CHEM
Fall 2016.pdf

# 5 pts a the thermodynamic standard states of indium

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(5 pts) (a). The thermodynamic standard states of indium and bromine are In(s) and Br 2 ( l ). Write a chemical equation for the formation reaction of InBr 3 (s). Be sure to include the correct physical states in your equation. In(s) + 3 2 Br 2 ( l ) InBr 3 (s) (8 pts) (b). The heat of formation ( Δ H f ° ) for InBr 3 (s) is –428.9 kJ/mol, and Δ H ° = –947.7 kJ for the following reaction: 2 In(s) + 3 Br 2 (g) 2 InBr 3 (s) Find the heat of vaporization for Br 2 . (Vaporization is a phase change from liquid to gas, so the heat of vaporization would be Δ H ° for the process Br 2 ( l ) Br 2 (g).) There are a couple of ways to do this. One is to use the formation reaction from (a) and the reaction above and do Hess’s law. The vaporization equation is: 2 3 × {In(s) + 3 2 Br 2 ( l ) InBr 3 (s)} 1 3 × {2 In(s) + 3 Br 2 (g) 2 InBr 3 (s)} So Δ H ° = 2 3 × (–428.9 kJ) 1 3 × (–947.7) = 29.97 kJ/mol Another (maybe easier?) way is to realize that the vaporization reaction is the formation reaction for Br 2 (g). So you can do: Δ H ° = –947.7 kJ = 2 Δ H f ° (InBr 3 ) – 2 Δ H f ° (In) – 3 Δ H f ° (Br 2 (g)) Δ H f ° (Br 2 (g)) = 1 3 { 2 Δ H f ° (InBr 3 ) – 2 Δ H f ° (In) + 947.7 } = 1 3 { 2 (–428.9) – 2 (0) + 947.7 } = 29.97 kJ/mol

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A8 © 2015 L.S. Brown (10 pts) 7. An insulated container holds 525 g of molten ( i.e., liquid) tin at 358 ° C. If 269 g of solid tin are heated just to the melting point (232 ° C) and then dropped into the molten tin, the system comes to thermal equilibrium at 232 ° C with only liquid present. (In other words, the solid melts and the liquid cools to a final temperature of 232 ° C.) Find the molar heat of fusion ( H ° fusion ) for tin. Some of the data below may be useful. Heat capacity of solid tin = 27.11 J mol –1 K –1 Melting point of tin = 232 ° C Heat capacity of liquid tin = 28.49 J mol –1 K –1 Boiling point of tin = 2602 ° C I’m going to convert both masses to moles first. Tin is 118.71 g/mol, so: n liq = 525 g/118.71 g/mol = 4.423 mol and n solid = 269 g/118.71 g/mol = 2.266 mol Now set up the heat balance. The solid melts, but never changes its T; the liquid changes its T, but stays liquid. So: q solid = –q liquid n solid H ° fusion = n liquid C liquid T liquid I’d now just solve for H ° fusion : H ° fusion = n
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