Chemistry
Calculating Enthalpy Change
Calculating enthalpy change involves determining the heat energy exchanged in a chemical reaction at constant pressure. This can be done using the equation ΔH = q / n, where ΔH is the enthalpy change, q is the heat energy, and n is the number of moles involved in the reaction. Enthalpy change is a key concept in understanding the energy changes associated with chemical reactions.
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eBook - ePub- Gavin Whittaker, Andy Mount, Matthew Heal(Authors)
- 2000(Publication Date)
- Taylor & Francis(Publisher)
n is the molar change in gaseous component. Properties of EnthalphyEnthalpy is a state function whose absolute value cannot be known.ΔH can be ascertained, either by direct methods, where feasible, or indirectly. An increase in the enthalpy of a system, for which ΔH is positive, is referred to as an endothermic process. Conversely, loss of heat from a system, for which ΔH has a negative value, is referred to as an exothermic process. The enthalpy change arising from a temperature change at constant pressure is given by the expression ΔH =C p ΔT , providing that C p does not appreciably change over the temperature range of interest. Where Cp does change, the integral form of the equation , is used. In a chemical reaction, the enthalpy change is equal to the difference in enthalpy between the reactants and products: Kirchhoff’s lawThe value of ΔH for a reaction varies considerably with temperature. Kirchhoff’s s equation, derived from the properties of enthalpy, quantifies this variation. Where C p does not appreciably change over the temperature range of interest, it may be expressed in the form ΔH T2 -ΔH T1 =ΔC p ΔT , or as , where ΔC p is a function of temperature. Related topicsThe first law (B1 )Entropy and change (B5 ) Thermochemistry (B3 )Free energy (B6 ) Entropy ()Statistical thermodynamics (G8 )Enthalpy
The majority of chemical reactions, and almost all biochemical processes in vivo , are performed under constant pressure conditions and involve small volume changes. When a process takes place under constant pressure, and assuming that no work other than pV
eBook - ePub- Atef Korchef(Author)
- 2022(Publication Date)
- CRC Press(Publisher)
Figure 5.9 ) and energy is used to overcome attractive forces between molecules.- 14. For a chemical reaction, when
the reaction is exothermic, whenΔH< 0rxnthe reaction is endothermic and whenΔH> 0rxnthe reaction is athermic (no heat is gained or lost).ΔH= 0rxn- 15. The standard enthalpy of formation, designated by
, is the change in enthalpy when one mole of a substance is formed under standard conditions (P = 1 atm and T = 25°C) from its pure elements under the same standard conditions. Conventionally, the standard enthalpy of formation of a pure element in its most stable form is zero.ΔH f °- 16. Bond breaking requires energy. The energy required to break one mole of that chemical bond is called the bond enthalpy or bond dissociation enthalpy. It is also a measure of the bond strength. Bond enthalpy values are always positive since bond breaking is an endothermic process. However, bond making is an exothermic process (it releases energy). Note that a chemical reaction can be described as the breaking of bonds in the reactants and the making of bonds in the products. Thus, if the bond enthalpies of the reactants and products are known, we can calculate the standard enthalpy of the reaction,
:ΔHrxn°ΔH= ∑ Δ Hrxn°− ∑ Δ Hbroken bondsformed bonds- 17. Hess’s law states that, if a chemical reaction is carried out in a series of steps, the enthalpy change ΔH for the overall reaction is equal to the sum of the enthalpy changes for the individual steps. Hess’s law is used to determine the enthalpy of chemical reactions and it can be applied using the Born–Haber cycle to determine the lattice energy for ionic compounds.
- Kevin Reel, Derrick C. Wood, Scott A. Best(Authors)
- 2014(Publication Date)
- Research & Education Association(Publisher)
According to the first law of thermodynamics, heat lost by a hotter object is gained by a colder object. In contrast, enthalpy (H) refers to the energy released or absorbed by a chemical reaction. Unless there is pressure–volume (PΔV) work performed by the system, changes in enthalpy (ΔH) are essentially the same as the heat exchanged during a reaction. ΔH is the difference between the enthalpy of the products and the enthalpy of the reactants. Exothermic processes describe a heat transfer from the system into the surroundings. Heat is given off during an exothermic reaction (ΔH < 0) and the products of the reaction have less enthalpy than the reactants. Endothermic processes undergo a transfer of energy from the surroundings into the system. Heat is absorbed during an endothermic reaction (ΔH > 0), because energy must be put into the reaction to move it from reactants to products. The products of the reaction have more enthalpy than the reactants. Energy diagrams for exothermic and endothermic reactions can be found in Chapter 11, Kinetics. Heat Capacity, Specific Heat, and Units Heat capacity is a measure of how much an object changes temperature when a given amount of heat is absorbed. For example, when a metal pan is placed on a stovetop it only takes a few moments for the pan to get really hot. This is because metals have a low heat capacity. The units for heat capacity are in J/°C (J/K) as determined by the equation: Specific heat (c) is related to heat capacity because it defines the quantity of heat required to raise 1 gram of substance by 1°C (or 1 Kelvin). The energy unit of the calorie was originally defined in the early 1800s from the specific heat of water, which is the amount of energy required to raise 1 gram of water by 1°C. Of course, calories are not the same as Calories. The “food calorie” is sometimes written with the capital “C” and is actually kilocalories. Since then, scientists have embraced the SI unit of the Joule for measuring heat
eBook - ePub- Jeffrey Gaffney, Nancy Marley(Authors)
- 2017(Publication Date)
- Elsevier(Publisher)
H ) between equilibrium states can be calculated and is given by;Δ H = Δ E + Δ PV= q + w + Δ PV(9)For chemical reactions at constant pressure, where the only work done by the system is pressure-volume work, w p is equal to − P ΔV and the change in enthalpy becomes;ΔH p=q p– P Δ V + P Δ VΔH p=q p(10)So, for the condition of constant pressure, the change in enthalpy is simply equal to the heat transferred to or from the system (q p ). Similar to ΔE v , ΔH p will also be negative for exothermic reactions as the heat released at constant pressure will move to the surroundings from the system, and ΔH p will be positive for endothermic reactions when heat must be supplied to the system for the reaction to occur. Heat capacities are also most important under conditions of constant pressure since most materials and chemical species are used under these conditions. The specific heat capacity of a substance under conditions of constant pressure (C p ) is related to the change in internal energy as;C p=q p/ m Δ T = ΔH p/ m Δ T(11)Changes in enthalpy are also associated with the phase transitions discussed in Chapter 1 . Since phase transitions are caused by heat added to or released from a system at constant pressure, they are accompanied with a change in enthalpy according to Eq. (10) . The enthalpy changes that occur in each phase transition are related to each other as shown in Fig. 8.3 . The length of the arrows in Fig. 8.3
eBook - ePub- Adrian Dingle, Derrick C. Wood(Authors)
- 2014(Publication Date)
- Research & Education Association(Publisher)
f .Enthalpy Change for a ReactionRemember, when an element is formed from itself, there is no change. The standard enthalpy of formation of elements in their standard states is zero.
A. Manipulating Chemical EquationsIII. Hess’s Law1. Hess’s Law states that the overall enthalpy change in a reaction is the sum of all the reactions for the process and is independent of the route taken.i. Rule 1: If you reverse the reactions, then change the sign of ΔH. For example,ii. Rule 2: If you multiply the reaction by a coefficient, multiply the value of ΔH by the same coefficient. For example,iii. Rule 1 and 2 can be combined. For example, if the first reaction is tripled and reversed,Practice Questions
1. Based on the equation below, what is the enthalpy change when 56.4 g of C2 H5 OH(l) decomposes?2. What is the standard enthalpy change, ΔH °rxn , for the reaction below? The enthalpies of formation:3. Given the data below, calculate the enthalpy change for the decomposition of phosphorous trichloride.Answers
1.2. Enthalpy Change for a Reaction =Enthalpy Change for a Reaction3. Reverse equation #2: Multiply equation #3 by 4:Sum the new equations and cancel out 4PCl5 (g) and 4Cl2 (g) from both sides:
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Chapter 21
Inter and Intra Forces and Physical and Chemical Change
A. Differences at the Particulate LevelI. Chemical and Physical Change1. When weak intermolecular forces (between molecules) are broken or formed, physical changes take place.2. When strong intra bonds (chemical bonds within compounds) are broken or formed, chemical changes take place.3. Some changes (e.g., dissolving an ionic salt in water) involve both intermolecular and chemical bonds changing, and, as such, can be classified as chemical and/or physical changes.4. In large molecules (particularly those encountered in biochemistry), intermolecular forces can occur between different parts of the same
eBook - ePub- Kevin R. Reel(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
reaction = (products) (reactants)Δ H°reaction= [(2 × −219.9) + (2 × −68.3)] − [(−66.3) + (2 − −194.5)]ΔH°reaction= −121.1 kJHess’s Law
• Hess’s Law states the ΔH° of a reaction that is composed of multiple steps is equal to the sum of the ΔH° from each step. Hess’s law is an offshoot of the first law of thermodynamics because energy must be conserved in order for the sum of the energies of component reactions to be equal to the energy of the total reaction.Example:Bond Energies
• Bond energies are the amount of energy given off when bonds are formed, or the amount of energy used when bonds are broken.• Bond energies deal with reactants and products in their gaseous state under standard conditions. • Breaking bonds is an exothermic process; making bonds is an endothermic process. • Heats of reaction can be estimated by finding the difference between the bond energies of the bonds made and the bond energies of the bonds broken. • Bond energies used in this way to find heats of reaction is an example of Hess’s Law. Example: Use the following table of bond energies to approximate the change in enthalpy when one mole of propane undergoes complete combustion, according to this reaction:C3 H8 + 5O2 →3CO2 + 4H2 O
Solution:Bond Δ H°bond (kJ/mol)C − H 413 C − C 347 O − H 467 C = O 794 O = O 495 ΔH°reaction= ∑H°bond (bonds broken) −∑H°bond(bonds made)Bond broken Bonds made 8 × C − H 8 × O − H 1 × C − C 6 ×C = O 5 × O = O ΔH°reaction= [(347) + (8 ×413) + (5 ×495)] −[(6 ×794) + (8 ×467)]ΔH°reaction= −2374 kJCalorimetry
• Calorimetry is the laboratory measurement of heats of reaction. The measured change in temperature of the calorimeter identifies how much heat is either absorbed or contributed to the reaction inside it.• The heat given off by the reaction equals the heat absorbed by the calorimeter. Likewise, the heat absorbed by the reaction equals the heat given off by the calorimeter.
eBook - ePub- Warren C. Strahle, William A. Sirignano, William A. Sirignano(Authors)
- 2020(Publication Date)
- Routledge(Publisher)
where the overall enthalpy change is that of the sum of the enthalpy changes for each step along the path. However, the decompression and compression processes of steps a→b and c→d require both heat transfer and work to be done on and by the fluid. In those processes the enthalpy change is not merely the heat added or subtracted.Path 1→a is a constant pressure heating process. The reactants are perfect gases. Looking at Table 2.3a , column 5 of the JANNAF Tables gives us the enthalpy change in going from 298 K to any other temperature. From Appendix A for hydrogen and fluorine, the reactants,Step b→c is a formation reaction at 1 bar and 298 K, for which Path d→2 is a constant pressure heating process for pure HF, which from Appendix A requires an enthalpy change of Paths a→b and c→d have no enthalpy change associated because the enthalpy only depends upon temperature for perfect gases. Consequently, the overall enthalpy change in going from 1 to 2 isNotice that this Qp is negative, so that the overall process is exothermic, even in view of the fact that the temperature has increased. These reactants, if heat is not transferred out of the system, would go to an even higher temperature than 2000 K.There is a simpler way of accomplishing the overall state change. Recalling that the enthalpy of a perfect gas independent of pressure, we see that the results will not depend upon the fact that we are at 5 bar. Figure 2.3d shows a simpler path for calculation. It is:1→a. Same as above.a→b. Reaction at 5 bar and 298 K. This is a formation reaction for HEb→2. Same as d→2 above.FIGURE 2.3d. Simplified perfect gas transition to the final state.For each step the Qp is the enthalpy change for that step and the overall Qp is merely the sum of the heat added or subtracted at each step. That is,which is the same as in the previous calculation.To generalize the steps above, the enthalpy for any substance i is written as(2.8)where the difference between the first two terms on the right side is called the sensible enthalpy and the last term is, of course, the heat of formation. This nomenclature is somewhat confusing, since Eq. (2.8) is an identity. The last two terms are, in fact, equal, but convention gives them different symbols. Shown here is the reference temperature of 298 K. Any other temperature could have been chosen, but the JANNAF Tables make the use of 298 K most convenient. The reader should verify that the placement of Eq. (2.8) in Eq. (2.7)
eBook - ePub- Andrew L. Dicks, David A. J. Rand(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
Appendix 1 Calculations of the Change in Molar Gibbs Free Energy A1.1 Hydrogen Fuel Cell This appendix shows how to calculate the change in molar Gibbs free energy for the reaction: (A1.1) The Gibbs free energy (G) of a system (also known as Gibbs energy or Gibbs function) is defined in terms of the enthalpy (H), temperature (T) and entropy (S) according to the relationship: (A1.2) Similarly, the molar Gibbs free energy of formation (), the molar enthalpy of formation () and the molar entropy () 1 are connected by the equation: (A1.3) In the case of the hydrogen oxidation reaction (A1.1), it is the change in energy that is important, i.e., the difference in energy between the reactants (hydrogen and oxygen) and the product (water or steam). Also, in a fuel cell, the temperature can be taken as constant 2 and, therefore, the following holds: (A1.4) The value of is the difference between of the products and of the reactants. Thus, for the hydrogen oxidation reaction: (A1.5) Similarly, Δ is the difference between of the products and of the reactants. Consequently, for the reaction under consideration: (A1.6) The values of and vary with temperature according to equations (A1.7) and (A1.8) given in the following text. These standard equations are derived using thermodynamic theory, and their proof can be found in textbooks on engineering thermodynamics. 3 The subscript to and is the temperature, and is the molar heat capacity at constant pressure. Standard temperature is taken as 298.15 K. The molar enthalpy of formation at temperature T is given by (A1.7) The molar entropy is given by: (A1.8) The values for both the molar enthalpy of formation and the molar entropy of formation at 298.15 K are obtainable from thermodynamics tables
