Thermochemistry Heats of Formation and Calculating Heats of Reaction

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. How much heat is released when a diamond changes into graphite? Diamonds are gemstones composed of carbon. Over a time period of millions and millions of years, diamond will break down into graphite, which is another form of carbon.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law How can you calculate the heat of reaction when it cannot be directly measured?

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law Hess’s law of heat summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. C(s, diamond) → C(s, graphite) Hess’s Law Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of diamond to graphite by using the following combustion reactions. a. C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law Write equation a in reverse to give: c. CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = kJ When you reverse a reaction, you must also change the sign of ΔH. a. C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ C(s, diamond) → C(s, graphite)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law If you add equations b and c, you get the equation for the conversion of diamond to graphite. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = kJ C(s, diamond) → C(s, graphite) b. C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ c. CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = kJ C(s, diamond) → C(s, graphite)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law If you also add the values of ΔH for equations b and c, you get the heat of reaction for this conversion. C(s, diamond) → C(s, graphite) C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = kJ C(s, diamond) → C(s, graphite)ΔH = –1.9 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law C(s, diamond) + O 2 (g) → CO 2 (g)ΔH = –395.4 kJ CO 2 (g) → C(s, graphite) + O 2 (g)ΔH = kJ C(s, diamond) → C(s, graphite)ΔH = –1.9 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. How can you determine ΔH for the conversion of diamond to graphite without performing the reaction? CHEMISTRY & YOU

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. How can you determine ΔH for the conversion of diamond to graphite without performing the reaction? CHEMISTRY & YOU You can use Hess’s law by adding thermochemical equations in which the enthalpy changes are known and whose sum will result in an equation for the conversion of diamond to graphite.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Suppose you want to determine the enthalpy change for the formation of carbon monoxide from its elements. Carrying out the reaction in the laboratory as written is virtually impossible. Hess’s Law Another case where Hess’s law is useful is when reactions yield products in addition to the product of interest. C(s, graphite)+ O 2 (g) → CO(g)ΔH = ? 1 2

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law You can calculate the desired enthalpy change by using Hess’s law and the following two reactions that can be carried out in the laboratory: C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ CO 2 (g) → CO(g) + O 2 (g)ΔH = kJ C(s, graphite)+ O 2 (g) → CO(g)ΔH = –110.5 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Hess’s Law C(s, graphite) + O 2 (g) → CO 2 (g)ΔH = –393.5 kJ CO 2 (g) → CO(g) + O 2 (g)ΔH = kJ C(s, graphite)+ O 2 (g) → CO(g)ΔH = –110.5 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation How can you calculate the heat of reaction when it cannot be directly measured?

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation Enthalpy changes generally depend on the conditions of the process. Scientists specify a common set of conditions as a reference point. These conditions, called the standard state, refer to the stable form of a substance at 25°C and kPa.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation The standard heat of formation (ΔH f °) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. The ΔH f ° of a free element in its standard state is arbitrarily set at zero. Thus, ΔH f ° = 0 for the diatomic molecules H 2 (g), N 2 (g), O 2 (g), F 2 (g), Cl 2 (g), Br 2 (l), and I 2 (s).

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation (ΔH f °) at 25°C and kPa Substance ΔH f ° (kJ/mol) Substance ΔH f ° (kJ/mol) Substance ΔH f ° (kJ/mol) Al 2 O 3 (s)–1676.0F2(g)F2(g) 0.0NO(g) Br 2 (g) 30.91Fe(s) 0.0NO 2 (g) Br 2 (l) 0.0Fe 2 O 3 (s)–822.1NaCl(s)–411.2 C(s, diamond) 1.9H2(g)H2(g) 0.0O2(g)O2(g) C(s, graphite) 0.0H 2 O(g)–241.8O3(g)O3(g) CH 4 (g) –74.86H 2 O(l)–285.8P(s, white) 0.0 CO(g) –110.5H2O2(l)H2O2(l)–187.8P(s, red) –18.4 CO 2 (g) –393.5I2(g)I2(g) 62.4S(s, rhombic) 0.0 CaCO 3 (s)–1207.0I2(s)I2(s) 0.0S(s, monoclinic) 0.30 CaO(s) –635.1N2(g)N2(g) 0.0SO 2 (g)–296.8 Cl 2 (g) 0.0NH 3 (g) –46.19SO 3 (g)–395.7

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation Such an enthalpy change is called the standard heat of reaction (ΔH°). For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation The standard heat of reaction is the difference between the standard heats of formation of all the reactants and products. ΔH° = ΔH f °(products) – ΔH f °(reactants) For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation This enthalpy diagram shows the standard heat of formation of water. The enthalpy difference between the reactants and products, –285.8 kJ/mol, is the standard heat of formation of liquid water from the gases hydrogen and oxygen. Notice that water has a lower enthalpy than the elements from which it is formed.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the standard heat of reaction (ΔH°) for the reaction of CO(g) with O 2 (g) to form CO 2 (g)? Calculating the Standard Heat of Reaction

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 KNOWNSUNKNOWN ΔH f °CO(g) = –110.5 kJ/mol ΔH f °O 2 (g) = 0 kJ/mol (free element) ΔH f °CO 2 (g) = –393.5 kJ/mol ΔH° = ? kJ Balance the equation of the reaction of CO(g) with O 2 (g) to form CO 2 (g). Then determine ΔH° using the standard heats of formation of the reactants and products.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. First write the balanced equation. Calculate Solve for the unknown. 2 2CO(g) + O 2 (g) → 2CO 2 (g)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Find and add ΔH f ° of all the reactants. Calculate Solve for the unknown. 2 Remember to take into account the number of moles of each reactant and product. ΔH f °(reactants) = 2 mol CO(g)  ΔH f °CO(g) + 1 mol O 2 (g)  ΔH f °O 2 (g) = 2 mol CO(g)  + 1 mol O 2 (g)  = –221.0 kJ –110.5 kJ 2 mol CO(g) 0 kJ 1 mol O 2 (g)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Find ΔH f ° of the product in a similar way. Calculate Solve for the unknown. 2 Remember to take into account the number of moles of each reactant and product. ΔH f °(products) = 2 mol CO 2 (g)  ΔH f °CO 2 (g) = 2 mol CO 2 (g)  = –787.0 kJ –393.5 kJ 1 mol CO 2 (g)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate ΔH° for the reaction. Calculate Solve for the unknown. 2 ΔH° = ΔH f °(products) – ΔH f °(reactants) = (–787.0 kJ) – (–221.0 kJ) = –566.0 kJ

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The ΔH° is negative, so the reaction is exothermic. This outcome makes sense because combustion reactions always release heat. Evaluate Does the result make sense? 3

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation Standard heats of formation are used to calculate the enthalpy change for the reaction of carbon monoxide and oxygen. 2CO(g) + O 2 (g) → 2CO 2 (g) The diagram shows the difference between ΔH f °(product) and ΔH f °(reactants) after taking into account the number of moles of each.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate the standard heat of reaction for the following: CH 4 (g) + Cl 2 (g) → C(s, diamond) + 4HCl(g) ΔH f ° (CH 4 (g)) = –74.86 kJ/mol ΔH f ° (C(s, diamond)) = 1.9 kJ/mol ΔH f ° (HCl(g)) = –92.3 kJ/mol

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate the standard heat of reaction for the following: CH 4 (g) + Cl 2 (g) → C(s, diamond) + 4HCl(g) ΔH f ° (CH 4 (g)) = –74.86 kJ/mol ΔH f ° (C(s, diamond)) = 1.9 kJ/mol ΔH f ° (HCl(g)) = –92.3 kJ/mol ΔH f °(reactants) = [1 mol CH 4 (g)  ΔH f °CH 4 (g)] + [1 mol Cl 2  ΔH f °Cl 2 (g)] = –74.86 kJ kJ = –74.86 kJ ΔH f °(products) = [1 mol C(s)  ΔH f °C(s, diamond)] + [4 mol HCl  ΔH f °HCl(g)] = 1.9 kJ + (4  –92.3 kJ) = –367.3 kJ ΔH° = ΔH f °(products) – ΔH f °(reactants) = –367.3 kJ – (–74.86 kJ) = –292.4 k

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Key Concepts & Key Equation Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations. For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. ΔH° = ΔH f °(products) – ΔH f °(reactants)

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Glossary Terms Hess’s law of heat summation: if you add two or more thermochemical equations to give a final equation, then you also add the heats of reaction to give the final heat of reaction standard heat of formation (ΔH f °): the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25°C