HW-Solns-Chap-6-1413

# Look at the first reactant pbcl 2 s the only given

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Look at the first reactant: PbCl 2 (s). The only given equation that has PbCl 2 (s) is the first one, but PbCl 2 (s) is a product in the equation, so the equation must be reversed. Therefore, we write the reverse of equation with the sign of its H° changed. Look at the second reactant: Cl 2 . This chemical is in both of the given reactions, so let’s skip it. Look at the product: PbCl 4 (s). The only given equation that has PbCl 4 (s) is the second one, so you must include that equation, as written, with its given H°. We have now planned how we will use all the given reactions, so let’s do it: Add all the reactions as planned, eliminate reactants that also show up as products: one of the Cl 2 (g) molecules. Add all the remaining reactants and products to form the net equation for the desired reaction, and add all the individual H° values to get the H° for the reaction: PbCl 2 (s) Pb(s) + Cl 2 (g) H° = – (–359.4 kJ) + Pb(s) + 2 Cl 2 (g) PbCl 4 (s) + H° = –329.3 kJ PbCl 2 (s) + Cl 2 (g) PbCl 4 (s) H° = 30.1 kJ Reasonable Answer Check: The net equation is the desired reaction. The enthalpy of formation of these two lead chloride compounds are similar to each other, so forming one from the other should have a small H°. 58. Answer: H° = –905.47 kJ; exothermic Strategy and Explanation: Given a balanced chemical equation for a reaction and a table of molar enthalpies of formation, determine the enthalpy change of the reaction and whether it is exothermic or endothermic. Use the stoichiometric coefficient of the balanced equation to describe the moles of each of

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