# 3.0 Calculating Enthalpy Change

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1 3.0 Calculating Enthalpy Change 3.1 Heats of Formation The amount of energy absorbed or released during the synthesis of 1 mole of a substance formed from its elements is referred to as the standard heat of formation, H f. For example, the formation reaction for liquid water is described by the following equation: H2 (g) + ½O2 (g) H2O(l) kj Since the reaction is exothermic we would indicate the heat of formation, ΔH o f, as kj. You will find the standard heats of formation listed for a variety of compounds in the Table of Thermochemical Data. (The heat of formation for pure elements, such as H2(g), O2(g), Al(s), etc. is 0 kj mole -1.) To write heat of formation reactions: Balance the equation so that one mole of the compound is produced. Remember the diatomic molecules and write them correctly (H2, N2, O2, F2, Cl2, Br2, I2). The reactants must be elements, not polyatomic ions. 3.2 Hess s Law Scientists have determined that enthalpy change is independent of how the system changes from beginning to end. Chemical reactions do not occur in one step but rather by a series of steps, referred to as the reaction mechanism. For instance, carbon and oxygen can react to form carbon monoxide: (1) C (s) + 1/2 O 2(g) CO (g) Δ H = kj Carbon monoxide can react further with oxygen to form carbon dioxide: (2) CO (g) + 1/2 O 2(g) CO 2(g) ΔH = kj

2 A third possible reaction involves carbon and oxygen reacting to form carbon dioxide directly: (3) C (s) + O 2(g) CO 2(g) Δ H = kj You will note that the heat of reaction from (1) plus the heat of reaction from (2) exactly equals the heat of reaction from (3). This makes sense when you consider what is going on; chemically, reactions (1) and (2) perform in two steps what reaction (3) does in one step. Note what happens when reactions (1) and (2) are added up: C (s) + 1/2 O 2(g) CO (g) CO (g) + 1/2 O 2(g) CO 2(g) ΔH = kj ΔH = kj C (s) + O 2(g) CO 2(g) ΔH = kj The end result is that it doesn't matter if the reaction proceeds all at once or in series of steps; the net energy change is the same. This illustrates Hess Law of Constant Heat Summation: The sum of enthalpies of the steps of a reaction will equal the enthalpy of the overall reaction. This law allows us to take several reactions from the heats of formation table, add them up and get ΔH values for reactions that do not appear directly on the table. H (net) = H (step 1) + H (step 2) + + H (step n) 3.3 Calculating ΔH using Hess Law In order to calculate the overall ΔH using Hess idea we need to be able to manipulate the steps of a reaction so that they can add up to the overall reaction. We can manipulate the steps of reactions in only 2 ways: 1. We can reverse the equation and write backwards from products to reactants. When we do this the value of ΔH changes signs. 2. We can multiply the equation by a common factor. When we do this we must also multiply the heat by the same factor.

3 Example 1: 2 Al (s) + 3 CuO (s) 3 Cu (s) + Al 2O 3(s) 1. Find an equation on the Heat of Formation Table to match each of the compounds in the given equation. 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = kj/mol Cu (s) + ½ O 2(g) CuO (s) ΔH = kj/mol 2. Flip the CuO equation to make the compound a reactant as well: Cu (s) + ½ O 2(g) CuO (s) ΔH = kj/mol Becomes CuO (s) Cu (s) + ½ O 2(g) ΔH = kj/mol 3. Multiply the CuO equation by 3 to have the same number of molecules as in the original equation. 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = kj/mol 3 x CuO (s) Cu (s) + ½ O 2(g) ΔH = 3 x ( kj/mol) Gives 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = kj/mol 3 CuO (s) 3 Cu (s) + 1½ O 2(g) ΔH = kj/mol 4. Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added: 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = kj/mol 3 CuO (s) 3 Cu (s) + 1½ O 2(g) ΔH = kj/mol 2 Al(s) + 1½ O2(g) + 3 CuO(s) Al2O3(s) + 3 Cu(s) + 1½ O2(g) ΔH = kj You will notice that there are similar terms on each side of the reaction equation. These can be treated again just like mathematical terms and cancelled out where you can: 2 Al (s) + 3 CuO (s) Al 2O 3(s) + 3 Cu (s) ΔH = kj

4 Example 2: SiO 2(s) + C (s) CO 2(g) + Si (s) Where the heat of formation of SiO 2(s) = kj/mol 1. Find an equation on the Heat of Formation Table to match each of the compounds in the given equation. C (s) + O 2(g) CO 2(g) ΔH = kj/mol The equation for SiO 2 is not on the table, but you are given the heat of formation of the substance from its elements: Si (s) + O 2(g) SiO 2(s) ΔH = kj/mol 2. Flip the SiO 2 equation to make the compound a reactant as well: Si (s) + O 2(g) SiO 2(s) ΔH = kj/mol Becomes SiO 2(s) Si (s) + O 2(g) ΔH = kj/mol 3. Since there is only one of each molecule in the original equation, no multiplication is necessary. 4. Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added: C (s) + O 2(g) CO 2(g) ΔH = kj/mol SiO 2(s) Si (s) + O 2(g) ΔH = kj/mol C (s) + O 2(g) + SiO 2(s) CO 2(g) + Si (s) + O 2(g) ΔH = kj Cancel like terms: C (s) + SiO 2(s) CO 2(g) + Si (s) ΔH = kj

5 3.4 Hess Law The Equation There is another way to calculate enthalpy changes that occur during chemical reactions that is based on the principal of Hess's Law. ΔH for a reaction may be calculated using the published ΔHf values and the equation: ΔH = ΔHproducts ΔHreactants That is, to find ΔH o for the reaction, add together all of the heats of formation, ΔH o f, for all of the products and subtract from that the sum of the heats of formation of all of the reactants. Pay close attention to balancing coefficients in the equation, as you must multiply the ΔHf values by these coefficients. Finally, be very careful with + and - values. Example 1: Calculate ΔH for the combustion of benzene, C 6H 6, as shown by the following reaction: Solution: C 6H 6 (l) + 15/2 O 2 (g) 6 CO 2 (g) + 3 H 2O (l) Remember that ΔH for any pure element = 0. (some exceptions) Look up Δ H values for C 6H 6 (l), CO 2(g), and H 2O(l). Remember, these ΔH values are given for 1 mole. In our final reaction; There are 6 mole of CO 2, so multiply ΔH by 6. There are 3 mole of H 2O, so multiply ΔH by 3. C 6H 6 (l) ΔH = CO 2 (g) ΔH = H 2O (l) ΔH = Using the formula ΔH = ΔH products - ΔH reactants C 6H 6 (l) + 15/2 O 2 (g) 6 CO 2 (g) + 3 H 2O (l) /2 (0) 6 (-393.5) + 3 (-285.8) ΔH = ΔH products ΔH reactants ΔH = (+49.0) ΔH = kj

6 Common Sources of Error Forgetting to multiply ΔH values by the appropriate coefficient. Using the wrong value of ΔH for water: ΔHf for H2O(l) = kj/mol; ΔHf for H2O(g) = kj/mol Solving for ΔH as "Reactants - Products" instead of "Products Reactants". Accidentally changing the sign for ΔH. Example 2: The standard heats of formation of HCl (g) and HBr (g) are kj/mol and kj/mol respectively. Using this information, calculate ΔH for the following reaction: Solution: Cl 2 (g) + 2 HBr (g) 2 HCl (g) + Br 2 (g) ΔH reaction = ΣΔH products - ΣΔH reactants It is helpful to write ΔH f values directly below the reaction participants, and find the sum of the reaction and product sides of the equation before using that formula: Cl HBr 2 HCl + Br (-36.4) 2 (-92.0) ΔH reaction = ΣΔH products - ΣΔH reactant (-184.0) - (-72.8) kj

7 3.5 Bond Enthalpy There is yet another way to calculate the heat of reaction, using bond enthalpies. Bond enthalpy refers to the amount of energy stored in the chemical bonds. Bond enthalpies have been experimentally determined and can be found in a Table of Bond Enthalpies. The net change in energy during a chemical reaction is the difference between how much energy it takes to break chemical bonds and how much energy is released when bonds form. So if we know how much energy is stored in these chemical bonds, we can calculate the change in energy for a reaction. ΔH = ΔHreactancts ΔHproducts It is important to realize that bonds are broken on the reactant side of the equation, and bonds are formed on the product side. This is different from Hess Law - we now subtract the products from the reactants. Example 1. Using bond enthalpies, provided in the table below, calculate the heat of reaction, ΔH, for Given the following bond enthalpies: ½ H 2(g) + ½ Cl 2(g) HCl(g) H H Cl Cl H Cl 436 kj 243 kj 433 kj Solution: Draw structural formulas for all molecules. Chemical formula Structural Formula H 2 Cl 2 HCl H H Cl Cl H Cl

8 Using the balancing coefficients in the balanced equation and the structural formulas, determine how much energy is required to break all of the bonds of the reactants, and how much energy is released when all of the product bonds form: REACTANTS: BOND BREAKING PRODUCTS: BOND FORMATION Bond No. bonds per molecule Bond Enthalpy Total Energy No. bonds per molecule Bond Enthalpy Total Energy H - H = 436 Cl - Cl = 243 H-Cl = 433 We still need to use the balancing coefficients from the balanced equation: For example, 1 mole of H - H bonds requires 436 kj but in our balanced equation only ½ mole of H - H bonds is involved, which will require only 218 kj ( ): ΔH = Σ (bonds broken) Σ (bonds formed) ΔH = [½(436) + ½(243)] [(433)] ΔH = = kj answer Example 2: Using the bond enthalpies provided, calculate the heat of reaction, ΔH, for: Given the following table of bond enthalpies: C 2H 4(g) + H 2(g) C 2H 6(g) C H C = C C C H H 413 kj 614 kj 348 kj 436 kj

9 Solution: Begin by drawing the structural formulas for all molecules. Chemical formula Structural Formula C 2H 4 C 2H 6 H 2 H H Determine how many bonds of each type are present in each molecule, and calculate the bond energies. For example, in C 2H 4 there is a single C=C bond (a double bond) but there are four C - H bonds. Bond REACTANTS: BOND BREAKING Number bonds per molecule bond enthalpy Total Energy PRODUCTS: BOND FORMATION bond enthalpy Number bonds per molecule Energy C - H = C = C = 614 H - H = 436 C - C = 348 Finally calculate ΔH for the reaction, being sure to check for coefficients from the balanced equation: C 2H 4(g) + H 2(g) C 2H 6(g) ΔH = Σ(reactant bonds) Σ(product bonds) ΔH = ( ) ( ) ΔH = = -124 kj answer

10 3.0 Calculating Enthalpy Change 3.1 Heats of Formation The amount of energy absorbed or released during the synthesis of 1 mole of a substance formed from its elements is referred to as the, H f. For example, the formation reaction for liquid water is described by the following equation: H2 (g) + ½O2 (g) H2O(l) kj Since the reaction is exothermic we would indicate the heat of formation, ΔH o f, as kj. You will find the standard heats of formation listed for a variety of compounds in the Table of Thermochemical Data. (The heat of formation for pure elements, such as H2(g), O2(g), Al(s), etc. is 0 kj mole -1.) To write heat of formation reactions: Balance the equation so that of the compound is produced. Remember the diatomic molecules and write them correctly (H2, N2, O2, F2, Cl2, Br2, I2). The reactants must be. 3.2 Hess s Law Scientists have determined that enthalpy change is independent of how the system changes from beginning to end. Chemical reactions do not occur in one step but rather by a series of steps, referred to as the. For instance, carbon and oxygen can react to form carbon monoxide: (1) C (s) + 1/2 O 2(g) CO (g) Δ H = kj Carbon monoxide can react further with oxygen to form carbon dioxide: (2) CO (g) + 1/2 O 2(g) CO 2(g) ΔH = kj

11 A third possible reaction involves carbon and oxygen reacting to form carbon dioxide directly: (3) C (s) + O 2(g) CO 2(g) Δ H = kj You will note that the heat of reaction from (1) plus the heat of reaction from (2) exactly equals the heat of reaction from (3). This makes sense when you consider what is going on; chemically, reactions (1) and (2) perform in two steps what reaction (3) does in one step. Note what happens when reactions (1) and (2) are added up: C (s) + 1/2 O 2(g) CO (g) CO (g) + 1/2 O 2(g) CO 2(g) ΔH = kj ΔH = kj C (s) + O 2(g) CO 2(g) ΔH = kj The end result is that it doesn't matter if the reaction proceeds all at once or in series of steps; the net energy change is the same. This illustrates :. This law allows us to take several reactions from the heats of formation table, add them up and get ΔH values for reactions that do not appear directly on the table. H (net) = H (step 1) + H (step 2) + + H (step n) 3.3 Calculating ΔH using Hess Law In order to calculate the overall ΔH using Hess idea we need to be able to of a reaction so that they can add up to the overall reaction. We can manipulate the steps of reactions in only 2 ways: 1. We can the equation and write backwards from products to reactants. When we do this the value of ΔH. 2. We can the equation by a common factor. When we do this we must also by the same factor.

12 Example 1: 2 Al (s) + 3 CuO (s) 3 Cu (s) + Al 2O 3(s) 1. Find an equation on the Heat of Formation Table to match each of the compounds in the given equation. 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = Cu (s) + ½ O 2(g) CuO (s) ΔH = 2. Flip the CuO equation to make the compound a reactant as well: Cu (s) + ½ O 2(g) CuO (s) ΔH = kj/mol Becomes CuO (s) Cu (s) + ½ O 2(g) ΔH = 3. Multiply the CuO equation by 3 to have the same number of molecules as in the original equation. 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = kj/mol 3 x CuO (s) Cu (s) + ½ O 2(g) ΔH = 3 x ( kj/mol) Gives 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = 3 CuO (s) 3 Cu (s) + 1½ O 2(g) ΔH = 4. Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added: 2 Al (s) + 1½ O 2(g) Al 2O 3(s) ΔH = 3 CuO (s) 3 Cu (s) + 1½ O 2(g) ΔH = 2 Al(s) + 1½ O2(g) + 3 CuO(s) Al2O3(s) + 3 Cu(s) + 1½ O2(g) ΔH = You will notice that there are similar terms on each side of the reaction equation. These can be treated again just like mathematical terms and cancelled out where you can: 2 Al (s) + 3 CuO (s) Al 2O 3(s) + 3 Cu (s) ΔH =

13 Example 2: SiO 2(s) + C (s) CO 2(g) + Si (s) Where the heat of formation of SiO 2(s) = kj/mol 1. Find an equation on the Heat of Formation Table to match each of the compounds in the given equation. The equation for SiO 2 is not on the table, but you are given the heat of formation of the substance from its elements: 2. Flip the SiO 2 equation to make the compound a reactant as well: Si (s) + O 2(g) SiO 2(s) ΔH = kj/mol Becomes 3. Since there is only one of each molecule in the original equation, no multiplication is necessary. 4. Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added: C (s) + O 2(g) CO 2(g) ΔH = kj/mol SiO 2(s) Si (s) + O 2(g) ΔH = kj/mol Cancel like terms:

14 3.4 Hess Law The Equation There is another way to calculate that occur during chemical reactions that is based on the principal of Hess's Law. ΔH for a reaction may be calculated using the published ΔHf values and the equation: That is, to find ΔH o for the reaction, add together all of the heats of formation, ΔH o f, for all of the products and subtract from that the sum of the heats of formation of all of the reactants. Pay close attention to balancing coefficients in the equation, as you must the ΔHf values by these coefficients. Finally, be very careful with + and - values. Example 1: Calculate ΔH for the combustion of benzene, C 6H 6, as shown by the following reaction: Solution: C 6H 6 (l) + 15/2 O 2 (g) 6 CO 2 (g) + 3 H 2O (l) Remember that ΔH for any pure element = 0. (some exceptions) Look up Δ H values for C 6H 6 (l), CO 2(g), and H 2O(l). Remember, these ΔH values are given for 1 mole. In our final reaction; There are 6 mole of CO 2, so multiply ΔH by 6. There are 3 mole of H 2O, so multiply ΔH by 3. C 6H 6 (l) ΔH = CO 2 (g) ΔH = H 2O (l) ΔH = Using the formula ΔH = ΔH products - ΔH reactants C 6H 6 (l) + 15/2 O 2 (g) 6 CO 2 (g) + 3 H 2O (l) /2 (0) 6 (-393.5) + 3 (-285.8) ΔH = ΔH products ΔH reactants ΔH = ΔH =

15 Common Sources of Error Forgetting to ΔH values by the appropriate coefficient. Using the wrong value of ΔH for water: ΔHf for H2O(l) = kj/mol; ΔHf for H2O(g) = kj/mol Solving for ΔH as "Reactants - Products" instead of " ". Accidentally for ΔH. Example 2: The standard heats of formation of HCl (g) and HBr (g) are kj/mol and kj/mol respectively. Using this information, calculate ΔH for the following reaction: Solution: Cl 2 (g) + 2 HBr (g) 2 HCl (g) + Br 2 (g) ΔH reaction = ΣΔH products - ΣΔH reactants It is helpful to write ΔH f values directly below the reaction participants, and find the sum of the reaction and product sides of the equation before using that formula: Cl HBr 2 HCl + Br 2

16 3.5 Bond Enthalpy There is yet another way to calculate the heat of reaction, using. Bond enthalpy refers to the. Bond enthalpies have been experimentally determined and can be found in a Table of Bond Enthalpies. The net change in energy during a chemical reaction is the difference between how much energy it takes to break chemical bonds and how much energy is released when bonds form. So if we know how much energy is stored in these chemical bonds, we can calculate the change in energy for a reaction. It is important to realize that bonds are broken on the reactant side of the equation, and bonds are formed on the product side. This is different from Hess Law - we now. Example 1. Using bond enthalpies, provided in the table below, calculate the heat of reaction, ΔH, for Given the following bond enthalpies: ½ H 2(g) + ½ Cl 2(g) HCl(g) H H Cl Cl H Cl 436 kj 243 kj 433 kj Solution: Draw structural formulas for all molecules. Chemical formula Structural Formula H 2 Cl 2 HCl H H Cl Cl H Cl

17 Using the balancing coefficients in the balanced equation and the structural formulas, determine how much energy is required to break all of the bonds of the reactants, and how much energy is released when all of the product bonds form: REACTANTS: BOND BREAKING PRODUCTS: BOND FORMATION Bond No. bonds per molecule Bond Enthalpy Total Energy No. bonds per molecule Bond Enthalpy Total Energy H - H = 436 Cl - Cl = 243 H-Cl = 433 We still need to use the balancing coefficients from the balanced equation: For example, 1 mole of H - H bonds requires 436 kj but in our balanced equation only ½ mole of H - H bonds is involved, which will require only 218 kj ( ): ΔH = Σ (bonds broken) Σ (bonds formed) ΔH = [½(436) + ½(243)] [(433)] ΔH = = kj answer Example 2: Using the bond enthalpies provided, calculate the heat of reaction, ΔH, for: Given the following table of bond enthalpies: C 2H 4(g) + H 2(g) C 2H 6(g) C H C = C C C H H 413 kj 614 kj 348 kj 436 kj

18 Solution: Begin by drawing the structural formulas for all molecules. Chemical formula Structural Formula C 2H 4 C 2H 6 H 2 Determine how many bonds of each type are present in each molecule, and calculate the bond energies. For example, in C 2H 4 there is a single C=C bond (a double bond) but there are four C - H bonds. Bond REACTANTS: BOND BREAKING Number bonds per molecule bond enthalpy Total Energy PRODUCTS: BOND FORMATION bond enthalpy Number bonds per molecule Energy C - H 413 = 413 = C = C 614 = H - H 436 = C - C 348 = Finally calculate ΔH for the reaction, being sure to check for coefficients from the balanced equation: C 2H 4(g) + H 2(g) C 2H 6(g)

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