ALE 27. Hess s Law. (Reference: Chapter 6 - Silberberg 5 th edition)

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1 Answer Key ALE 27. Hess s Law (Reference: Chapter 6 - Silberberg 5 th edition) Important!! For answers that involve a calculation you must show your work neatly using dimensional analysis with correct significant figures and units to receive full credit. No work, no credit. Report numerical answers to the correct number of significant figures. CIRCLE ALL NUMERICAL RESPONSES. What are thermochemical equations and why can they be added algebraically? The Model: Heats of Reaction H is the symbol for enthalpy (i.e., heat content ) of a chemical substance. When H is the difference in enthalpies between products and reactants ( H = H products H reactants ), H is called the heat of reaction. Atomization reactions: written as a Thermochemical Equation written with the heat of reaction separate and to the side C(s, graphite) kj C(g) C(s, graphite) C(g) H = 715 kj H 2 (g) kj 2 H(g) H 2 (g) 2 H(g) H = 436 kj CH 4 (g) kj C(g) + 4 H(g) CH 4 (g) C(g) + 4 H(g) H = 1662 kj Key Questions 1. What is a thermochemical equation? A thermochemical equation is a balanced chemical equation that shows not only the stoichiometric relationship between reactants and products but also shows how much heat is absorbed / released per mole of limiting reagent when the reaction takes place. 2. What is an atomization reaction? An atomization equation is a balanced chemical equation that shows the stoichiometric relationship when a substance is broken down to its component atoms in the gaseous state. 3. When a reaction is endothermic (circle the correct answer): i. H < 0 and heat is a reactant iii. H < 0 and heat is a product ii. H > 0 and heat is a reactant iv. H > 0 and heat is a product 4. The symbol kj stands for heat just as CH 4 (g) stands for gaseous methane. Thermochemical equations are best interpreted on a macroscopic scale. Complete the macroscopic interpretation of the following thermochemical equation. CH 4 (g) kj C(g) + 4 H(g) One mole of gaseous methane reacts with 1662 kj of heat to produce 1 mole of gaseous monatomic carbon and 4 moles of gaseous monatomic hydrogen. Page 1 of 5

2 5. When 1 mole of gaseous carbon is allowed to combine with 4 moles of gaseous atomic hydrogen, 1 mole of gaseous methane is produced and (circle the correct answer):: i kj of heat is consumed ii kj of heat is liberated 6. When 2 moles of gaseous molecular hydrogen are atomized, 4 moles of gaseous atomic hydrogen are produced and (circle the correct answer):: i. 218 kj of heat are consumed iv. 218 kj of heat are liberated ii. 436 kj of heat are consumed iii. 872 kj of heat are consumed The Model: Hess s Law v. 436 kj of heat are liberated vi. 872 kj of heat are liberated Just like algebraic equations can be added together, chemical equations can be added together. A species that is on the reactant side of one reaction algebraically cancels with the same species in the same physical state on the product side of another reaction. A set of chemical equations that we will add together in an attempt to get equation 6, below: eqn 1 C(s, graphite) C(g) H = 715 kj eqn 2 H 2 (g) 2 H(g) H = 436 kj eqn 3 CH 4 (g) C(g) + 4 H(g) H = 1662 kj We ll start out with eqn 1 as is. But we ll have to get rid of the 1 mole of gaseous carbon atoms on the product side. Equation 3 has 1 mole of gaseous carbon as a product. So we ll reverse eqn 3 (i.e., we ll make the reactants products, and the products reactants) and label the reverse of eqn 3 as eqn 4: eqn 4 C(g) + 4 H(g) CH 4 (g) H = kj Now if we add eq 1 and eq 4 we ll be able to cancel the gaseous carbon atoms, but we ll still have 4 moles of gaseous atomic hydrogen on the reactant side. If we define eq 5 as two times eq 2, eqn 5 2 H 2 (g) 4 H(g) H = 872 kj we have an equation which we can add to eq 1 and eq 4 to eliminate the gaseous atomic hydrogen. Hess s Law says that when we add chemical equations together, the heat of the resultant reaction is a sum of the heats of the individual reactions. So... eqn 1 C(s, graphite) C(g) H = 715 kj eqn 4 C(g) + 4 H(g) CH 4 (g) H = kj eqn 5 2 H 2 (g) 4 H(g) H = 872 kj eqn 6 C(s, graphite) + 2 H 2 (g) CH 4 (g) H = ( ) kj = -75 kj Equation 6, above, is an example of a formation reaction because it shows the production of 1 mole of a compound of interest (e.g., gaseous methane) from elements in their naturally-occurring states (e.g., solid carbon graphite and gaseous molecular hydrogen). ALE 27 - Page 2 of 5

3 Key Questions 7. What is a formation reaction? A formation reaction involves the production of 1 mole of a compound of from elements in their naturally-occurring states. 8. What is the heat of formation of any element in its naturally-occurring state? Explain why. A formation reaction shows 1 mole of the compound of interest being produced from elements in their naturally-occurring states. If the product is an element in its naturallyoccurring state, then the reaction would be something like: X( ) X( ) H =? There is no change in going from the reactant to the product, so there is no heat exchanged between the chemical system and its surroundings. The heat of formation of any element in its naturally-occurring state is, therefore, defined to be 0 kj/mol. Exercise A. A combustion reaction is one in which a substance reacts with gaseous molecular oxygen. The complete combustion reactions of graphite, hydrogen, and methane are: C(s, graphite) + O 2 (g) CO 2 (g) H = -394 kj H 2 (g) + ½ O 2 (g) H 2 O(l) H = -286 kj CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) H = -891 kj Use the above combustion reactions and Hess s Law to determine the heat of the following reaction: C(s, graphite) + 2 H 2 (g) CH 4 (g) H =? eq 1 C(s,graphite) + O 2 (g) CO 2 (g) H 1 = -394 kj eq 4 2 H 2 (g) + O 2 (g) 2 H 2 O(l) H 4 = -572 kj eq 5 CO 2 (g) + 2 H 2 O(l) CH 4 (g) + 2 O 2 (g) H 5 = +891 kj C(s,graphite) + 2 H 2 (g) CH 4 (g) H = H 1 + H 4 + H 5 = ( ) kj = -75 kj Key Question 9. Compare the heat of the formation reaction of methane you determined in Exercise A (i.e., when combustion reactions were added together) with that which was presented in the Model (i.e., when atomization reaction were added together). Explain the relevance of this discovery. They are both -75 kj/mol (i.e., 75 kj of heat energy are released for every mole of methane that is produced). It does not matter if we added atomization thermochemical equations or if we added combustion thermochemical equations together, we get the same H net. The reason why is that we have the same reactanst and the same products in the two net chemical equations. It doesn t matter HOW we go from 1 mole of graphite and 2 moles of molecular hydrogen to become 1 mole of methane, H net will always be the same value because H is the enthalpy of the products minus the enthalpy of the reactants. ALE 27 - Page 3 of 5

4 Exercise B. Write the balanced formation reaction for Fe 3 O 4 (s). Hint: What are the naturally-occurring states of iron and oxygen? 3 Fe(s) + 2 O 2 (g) Fe 3 O 4 (s) Calculate the heat of formation ( H f ) for Fe 3 O 4 (s) given the following bank of thermochemical equations: EQN 1 Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) H = -28 kj EQN 2 3 Fe 2 O 3 (s) + CO(g) 2 Fe 3 O 4 (s) + CO 2 (g) H = -59 kj EQN 3 2 CO(g) + O 2 (g) 2 CO 2 (g) H = -568 kj 3/2 x (Reverse eqn 1): 3 Fe(s) + 9 / 2 CO 2 (g) Fe 2 O 3 (s) + 9 / 2 CO(g) H 1 = 3/2(28 kj) = +42 kj 1/2 x EQN 2: 3 / 2 Fe 2 O 3 (s) + 3 / 2 CO(g) Fe 3 O 4 (s) + 3 / 2 CO 2 (g) H 2 = ½ (59 kj) = kj 2 x EQN 3: 4 CO(g) + 2 O 2 (g) 4 CO 2 (g) H 3 = 2(-568 kj) = kj 3 Fe(s) + 2 O 2 (g) Fe 3 O 4 (s) H = H 1 + H 2 + H 3 = ( ) kj = kj = kj C. Problem 6.48: Would you expect O 2(g) 2 O (g) to have a positive or negative H rxn? Explain your reasoning. Energy is required to break the O O bond. Energy + O 2 (g) 2 O(g) H rxn = (+) D. Problem 6.51: Consider the following balanced thermochemical equation for the decomposition of the mineral magnesite: MgCO 3(g) MgO (s) + CO 2(g) H rxn = kj a.) Is heat absorbed or released in the reaction? Explain your reasoning. Heat is absorbed since H is positive: b.) What is H rxn of the reverse reaction? kj + MgCO 3(g) MgO (s) + CO 2(g) H rxn (reverse) = kj: MgO (s) + CO 2(g) MgCO 3(g) kj c.) What is H when 5.35 mol of CO 2 reacts with excess MgO? Show work and circle your answer kj H rxn = x 5.35 mol CO 2 = 628 k 1 mol CO 2 d.) What is H when 35.5 g of CO 2 reacts with excess MgO? Show work and circle your answer kj H rxn = x 35.5 g CO 2 = 94.6 kj g CO 2 ALE 27 - Page 4 of 5

5 E. Problem 6.53: When 1 mol of solid potassium bromide, KBr, decomposes to its elements, 394 kj of heat is absorbed. a.) Write the balanced thermochemical equation for this reaction. KBr (s) K (s) + ½ Br 2(l) H rxn = 394 kj b.) How much heat is released when 10.0 kg of KBr forms from its elements? Show work and circle your answer. 394 kj 1 mol KBr 4 q(kj) = x x 1.00 x 10 g KBr 1 mol KBr 119 g KBr 4 = 3.31 x 10 kj, therefore, 3.31 x 10 4 kj would be released. F. Problem 6.59: Sucrose (table sugar, C 12 H 22 O 11 ) is oxidized in the body by O 2 via a complex set of reactions (glycolysis, Krebs cycle and electron transport chain) that ultimately produces CO 2(g) and H 2 O (g) and releases 5.64 x 10 3 kj/mol sucrose. a.) Write the balanced thermochemical equation for this reaction. C 12 H 22 O 11 (s) + 12 O 2 (g) 12 CO 2 (g) + 11 H 2 O(l) H rxn = 5.64 x 10 3 kj b.) How much heat is released per gram of sucrose oxidized? Show work and circle your answer. b.) How much heat is released per gram of sucrose oxidized? Show your work x 10 kj 1 mol C12H22O11 q(kj) = x x 1.00 g C12H22O11 = 16.5 k 1 mol C H O g C H O G. Problem 6.64: Given the thermochemical reactions 1 and 2, below, calculate H rxn for reaction #3. Show work and circle your answer. Reaction #1: ½ N 2(g) + ½ O 2(g) NO (g) H rxn = 90.3 kj Reaction #2: NO (g) + ½ Cl 2(g) NOCl (g) H rxn = kj Reaction #3: 2 NOCl (g) N 2(g) + O 2(g) + Cl 2(g) H rxn =??? NOCl(g) 2 NO(g) + Cl 2 (g) H = 2( 38.6 kj) 2 NO(g) N 2 (g) + O 2 (g) H = 2(90.3 kj) 2 NOCl(g) N 2 (g) + O 2 (g) + Cl 2 (g) H = kj ALE 27 - Page 5 of 5