how to calculate enthalpy change per mole

Learning Objectives

By the end of this section, you lot will exist able to:

• State the starting time police force of thermodynamics
• Define enthalpy and explain its classification as a state function
• Write and rest thermochemical equations
• Summate enthalpy changes for various chemical reactions
• Explain Hess'south law and use information technology to compute reaction enthalpies

Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships betwixt heat, work, and other forms of energy in the context of chemical and physical processes. Equally we concentrate on thermochemistry in this chapter, nosotros need to consider some widely used concepts of thermodynamics.

Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic free energy falls. The total of all possible kinds of energy present in a substance is called the internal free energy (U), sometimes symbolized as Eastward.

As a organisation undergoes a modify, its internal energy tin can change, and energy tin exist transferred from the system to the surround, or from the environment to the system. Free energy is transferred into a system when it absorbs estrus (q) from the surroundings or when the surroundings practise work (west) on the organisation. For instance, energy is transferred into room-temperature metal wire if it is immersed in hot h2o (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer considering of the work done on it). Both processes increase the internal free energy of the wire, which is reflected in an increase in the wire's temperature. Conversely, energy is transferred out of a system when rut is lost from the system, or when the organisation does work on the environment.

The relationship between internal free energy, oestrus, and piece of work can be represented past the equation:

[latex]\Delta U=q+west[/latex]

as shown in Effigy i. This is i version of the first police force of thermodynamics, and information technology shows that the internal energy of a system changes through heat flow into or out of the system (positive q is oestrus flow in; negative q is heat period out) or work washed on or by the arrangement. The piece of work, due west, is positive if it is done on the system and negative if it is washed past the system.

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Figure 1. The internal energy, U, of a system tin can be changed by estrus catamenia and piece of work. If heat flows into the system, qin, or piece of work is done on the system, won, its internal free energy increases, ΔU < 0. If heat flows out of the organisation, qout, or work is done past the system, wby, its internal free energy decreases, ΔU > 0.

A type of work called expansion work (or pressure level-volume work) occurs when a organization pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the organisation, and the engine and the rest of the universe are the surroundings. The system loses free energy by both heating and doing work on the surround, and its internal free energy decreases. (The engine is able to continue the machine moving considering this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical modify in the chapter on thermodynamics.

This view of an internal combustion engine on How Stuff Works illustrates the conversion of free energy produced by the exothermic combustion reaction of a fuel such as gasoline into free energy of motion.

As discussed, the relationship between internal free energy, heat, and work can be represented equally ΔU = q + w. Internal energy is a type of quantity known as a state function (or state variable), whereas estrus and piece of work are not state functions. The value of a state function depends only on the country that a arrangement is in, and not on how that country is reached. If a quantity is not a land function, and so its value does depend on how the country is reached. An case of a state function is altitude or elevation. If you stand up on the elevation of Mt. Kilimanjaro, yous are at an altitude of 5895 1000, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, nonetheless, is non a state function. Yous could climb to the summit past a directly route or by a more roundabout, circuitous path (Figure 2). The distances traveled would differ (distance is not a state part) just the elevation reached would be the same (altitude is a land part).

Figure two. Paths 10 and Y represent ii different routes to the summit of Mt. Kilimanjaro. Both accept the same change in elevation (altitude or summit on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)

Chemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a organisation's internal energy (U) and the mathematical product of its pressure (P) and volume (V):

[latex]H=U+PV[/latex]

Since information technology is derived from iii state functions (U, P, and V), enthalpy is besides a country function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take identify at constant pressure (a common condition for many chemical and physical changes), the enthalpy alter (ΔH) is:

[latex]\Delta H=\Delta U+P\Delta Five[/latex]

The mathematical production PΔFive represents piece of work (westward), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetics signs of ΔV and w will ever exist opposite:

[latex]P\Delta V=-w[/latex]

Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:

[latex]\begin{assortment}{rll}\Delta H&=&\Delta U+P\Delta V\\&=&{q}_{\text{p}}+westward-westward\\&=&{q}_{\text{p}}\end{assortment}[/latex]

where q_p is the rut of reaction nether conditions of constant pressure.

And and so, if a chemical or physical process is carried out at abiding force per unit area with the only work done caused past expansion or contraction, then the heat menstruum (q_p ) and enthalpy change (ΔH) for the procedure are equal.

The heat given off when yous operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since information technology occurs at the essentially abiding pressure of the atmosphere. On the other hand, the heat produced past a reaction measured in a flop calorimeter is not equal to ΔH, considering the airtight, constant-volume metal container prevents expansion piece of work from occurring. Chemists usually perform experiments nether normal atmospheric weather condition, at constant external pressure with q = ΔH, which makes enthalpy the most convenient choice for determining heat.

The following conventions apply when we utilise ΔH:

1. Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy alter of a reaction is shown as a ΔH value post-obit the equation for the reaction. This ΔH value indicates the amount of estrus associated with the reaction involving the number of moles of reactants and products equally shown in the chemic equation. For example, consider this equation:

   [latex]{\text{H}}_{ii}\left(g\right)+\frac{i}{2}{\text{O}}_{two}\left(g\right)\rightarrow{\text{H}}_{ii}\text{O}\left(50\right)\Delta\text{H}=-286\text{kJ}[/latex]

   This equation indicates that when 1 mole of hydrogen gas and [latex]\frac{ane}{ii}[/latex] mole of oxygen gas at some temperature and pressure alter to one mole of liquid water at the aforementioned temperature and pressure level, 286 kJ of heat are released to the surroundings. If the coefficients of the chemic equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (ΔH is an all-encompassing property):

   [latex]\brainstorm{array}{l}\left(\text{two-fold increase in amounts}\right)\\ two{\text{H}}_{two}\left(g\correct)+{\text{O}}_{2}\left(g\correct)\rightarrow ii{\text{H}}_{2}\text{O}\left(50\right)\Delta\text{H}=2\times \left(-286\text{kJ}\right)=-572\text{kJ}\\ \left(\text{ii-fold decrease in amounts}\right)\\ \frac{ane}{2}{\text{H}}_{ii}\left(chiliad\right)+\frac{one}{four}{\text{O}}_{2}\left(g\right)\rightarrow\frac{one}{ii}{\text{H}}_{ii}\text{O}\left(50\right)\Delta\text{H}=\frac{1}{2}\times \left(-286\text{kJ}\correct)=-143\text{kJ}\end{array}[/latex]

2. The enthalpy alter of a reaction depends on the physical state of the reactants and products of the reaction (whether we accept gases, liquids, solids, or aqueous solutions), so these must exist shown. For example, when 1 mole of hydrogen gas and [latex]\frac{1}{ii}[/latex] mole of oxygen gas change to 1 mole of liquid h2o at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, just 242 kJ of oestrus are released.

   [latex]{\text{H}}_{2}\left(g\right)+\frac{i}{ii}{\text{O}}_{2}\left(k\right)\rightarrow{\text{H}}_{two}\text{O}\left(g\right)\Delta\text{H}=-242\text{kJ}[/latex]

3. A negative value of an enthalpy change, ΔH, indicates an exothermic reaction; a positive value of ΔH indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its ΔH is changed (a process that is endothermic in one direction is exothermic in the opposite management).

Example 1:Measurement of an Enthalpy Modify

When 0.0500 mol of HCl(aq) reacts with 0.0500 mol of NaOH(aq) to grade 0.0500 mol of NaCl(aq), 2.9 kJ of heat are produced. What is ΔH, the enthalpy change, per mole of acid reacting, for the acid-base of operations reaction run under the atmospheric condition described in Example iii of Calorimetry?

[latex]\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\rightarrow\text{NaCl}\left(aq\correct)+{\text{H}}_{two}\text{O}\left(l\right)[/latex]

Cheque Your Learning

When 1.34 g Zn(due south) reacts with sixty.0 mL of 0.750 K HCl(aq), iii.14 kJ of heat are produced. Determine the enthalpy modify per mole of zinc reacting for the reaction:

[latex]\text{Zn}\left(southward\right)+2\text{HCl}\left(aq\right)\rightarrow{\text{ZnCl}}_{2}\left(aq\correct)+{\text{H}}_{2}\left(g\right)[/latex]

Be certain to take both stoichiometry and limiting reactants into business relationship when determining the ΔH for a chemical reaction.

Instance 2:Some other Example of the Measurement of an Enthalpy Change

A viscid bear contains 2.67 g sucrose, C₁₂H₂₂O₁₁. When it reacts with 7.19 g potassium chlorate, KClO₃, 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

[latex]{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(aq\right)+viii{\text{KClO}}_{three}\left(aq\correct)\rightarrow 12{\text{CO}}_{2}\left(1000\correct)+11{\text{H}}_{two}\text{O}\left(l\right)+eight\text{KCl}\left(aq\right)[/latex]

Bank check Your Learning

When 1.42 g of iron reacts with ane.fourscore thou of chlorine, iii.22 thousand of FeCl_two(s) and viii.sixty kJ of heat is produced. What is the enthalpy alter for the reaction when 1 mole of FeCl₂(s) is produced?

Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A standard state is a commonly accepted set of conditions used as a reference point for the determination of properties under other unlike atmospheric condition. For chemists, the IUPAC standard state refers to materials nether a pressure of 1 bar and solutions at 1 M, and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm. Because the ΔH of a reaction changes very petty with such minor changes in pressure level (ane bar = 0.987 atm), ΔH values (except for the about precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted "o" in the enthalpy change symbol to designate standard country. Since the usual (merely non technically standard) temperature is 298.fifteen Grand, we volition use a subscripted "298" to designate this temperature. Thus, the symbol [latex]\left(\Delta{H}_{298}^{\circ}\right)[/latex] is used to indicate an enthalpy alter for a process occurring under these conditions. (The symbol ΔH is used to betoken an enthalpy modify for a reaction occurring under nonstandard conditions.)

The enthalpy changes for many types of chemic and physical processes are available in the reference literature, including those for combustion reactions, stage transitions, and formation reactions. As we discuss these quantities, it is important to pay attending to the extensive nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.eastward., as the ΔH for specific amounts of reactants). However, we oft find it more useful to divide one extensive property (ΔH) by another (amount of substance), and report a per-amount intensive value of ΔH, often "normalized" to a per-mole basis. (Annotation that this is similar to determining the intensive property specific estrus from the extensive property heat capacity, as seen previously.)

Exercises

1. Explicate how the heat measured in Example 3 of Calorimetry differs from the enthalpy change for the exothermic reaction described by the post-obit equation: [latex]\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\correct)\rightarrow\text{NaCl}\left(aq\right)+{\text{H}}_{two}\text{O}\left(fifty\right)[/latex]
2. Using the data in the bank check your learning section of Example 3 of Calorimetry, calculate ΔH in kJ/mol of AgNO3(aq) for the reaction: [latex]\text{NaCl(}aq\text{)}+{\text{AgNO}}_{three}\text{(}aq\text{)}\rightarrow\text{AgCl}\text{(}s\text{)}+{\text{NaNO}}_{three}\text{(}aq\text{).}[/latex]
3. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH₄NO₃ nether the conditions described in Figure 6 in Calorimetry.
4. Summate ΔH for the reaction described past the equation. [latex]\text{Ba}{\left(\text{OH}\correct)}_{ii}\cdot 8{\text{H}}_{2}\text{O}\left(s\correct)+ii{\text{NH}}_{4}\text{SCN}\left(aq\right)\rightarrow\text{Ba}{\left(\text{SCN}\correct)}_{ii}\left(aq\right)+ii{\text{NH}}_{three}\left(aq\right)+10{\text{H}}_{2}\text{O}\left(fifty\right)[/latex]
5. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of CaCl_two.

Enthalpy of Combustion

Standard enthalpy of combustion [latex]\left(\Delta{H}_{C}^{\circ }\right)[/latex] is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called "heat of combustion." For example, the enthalpy of combustion of ethanol, -1366.eight kJ/mol, is the amount of heat produced when one mole of ethanol undergoes consummate combustion at 25 °C and ane atmosphere force per unit area, yielding products as well at 25 °C and one atm.

[latex]{\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)+3{\text{O}}_{2}\left(yard\right)\rightarrow ii{\text{CO}}_{ii}+3{\text{H}}_{2}\text{O}\left(50\right)\Delta{H}_{298}^{\circ }=-\text{1366.viii}\text{kJ}[/latex]

Enthalpies of combustion for many substances accept been measured; a few of these are listed in Table 1. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.

Tabular array 1. Standard Tooth Enthalpies of Combustion
Substance	Combustion Reaction	Enthalpy of Combustion, [latex]\Delta{H}_{c}^{\circ }[/latex] [latex]\left(\frac{\text{kJ}}{\text{mol}}\text{ at }25^{\circ}\text{ C}\correct)[/latex]
carbon	[latex]\text{C}\left(s\correct)+{\text{O}}_{ii}\left(one thousand\correct)\rightarrow{\text{CO}}_{two}\left(m\right)[/latex]	−393.5
hydrogen	[latex]{\text{H}}_{2}\left(chiliad\right)+\frac{one}{ii}{\text{O}}_{2}\left(g\right)\rightarrow{\text{H}}_{ii}\text{O}\left(l\correct)[/latex]	−285.viii
magnesium	[latex]\text{Mg}\left(s\right)+\frac{1}{two}{\text{O}}_{2}\left(k\right)\rightarrow\text{MgO}\left(s\correct)[/latex]	−601.vi
sulfur	[latex]\text{S}\left(s\correct)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{And then}}_{2}\left(g\right)[/latex]	−296.8
carbon monoxide	[latex]\text{CO}\left(g\correct)+\frac{i}{two}{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)[/latex]	−283.0
methane	[latex]{\text{CH}}_{iv}\left(g\right)+2{\text{O}}_{two}\left(m\right)\rightarrow{\text{CO}}_{two}\left(thousand\right)+2{\text{H}}_{2}\text{O}\left(50\right)[/latex]	−890.8
acetylene	[latex]{\text{C}}_{2}{\text{H}}_{2}\left(thousand\right)+\frac{5}{two}{\text{O}}_{ii}\left(g\right)\rightarrow 2{\text{CO}}_{ii}\left(chiliad\right)+{\text{H}}_{2}\text{O}\left(fifty\right)[/latex]	−1301.1
ethanol	[latex]{\text{C}}_{2}{\text{H}}_{v}\text{OH}\left(fifty\right)+2{\text{O}}_{2}\left(grand\right)\rightarrow{\text{CO}}_{ii}\left(g\correct)+3{\text{H}}_{2}\text{O}\left(l\right)[/latex]	−1366.eight
methanol	[latex]{\text{CH}}_{3}\text{OH}\left(l\right)+\frac{iii}{ii}{\text{O}}_{2}\left(chiliad\correct)\rightarrow{\text{CO}}_{2}\left(g\correct)+2{\text{H}}_{2}\text{O}\left(l\right)[/latex]	−726.1
isooctane	[latex]{\text{C}}_{8}{\text{H}}_{18}\left(l\right)+\frac{25}{2}{\text{O}}_{2}\left(g\right)\rightarrow viii{\text{CO}}_{two}\left(k\right)+9{\text{H}}_{2}\text{O}\left(l\right)[/latex]	−5461

Example iii:Using Enthalpy of Combustion

As Figure 3 suggests, the combustion of gasoline is a highly exothermic process. Let us determine the judge amount of heat produced by burning 1.00 50 of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g/mL.

Figure 3. The combustion of gasoline is very exothermic. (credit: modification of piece of work by "AlexEagle"/Flickr)

Check Your Learning

How much heat is produced by the combustion of 125 g of acetylene?

Show Answer

six.25 × ten³ kJ

Emerging Algae-Based Free energy Technologies (Biofuels)

As reserves of fossil fuels diminish and get more than plush to excerpt, the search is ongoing for replacement fuel sources for the futurity. Amidst the most promising biofuels are those derived from algae (Effigy iv). The species of algae used are nontoxic, biodegradable, and among the globe's fastest growing organisms. About 50% of algal weight is oil, which can be readily converted into fuel such equally biodiesel. Algae can yield 26,000 gallons of biofuel per hectare—much more free energy per acre than other crops. Some strains of algae tin flourish in brackish h2o that is non usable for growing other crops. Algae tin produce biodiesel, biogasoline, ethanol, butanol, methyl hydride, and even jet fuel.

Effigy 4. (a) Tiny algal organisms can be (b) grown in big quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of piece of work past Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)

According to the US Department of Energy, only 39,000 foursquare kilometers (about 0.4% of the country mass of the US or less than [latex]\frac{ane}{7}[/latex] of the surface area used to grow corn) can produce plenty algal fuel to replace all the petroleum-based fuel used in the United states. The cost of algal fuels is becoming more competitive—for example, the Us Air Forcefulness is producing jet fuel from algae at a total cost of under $v per gallon. The process used to produce algal fuel is as follows: grow the algae (which use sunlight equally their energy source and CO₂ as a raw textile); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to brand biodiesel); purify; and distribute (Figure 5).

Figure v. Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.

Spotter this video to larn more about the process of creating algae biofuel.

Exercises

1. Although the gas used in an oxyacetylene torch (Figure 6) is essentially pure acetylene, the heat produced past combustion of ane mole of acetylene in such a torch is probable non equal to the enthalpy of combustion of acetylene listed in Table 1. Considering the conditions for which the tabulated data are reported, suggest an explanation.

   

   Figure half-dozen. (a) An oxyacetylene torch. (b) A cold pack. (credit a: modification of work by "Skatebiker"/Wikimedia commons)

2. How much estrus is produced by burning iv.00 moles of acetylene under standard state conditions?
3. How much heat is produced by combustion of 125 yard of methanol under standard state conditions?
4. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state atmospheric condition?
5. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state weather?
6. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these weather?
7. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, ane.00 1000/mL) are mixed?[latex]\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\correct)\rightarrow\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\Delta{H}_{298}^{\circ }=-58\text{kJ}[/latex] If both solutions are at the aforementioned temperature and the heat capacity of the products is four.nineteen J/k °C, how much will the temperature increase? What assumption did y'all make in your adding?
8. A sample of 0.562 one thousand of carbon is burned in oxygen in a flop calorimeter, producing carbon dioxide. Presume both the reactants and products are under standard state conditions, and that the heat released is direct proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the rut capacity of the calorimeter and its contents?
9. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO₂ must exist evaporated to remove every bit much estrus as evaporation of 1.00 kg of CCl₂F_ii (enthalpy of vaporization is 17.iv kJ/mol)? The vaporization reactions for And so₂ and CCl_twoF₂ are [latex]{\text{SO}}_{2}\left(l\correct)\rightarrow{\text{And so}}_{two}\left(g\correct)[/latex] and [latex]{\text{CCl}}_{two}\text{F}\left(l\right)\rightarrow{\text{CCl}}_{ii}{\text{F}}_{2}\left(g\right),[/latex] respectively.
10. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same corporeality of heat when cooled from 95.0 to 35.0 °C, as the rut provided when 100 g of steam is cooled from 110 °C to 100 °C.

Standard Enthalpy of Formation

A standard enthalpy of formation [latex]\Delta{H}_{\text{f}}^{\circ}[/latex] is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their nearly stable states under standard country conditions. These values are particularly useful for computing or predicting enthalpy changes for chemic reactions that are impractical or dangerous to carry out, or for processes for which it is hard to brand measurements. If we accept values for the appropriate standard enthalpies of germination, we can determine the enthalpy change for any reaction, which we volition do in the adjacent section on Hess'due south police force.

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

[latex]\text{C}\left(s\right)+{\text{O}}_{2}\left(g\correct)\rightarrow{\text{CO}}_{two}\left(g\correct)\Delta{H}_{\text{f}}^{\circ }=\Delta{H}_{298}^{\circ }=-393.5\text{kJ}[/latex]

starting with the reactants at a pressure level of one atm and 25°C (with the carbon present as graphite, the virtually stable class of carbon under these weather condition) and ending with one mole of CO₂, likewise at 1 atm and 25°C. For nitrogen dioxide, NO₂(g), [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] is 33.two kJ/mol. This is the enthalpy modify for the reaction:

[latex]\frac{i}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{NO}}_{ii}\left(g\right)\Delta{H}_{\text{f}}^{\circ }=\Delta{H}_{\text{298}}^{\circ }=\text{+33.2}\text{kJ}[/latex]

A reaction equation with [latex]\frac{1}{two}[/latex] mole of Northward₂ and 1 mole of O₂ is correct in this case considering the standard enthalpy of formation always refers to 1 mole of production, NO₂(k).

Yous will find a table of standard enthalpies of formation of many mutual substances in Standard Thermodynamic Properties for Selected Substances. These values bespeak that formation reactions range from highly exothermic (such equally −2984 kJ/mol for the formation of P_fourO_x) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C_twoH₂). By definition, the standard enthalpy of formation of an element in its almost stable form is equal to zero under standard conditions, which is i atm for gases and 1 One thousand for solutions.

Example 4:Evaluating an Enthalpy of Formation

Ozone, O_three(1000), 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, make up one's mind the standard enthalpy of formation, [latex]\Delta{H}_{\text{f}}^{\circ}[/latex] of ozone from the following information:

[latex]3{\text{O}}_{2}\left(g\correct)\rightarrow ii{\text{O}}_{three}\left(thou\correct)\Delta{H}_{298}^{\circ }=\text{+286}\text{kJ}[/latex]

Cheque Your Learning

Hydrogen gas, H₂, reacts explosively with gaseous chlorine, Cl₂, to form hydrogen chloride, HCl(grand). What is the enthalpy change for the reaction of 1 mole of H₂(g) with i 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.

Evidence Answer

For the reaction [latex]{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow 2\text{HCl}\left(g\correct)\Delta{H}_{298}^{\circ }=-184.half-dozen\text{kJ}[/latex]

Example five:Writing Reaction Equations for [latex]\Delta{H}_{\text{f}}^{\circ }[/latex]

Write the estrus of germination reaction equations for:

1. C_iiH₅OH(fifty)
2. Ca₃(PO₄)₂(due south)

Check Your Learning

Write the oestrus of formation reaction equations for:

1. C₂H₅OC₂H₅(l)
2. Na₂CO_iii(s)

Exercises

1. Does the standard enthalpy of formation of H_iiO(g) differ from ΔH° for the reaction [latex]{\text{2H}}_{2}\left(g\correct)+{\text{O}}_{ii}\left(g\right)\rightarrow 2{\text{H}}_{2}\text{O}\left(g\right)?[/latex]
2. Joseph Priestly prepared oxygen in 1774 by heating ruby mercury(2) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(50) and O₂(g) under standard conditions?
3. How many kilojoules of oestrus will be released when exactly 1 mole of manganese, Mn, is burned to class Mn₃O_four(s) at standard state conditions?
4. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Atomic number 26_twoO_iii(s) at standard state conditions?
5. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:[latex]4{\text{NH}}_{three}\left(chiliad\correct)+five{\text{O}}_{two}\left(g\right)\rightarrow 4\text{NO}\left(thou\right)+half dozen{\text{H}}_{2}\text{O}\left(fifty\right)\Delta\text{H}=-907\text{kJ}[/latex][latex]ii\text{NO}\left(g\right)+{\text{O}}_{ii}\left(grand\right)\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta\text{H}=-113\text{kJ}[/latex] [latex]iii{\text{NO}}_{ii}+{\text{H}}_{two}\text{O}\left(l\right)\rightarrow 2{\text{HNO}}_{2}\left(aq\right)+\text{NO}\left(g\correct)\Delta\text{H}=-139\text{kJ}[/latex]
   Determine the total free energy change for the production of i mole of aqueous nitric acid by this process.
6. Both graphite and diamond burn down. [latex]\text{C}\left(southward,\text{diamond}\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{two}\left(g\right)[/latex] For the conversion of graphite to diamond:[latex]\text{C}\left(s,\text{graphite}\right)\rightarrow\text{C}\left(southward,\text{diamond}\right)\Delta{H}_{298}^{\circ }=ane.ninety\text{kJ}[/latex] Which produces more heat, the combustion of graphite or the combustion of diamond?
7. From the molar heats of formation in Standard Thermodynamic Properties for Selected Substances, determine how much rut is required to evaporate i mole of water: [latex]{\text{H}}_{2}\text{O}\left(50\correct)\rightarrow{\text{H}}_{two}\text{O}\left(g\right)[/latex]
8. Which produces more heat? [latex]\text{Bone}\left(s\right)\rightarrow two{\text{O}}_{two}\left(m\right)\rightarrow{\text{OsO}}_{4}\left(s\correct)[/latex] or [latex]\text{Os}\left(s\correct)\rightarrow 2{\text{O}}_{two}\left(1000\right)\rightarrow{\text{OsO}}_{four}\left(g\correct)[/latex] for the phase change [latex]{\text{OsO}}_{4}\left(southward\correct)\rightarrow{\text{OsO}}_{four}\left(g\right)\Delta\text{H}=56.4\text{kJ}[/latex]
9. Summate [latex]\Delta{H}_{298}^{\circ }[/latex] for the process [latex]\text{Sb}\left(s\right)+\frac{5}{ii}{\text{Cl}}_{two}\left(g\correct)\rightarrow{\text{SbCl}}_{5}\left(one thousand\right)[/latex] from the post-obit data: [latex]\begin{array}{fifty}\text{Sb}\left(s\right)+\frac{3}{ii}{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{SbCl}}_{three}\left(g\right)\Delta{H}_{298}^{\circ }=-314\text{kJ}\\ {\text{SbCl}}_{three}\left(south\right)+{\text{Cl}}_{ii}\left(chiliad\right)\rightarrow{\text{SbCl}}_{5}\left(g\right)\Delta{H}_{298}^{\circ }=-fourscore\text{kJ}\end{array}[/latex]
10. Calculate [latex]\Delta{H}_{298}^{\circ }[/latex] for the process [latex]\text{Zn}\left(s\correct)+\text{Due south}\left(s\right)+2{\text{O}}_{2}\left(chiliad\right)\rightarrow{\text{ZnSO}}_{4}\left(s\right)[/latex] from the following information: [latex]\brainstorm{array}{fifty}\text{Zn}\left(s\right)+\text{S}\left(s\right)\rightarrow\text{ZnS}\left(due south\right)\Delta{H}_{298}^{\circ }=-206.0\text{kJ}\\ \text{ZnS}\left(s\right)+{\text{2O}}_{\text{ii}}\left(thou\correct)\rightarrow{\text{ZnSO}}_{4}\left(s\correct)\Delta{H}_{298}^{\circ }=-776.8\text{kJ}\cease{assortment}[/latex]

Hess'due south Constabulary

In that location are two ways to determine the corporeality of heat involved in a chemical alter: measure out it experimentally, or calculate it from other experimentally adamant enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not difficult 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 normally involves the use of Hess's law, which states: If a process can exist written as the sum of several stepwise processes, the enthalpy alter 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 simply on where a chemic process starts and ends, only not on the path it takes from first to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either straight or by a ii-step process. The direct process is written:

[latex]\text{C}\left(southward\right)+{\text{O}}_{2}\left(thou\right)\rightarrow{\text{CO}}_{2}\left(yard\right)\Delta{H}_{298}^{\circ }=-394\text{kJ}[/latex]

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

[latex]\text{C}\left(s\correct)+\frac{1}{2}{\text{O}}_{two}\left(g\right)\rightarrow\text{CO}\left(yard\right)\Delta{H}_{298}^{\circ }=-111\text{kJ}[/latex]

Then, carbon monoxide reacts further to form carbon dioxide:

[latex]\text{CO}\left(chiliad\right)+\frac{i}{2}{\text{O}}_{ii}\left(g\right)\rightarrow\text{CO}\left(g\correct)\Delta{H}_{298}^{\circ }=-283\text{kJ}[/latex]

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

[latex]\brainstorm{array}{l}\\ \text{Step one: }\text{C}\left(s\correct)+\frac{1}{2}{\text{O}}_{2}\left(yard\right)\rightarrow\text{CO}\left(g\correct)\\ \underline{\text{Step two: }\text{CO}\left(g\correct)+\frac{one}{two}{\text{O}}_{two}\left(grand\correct)\rightarrow{\text{CO}}_{two}\left(g\right)}\\ \text{Sum: }\text{C}\left(s\correct)+\frac{one}{two}{\text{O}}_{2}\left(grand\right)+\text{CO}\left(one thousand\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(thousand\right)+{\text{CO}}_{ii}\left(g\right)\terminate{array}[/latex]

Because the CO produced in Stride 1 is consumed in Pace two, the internet change is:

[latex]\text{C}\left(s\correct)+{\text{O}}_{2}\left(yard\right)\rightarrow{\text{CO}}_{two}\left(one thousand\correct)[/latex]

According to Hess's police force, 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 ane to discover the enthalpy change of the entire reaction from its two steps:

[latex]\brainstorm{array}{ll}\text{C}\left(southward\right)+\frac{ane}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(yard\correct)\hfill & \Delta{H}_{298}^{\circ }=-111\text{kJ}\hfill \\ \underline{\text{CO}\left(thou\correct)+\frac{i}{2}{\text{O}}_{2}\left(yard\right)\rightarrow{\text{CO}}_{ii}\left(k\right)}& \underline{\Delta{H}_{298}^{\circ }=-283\text{kJ}}\\{\text{C}\left(southward\right)+{\text{O}}_{2}\left(chiliad\right)\rightarrow{\text{CO}}_{2}\left(g\right)}&{\Delta{H}_{298}^{\circ }=-394\text{kJ}}\hfill \end{array}[/latex]

The consequence is shown in Figure 7. Nosotros see that ΔH of the overall reaction is the aforementioned whether it occurs in one pace or ii. 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 concrete processes.

Figure vii. The germination of CO_ii(g) from its elements tin be thought of equally occurring in two steps, which sum to the overall reaction, as described past Hess's constabulary. The horizontal blue lines represent enthalpies. For an exothermic procedure, the products are at lower enthalpy than are the reactants.

Before we further exercise using Hess'due south law, allow us call up two important features of ΔH.

1. ΔH is straight proportional to the quantities of reactants or products. For example, the enthalpy alter for the reaction forming 1 mole of NO₂(grand) is +33.ii kJ:

   [latex]\frac{1}{ii}{\text{N}}_{2}\left(one thousand\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{NO}}_{ii}\left(grand\right)\Delta\text{H}=\text{+33.2}\text{kJ}[/latex]

   When two moles of NO₂ (twice as much) are formed, the ΔH volition exist twice as big:

   [latex]{\text{N}}_{ii}\left(g\right)+2{\text{O}}_{ii}\left(g\correct)\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta\text{H}=\text{+66.4}\text{kJ}[/latex]

   In full general, if we multiply or divide an equation by a number, then the enthalpy change should also exist 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 opposite direction. For example, given that:

   [latex]{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow 2\text{HCl}\left(g\right)\Delta\text{H}=-184.6\text{kJ}[/latex]

   Then, for the "reverse" reaction, the enthalpy change is also "reversed:"

   [latex]2\text{HCl}\left(one thousand\right)\rightarrow{\text{H}}_{2}\left(1000\right)+{\text{Cl}}_{2}\left(g\correct)\Delta\text{H}=\text{+184.six}\text{kJ}[/latex]

Example 6:Stepwise Adding of [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] Using Hess'due south Law

Determine the enthalpy of formation, [latex]\Delta{H}_{\text{f}}^{\circ }[/latex], of FeCl₃(s) from the enthalpy changes of the following two-stride process that occurs under standard land conditions:

[latex]\text{Atomic number 26}\left(due south\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{FeCl}}_{two}\left(s\right)\Delta\text{H}^{\circ }=-341.8\text{kJ}[/latex]

[latex]{\text{FeCl}}_{2}\left(s\right)+\frac{one}{ii}{\text{Cl}}_{2}\left(chiliad\right)\rightarrow{\text{FeCl}}_{3}\left(southward\right)\Delta\text{H}^{\circ }=-57.vii\text{kJ}[/latex]

Check Your Learning

Summate ΔH for the process:

[latex]{\text{Due north}}_{2}\left(g\right)+2{\text{O}}_{ii}\left(g\right)\rightarrow two{\text{NO}}_{2}\left(g\correct)[/latex]

from the following data:

[latex]{\text{N}}_{2}\left(k\right)+{\text{O}}_{2}\left(g\right)\rightarrow ii\text{NO}\left(g\right)\Delta\text{H}=180.5\text{kJ}[/latex]
[latex]\text{NO}\left(grand\right)+\frac{i}{2}{\text{O}}_{2}\left(g\right)\rightarrow{\text{NO}}_{ii}\left(one thousand\right)\Delta\text{H}=-57.06\text{kJ}[/latex]

Hither is a less straightforward instance that illustrates the thought procedure involved in solving many Hess's police issues. It shows how nosotros tin can discover many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.

Example vii:A More Challenging Problem Using Hess's Law

Chlorine monofluoride tin can react with fluorine to form chlorine trifluoride:

(i) [latex]\text{ClF}\left(g\correct)+{\text{F}}_{2}\left(grand\correct)\rightarrow{\text{ClF}}_{3}\left(grand\right)\Delta\text{H}^{\circ }=?[/latex]

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

(ii) [latex]two{\text{OF}}_{2}\left(g\right)\rightarrow{\text{O}}_{2}\left(g\right)+2{\text{F}}_{2}\left(thousand\right)\Delta{H}_{\left(2\right)}^{\circ }=-49.iv\text{kJ}[/latex]

(3) [latex]two\text{ClF}\left(g\correct)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{Cl}}_{two}\text{O}\left(g\right)+{\text{OF}}_{2}\left(m\right)\Delta{H}_{\left(iii\right)}^{\circ }=\text{+205.6}\text{kJ}[/latex]

(iv) [latex]{\text{ClF}}_{3}\left(thou\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{Cl}}_{2}\text{O}\left(g\correct)+\frac{3}{two}{\text{OF}}_{2}\left(chiliad\correct)\Delta{H}_{\left(4\correct)}^{\circ }=\text{+266.7}\text{kJ}[/latex]

Check Your Learning

Aluminum chloride can be formed from its elements:

(i) [latex]2\text{Al}\left(s\right)+3{\text{Cl}}_{two}\left(g\right)\rightarrow 2{\text{AlCl}}_{iii}\left(s\right)\Delta H^{\circ }=?[/latex]

Utilise the reactions here to make up one's mind the ΔH° for reaction (i):

(ii) [latex]\text{HCl}\left(g\right)\rightarrow\text{HCl}\left(aq\right)\Delta{H}_{\left(2\correct)}^{\circ }=-74.8\text{kJ}[/latex]

(3) [latex]{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{ii}\left(yard\right)\rightarrow 2\text{HCl}\left(chiliad\correct)\Delta{H}_{\left(three\right)}^{\circ }=-185\text{kJ}[/latex]

(iv) [latex]{\text{AlCl}}_{3}\left(aq\correct)\rightarrow{\text{AlCl}}_{3}\left(s\right)\Delta{H}_{\left(four\right)}^{\circ }=+323\text{kJ/mol}[/latex]

(v) [latex]\text{2Al}\left(due south\right)+6\text{HCl}\left(aq\right)\rightarrow ii{\text{AlCl}}_{3}\left(aq\right)+iii{\text{H}}_{2}\left(g\right)\Delta{H}_{\left(v\correct)}^{\circ }=-1049\text{kJ}[/latex]

We also tin use Hess'south law to determine the enthalpy modify of any reaction if the corresponding enthalpies of germination of the reactants and products are available. The stepwise reactions nosotros 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 requite 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: (2) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of germination of the reactants. This is commonly rearranged slightly to exist written equally follows, with ∑ representing "the sum of" and n standing for the stoichiometric coefficients:

[latex]\Delta{H}_{\text{reaction}}^{\circ }=\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{products}\correct)-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]

The following case shows in detail why this equation is valid, and how to apply it to calculate the enthalpy alter for a reaction of involvement.

Case eight:Using Hess's Law

What is the standard enthalpy modify for the reaction:

[latex]3{\text{NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightarrow 2{\text{HNO}}_{iii}\left(aq\correct)+\text{NO}\left(g\right)\Delta\text{H}^{\circ }=?[/latex]

At that place are 2 possible solutions for this reaction. The first solution supports why the full general equation is valid, the second solution uses the equation.

Check Your Learning

Calculate the estrus of combustion of ane mole of ethanol, C₂H₅OH(l), when H₂O(l) and CO₂(g) are formed. Use the post-obit enthalpies of formation: C₂H_vOH(l), −278 kJ/mol; H₂O(l), −286 kJ/mol; and CO₂(m), −394 kJ/mol.

Exercises

1. Calculate the standard molar enthalpy of germination of NO(g) from the post-obit data: [latex]\begin{array}{50}{\text{N}}_{2}\left(chiliad\correct)+two{\text{O}}_{ii}\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=66.4\text{kJ}\\ \text{2NO}\left(one thousand\right)+{\text{O}}_{2}\rightarrow ii{\text{NO}}_{two}\left(thousand\right)\Delta{H}_{298}^{\circ }=-114.1\text{kJ}\end{assortment}[/latex]
2. Using the data in Standard Thermodynamic Backdrop for Selected Substances, calculate the standard enthalpy modify for each of the post-obit reactions:
   1. [latex]{\text{Northward}}_{2}\left(g\right)+{\text{O}}_{two}\left(g\right)\rightarrow two\text{NO}\left(thousand\correct)[/latex]
   2. [latex]\text{Si}\left(southward\right)+two{\text{Cl}}_{ii}\left(one thousand\right)\rightarrow{\text{SiCl}}_{4}\left(g\right)[/latex]
   3. [latex]{\text{Fe}}_{ii}\text{O}\left(s\right)+3{\text{H}}_{ii}\left(g\correct)\rightarrow 2\text{Fe}\left(southward\right)+iii{\text{H}}_{ii}\text{O}\left(l\correct)[/latex]
   4. [latex]ii\text{LiOH}\left(due south\right)+{\text{CO}}_{2}\left(g\right)\rightarrow{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{two}\text{O}\left(g\right)[/latex]
3. Using the data in Standard Thermodynamic Backdrop for Selected Substances, calculate the standard enthalpy change for each of the following reactions:
   1. [latex]\text{Si}\left(south\right)+2{\text{F}}_{2}\left(thou\right)\rightarrow{\text{SiF}}_{four}\left(g\right)[/latex]
   2. [latex]2\text{C}\left(south\correct)+2{\text{H}}_{2}\left(thousand\right)+{\text{O}}_{2}\left(grand\right)\rightarrow{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(l\right)[/latex]
   3. [latex]{\text{CH}}_{4}\left(1000\right)+{\text{Due north}}_{2}\left(grand\right)\rightarrow\text{HCN}\left(m\right){\text{+NH}}_{three}\left(thou\right)\text{;}[/latex]
   4. [latex]{\text{Cs}}_{2}\left(g\correct)+3{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{CCl}}_{iv}\left(g\right){\text{+Due south}}_{2}{\text{Cl}}_{2}\left(yard\right)[/latex]
4. The following reactions tin be used to prepare samples of metals. Decide the enthalpy change under standard state conditions for each.
   1. [latex]ii{\text{Ag}}_{2}\text{O}\left(s\right)\rightarrow 4\text{Ag}\left(s\right)+{\text{O}}_{two}\left(1000\correct)[/latex]
   2. [latex]\text{SnO}\left(s\correct)+\text{CO}\left(g\right)\rightarrow\text{Sn}\left(south\right)+{\text{CO}}_{two}\left(g\right)[/latex]
   3. [latex]{\text{Cr}}_{two}{\text{O}}_{3}\left(s\right)+iii{\text{H}}_{2}\left(g\right)\rightarrow 2\text{Cr}\left(s\right)+three{\text{H}}_{2}\text{O}\left(l\correct)[/latex]
   4. [latex]ii\text{Al}\left(s\right)+{\text{Fe}}_{ii}{\text{O}}_{iii}\left(s\right)\rightarrow{\text{Al}}_{2}{\text{O}}_{3}\left(south\right)+2\text{Atomic number 26}\left(s\right)[/latex]
5. The decomposition of hydrogen peroxide, H₂O_ii, has been used to provide thrust in the command jets of various space vehicles. Using the data in Standard Thermodynamic Backdrop for Selected Substances, make up one's mind how much heat is produced by the decomposition of exactly 1 mole of H₂O_two under standard weather condition. [latex]2{\text{H}}_{ii}{\text{O}}_{2}\left(fifty\right)\rightarrow ii{\text{H}}_{2}\text{O}\left(g\right)+{\text{O}}_{ii}\left(chiliad\correct)[/latex]

Key Concepts and Summary

If a chemic change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the alter is called the enthalpy alter with the symbol ΔH, or [latex]\Delta{H}_{\text{298}}^{\circ }[/latex] for reactions occurring under standard land conditions. The value of ΔH for a reaction in one management is equal in magnitude, just opposite in sign, to ΔH for the reaction in the reverse direction, and ΔH is straight 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, [latex]\Delta{H}_{\text{f}}^{\circ },[/latex] is the enthalpy change accompanying the germination of 1 mole of a substance from the elements in their near stable states at 1 bar (standard land). Many of the processes are carried out at 298.15 1000. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy alter can be calculated using Hess'south law: If a process can be written as the sum of several stepwise processes, the enthalpy modify of the total process equals the sum of the enthalpy changes of the various steps.

Fundamental Equations

• [latex]\Delta U=q+w[/latex]
• [latex]\Delta{H}_{\text{reaction}}^{\circ }=\sum north\times \Delta{H}_{\text{f}}^{\circ }\text{(products)}-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]

Exercises

1. Which of the enthalpies of combustion in Table ane are also standard enthalpies of formation?
2. Calculate ΔH for the procedure [latex]{\text{Hg}}_{ii}{\text{Cl}}_{2}\left(south\right)\rightarrow two\text{Hg}\left(50\right)+{\text{Cl}}_{ii}\left(g\right)[/latex] from the post-obit information:[latex]\begin{array}{l}\\ \text{Hg}\left(fifty\right)+{\text{Cl}}_{\text{2}}\left(yard\correct)\rightarrow{\text{HgCl}}_{\text{2}}\left(s\right)\Delta H=-224\text{kJ}\\ \text{Hg}\left(fifty\right)+{\text{HgCl}}_{\text{2}}\left(southward\right)\rightarrow{\text{Hg}}_{\text{2}}{\text{Cl}}_{2}\left(s\right)\Delta H=-41.two\text{kJ}\stop{array}[/latex]
3. Summate [latex]\Delta{H}_{298}^{\circ }[/latex] for the process [latex]{\text{Co}}_{3}{\text{O}}_{4}\left(due south\right)\rightarrow iii\text{Co}\left(south\right)+2{\text{O}}_{2}\left(thousand\right)[/latex] from the post-obit information: [latex]\brainstorm{array}{l}\\ \text{Co}\left(due south\right)+\frac{ane}{2}{\text{O}}_{ii}\left(1000\right)\rightarrow\text{CoO}\left(s\right)\Delta{H}_{298}^{\circ }=-237.nine\text{kJ}\\ \text{3Co}\left(southward\right)+{\text{O}}_{2}\left(grand\right)\rightarrow{\text{Co}}_{iii}{\text{O}}_{4}\left(s\right)\Delta{H}_{298}^{\circ }=-177.5\text{kJ}\end{assortment}[/latex]
4. Calculate the enthalpy of combustion of propane, C₃H₈(grand), for the formation of H_twoO(k) and CO₂(g). The enthalpy of germination of propane is −104 kJ/mol.
5. Calculate the enthalpy of combustion of butane, C₄H₁₀(g) for the formation of H₂O(g) and CO_two(g). The enthalpy of germination of butane is −126 kJ/mol.
6. Both propane and butane are used as gaseous fuels. Which compound produces more oestrus per gram when burned?
7. The white pigment TiO₂ is prepared past the reaction of titanium tetrachloride, TiCl₄, with water vapor in the gas phase: [latex]{\text{TiCl}}_{4}\left(g\right)+2{\text{H}}_{2}\text{O}\left(g\right)\rightarrow{\text{TiO}}_{2}\left(s\correct)+4\text{HCl}\left(thousand\right)[/latex]. How much estrus is evolved in the production of exactly 1 mole of TiO₂(s) under standard state conditions?
8. H2o gas, a mixture of H₂ and CO, is an of import industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: [latex]\text{C}\left(s\right)+{\text{H}}_{2}\text{O}\left(k\right)\rightarrow\text{CO}\left(g\correct)+{\text{H}}_{ii}\left(thou\correct).[/latex]
   1. Assuming that coke has the aforementioned enthalpy of formation as graphite, calculate [latex]\Delta{H}_{298}^{\circ }[/latex] for this reaction.
   2. Methanol, a liquid fuel that could possibly replace gasoline, tin be prepared from h2o gas and boosted hydrogen at high temperature and pressure in the presence of a suitable catalyst: [latex]2{\text{H}}_{2}\left(thou\right)+\text{CO}\left(g\right)\rightarrow{\text{CH}}_{three}\text{OH}\left(m\right)[/latex]. Under the weather condition of the reaction, methanol forms as a gas. Calculate [latex]\Delta{H}_{298}^{\circ }[/latex] for this reaction and for the condensation of gaseous methanol to liquid methanol.
   3. Calculate the heat of combustion of 1 mole of liquid methanol to H_twoO(g) and CO₂(g).
9. In the early days of automobiles, illumination at dark was provided by called-for acetylene, C_twoH₂. Though no longer used every bit auto headlamps, acetylene is yet used as a source of lite past some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC_two: [latex]{\text{CaC}}_{ii}\left(southward\right)+{\text{H}}_{two}\text{O}\left(l\right)\rightarrow\text{Ca}{\left(\text{OH}\correct)}_{two}\left(s\correct)+{\text{C}}_{two}{\text{H}}_{ii}\left(chiliad\right).[/latex] Calculate the standard enthalpy of the reaction. The [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of CaC₂ is −15.fourteen kcal/mol.
10. From the data in Tabular array 1, determine which of the following fuels produces the greatest corporeality of rut per gram when burned under standard conditions: CO(thousand), CH₄(chiliad), or C₂H₂(1000).
11. The enthalpy of combustion of hard coal averages −35 kJ/yard, that of gasoline, 1.28 × x⁵ kJ/gal. How many kilograms of difficult coal provide the aforementioned amount of heat as is available from ane.0 gallon of gasoline? Assume that the density of gasoline is 0.692 thou/mL (the same as the density of isooctane).
12. Ethanol, C_twoH₅OH, is used as a fuel for motor vehicles, specially in Brazil.
    1. Write the counterbalanced equation for the combustion of ethanol to CO₂(g) and H_iiO(one thousand), and, using the data in Standard Thermodynamic Properties for Selected Substances, calculate the enthalpy of combustion of ane mole of ethanol.
    2. The density of ethanol is 0.7893 thou/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol.
    3. Assuming that an car's mileage is direct proportional to the heat of combustion of the fuel, calculate how much further an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C_eightH₁₈ [latex]\left(\Delta{H}_{\text{f}}^{\circ }=-208.iv\text{kJ/mol}\right)[/latex] density = 0.7025 g/mL).
13. Among the substances that react with oxygen and that have been considered every bit potential rocket fuels are diborane [B_iiH₆, produces B_iiO_iii(s) and H_iiO(1000)], methane [CH₄, produces CO_ii(one thousand) and H₂O(g)], and hydrazine [N₂H₄, produces Due north_two(g) and H₂O(g)]. On the basis of the rut released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility equally a rocket fuel? The [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of B_iiH_vi(chiliad), CH₄(g), and N₂H₄(50) may be found in Standard Thermodynamic Properties for Selected Substances.
14. How much heat is produced when i.25 g of chromium metallic reacts with oxygen gas under standard conditions?
15. Ethylene, C_twoH₂, a byproduct from the fractional distillation of petroleum, is quaternary amongst the l chemical compounds produced commercially in the largest quantities. Nearly 80% of constructed ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. [latex]{\text{C}}_{2}{\text{H}}_{4}\left(m\right)+{\text{H}}_{2}\text{O}\left(yard\right)\rightarrow{\text{C}}_{ii}{\text{H}}_{5}\text{OH}\left(fifty\correct)[/latex] Using the data in the table in Standard Thermodynamic Properties for Selected Substances, calculate ΔH° for the reaction.
16. The oxidation of the sugar glucose, C₆H₁₂O₆, is described by the following equation: [latex]{\text{C}}_{six}{\text{H}}_{12}{\text{O}}_{6}\left(due south\correct)+vi{\text{O}}_{ii}\left(g\right)\rightarrow 6{\text{CO}}_{2}\left(g\right)+six{\text{H}}_{two}\text{O}\left(fifty\right)\Delta\text{H}=-2816\text{kJ}[/latex] The metabolism of glucose gives the aforementioned products, although the glucose reacts with oxygen in a serial of steps in the body.
    1. How much heat in kilojoules can be produced by the metabolism of one.0 g of glucose?
    2. How many Calories can be produced past the metabolism of i.0 g of glucose?
17. Propane, C_iiiH_eight, is a hydrocarbon that is usually used as a fuel.
    1. Write a balanced equation for the consummate combustion of propane gas.
    2. Calculate the book of air at 25 °C and ane.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percentage O₂ by book. (Hint: we will see how to do this adding in a afterward chapter on gases—for at present use the data that ane.00 L of air at 25 °C and 1.00 atm contains 0.275 grand of O₂ per liter.)
    3. The heat of combustion of propane is -2,219.2 kJ/mol. Summate the heat of formation, [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of propane given that [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of H_iiO(50) = −285.8 kJ/mol and [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of CO_ii(1000) = −393.5 kJ/mol.
    4. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of h2o, summate the increase in temperature of the h2o.
18. During a recent wintertime calendar month in Sheboygan, Wisconsin, information technology was necessary to obtain 3500 kWh of estrus provided by a natural gas furnace with 89% efficiency to keep a small firm warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).
    1. Assume that natural gas is pure methane and determine the book of natural gas in cubic feet that was required to heat the house. The boilerplate temperature of the natural gas was 56 °F; at this temperature and a pressure of i atm, natural gas has a density of 0.681 one thousand/L.
    2. How many gallons of LPG (liquefied petroleum gas) would be required to supersede the natural gas used? Presume the LPG is liquid propane [C_threeH₈: density, 0.5318 m/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO₂(g) and H₂O(l)] and the furnace used to burn the LPG has the same efficiency every bit the gas furnace.
    3. What mass of carbon dioxide is produced by combustion of the methane used to oestrus the firm?
    4. What mass of water is produced past combustion of the methane used to heat the house?
    5. What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the calendar month was ane.22 g/50.
    6. How many kilowatt–hours (1 kWh = 3.6 × 10^{half dozen} J) of electricity would exist required to provide the heat necessary to estrus the house? Note electricity is 100% efficient in producing heat inside a house.
    7. Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is nearly 40%. A certain type of coal provides two.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical free energy necessary to heat the house if the efficiency of generation and distribution is xl%?

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 arrangement's internal energy and the mathematical product of its pressure and volume

enthalpy change (ΔH): rut released or absorbed by a system under constant pressure during a chemical or physical process

expansion piece of work (pressure-volume work): piece of work done equally a system expands or contracts confronting 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 past the system

Hess'due south 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: chemical compound composed only of hydrogen and carbon; the major component of fossil fuels

internal free energy (U): total of all possible kinds of energy present in a substance or substances

standard enthalpy of combustion [latex]\text{(}\Delta{H}_{\text{c}}^{\circ }\text{)}[/latex]: estrus released when 1 mole of a compound undergoes complete combustion under standard weather

standard enthalpy of formation [latex]\text{(}\Delta{H}_{\text{f}}^{\circ }\text{)}[/latex]: enthalpy change of a chemical reaction in which i mole of a pure substance is formed from its elements in their almost stable states under standard country atmospheric condition

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

country part: property depending but on the country of a organisation, and not the path taken to achieve that state

Source: https://courses.lumenlearning.com/suny-albany-chemistry/chapter/enthalpy/

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