of energy leads to the formulation of the Hess's Law. Its application makes the enthalpy of reaction and standard enthalpy of formation very useful.

    Confidence Building Questions

    1. Which one of the following will you buy for its (potential) chemical energy: gasoline, beer, diamond, dry ice, a music CD, or a chemistry text book?

      Hint: gasoline

      Identify compounds that provide energy.
      A bottle of beer also provides a bit of energy, but the energy aspect is insignificant.

    2. Which of the following processes is endothermic?
      a. condensing steam into water
      b. burning a candle
      c. melting ice cream
      d. cooling hot coffee
      e. formation of snow flakes

      Hint: c.

      Identify process as endothermic or exothermic reaction.

    3. Which one of the following compounds, when formed from its elements at the standard condition, is endothermic?
      b. \(\ce{PCl3}\)
      c. \(\ce{PCl5}\)
      d. \(\ce{H2O}\)
      e. \(\ce{NO2}\)

      Hint: e.

      Common sense tells us that the formation of a through d are exothermic. Check the standard enthalpy of formation. A lot of energy is required to break the triple bond in \(\mathrm{0.5\, N_2 + O_2 \rightarrow NO_2}\)

    4. A 10.0-L tire containing air at 3.0 atm exploded on a freeway. How much energy is available for the explosion?

      Hint: 20.0 atm L (J)

      To evaluate pressure-volume work of a system. The final volume should be 30.0 L at 1.0 atm, a net gain of 20.0 L in volume.

    5. Calculate the pressure volume work for the reaction:

      \(\mathrm{ZnO_{\large{(s)}} \rightarrow Zn_{\large{(s)}} + 0.5\, O_2}\)

      Hint: 1.24 kJ

      Apply the ideal gas law to evaluate the P V work.

    6. In a chemical reaction, an unknown amount of propane is burned. It released 500000 kJ for the BBQ, and performed 9999 kJ of pressure work. Which one can you calculate from the information given above?
      a. internal energy change
      b. standard enthalpy change
      c. molar enthalpy change of
      d. enthalpy of formation of propane
      e. enthalpy of reaction

      Hint: e.

      Define enthalpy of reaction as the energy release in a reaction and the work done by the system.

    7. The enthalpy of formation is as indicated:

      \(\mathrm{H_{2\large{(g)}} + 0.5\, O_2 \rightarrow H_2O_{\large{(l)}}, \hspace{20px} \mathit{dH}_f = -285.8\: kJ/mol}\)

      Calculate the enthalpy of reaction when 1 mole of \(\ce{O2}\) reacts with 4 moles of \(\ce{H2}\) at the standard conditions.

      Hint: -571.6 kJ

      Calculate the energy change using standard enthalpy of reaction and standard enthalpy of formation.

    8. The enthalpy of formation is as indicated:

      \(\mathrm{H_{2\large{(g)}} + 0.5\, O_2 \rightarrow H_2O_{\large{(l)}}, \hspace{20px} \mathit{dH}_f = -285.8\: kJ/mol}\)

      Calculate the enthalpy required to decompose 1 mole of water by electrolysis, all products and reactant at their standard conditions.

      Hint: 285.8 kJ

      Calculate the energy change using standard enthalpy of formation.

    9. From the following data

      \(\mathrm{P_{\large{(s)}} + \dfrac{3}{2} Cl_{2\large{(g)}} \rightarrow PCl_{3\large{(g)}} \hspace{20px} \mathit{dH}^\circ_f = -320\: kJ/mol\\
      P_{\large{(s)}} + \dfrac{5}{2} Cl_{2\large{(g)}} \rightarrow PCl_{5\large{(g)}} \hspace{20px} \mathit{dH}^\circ_f = - 440\: kJ/mol}\)

      Evaluate the standard enthalpy of the reaction:

      \(\mathrm{PCl_{3\large{(g)}} + Cl_{2\large{(g)}} \rightarrow PCl_{5\large{(g)}} \hspace{20px} \mathit{dH}^\circ_f =\: ?\: kJ/mol}\)

      Hint: -120 kJ/mol

      Apply the principle of conservation of energy. Solving this problem is a prelude to Hess's Law.