5 pts a the thermodynamic standard states of arsenic

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(5 pts) (a). The thermodynamic standard states of arsenic and chlorine are As(s) and Cl 2 (g). Write a chemical equation for the formation reaction of AsCl 3 ( l ). Be sure to include the correct physical states in your equation. As(s) + 3 2 Cl 2 (g) AsCl 3 ( l ) (8 pts) (b). The heat of formation ( Δ H f ° ) for AsCl 3 ( l ) is –305.0 kJ/mol, and Δ H ° = –548.8 kJ for the following reaction: 2 As(s) + 3 Cl 2 ( l ) 2 AsClr 3 ( l ) 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 Cl 2 ( l ) Cl 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 × {As(s) + 3 2 Cl 2 (g) AsCl 3 ( l )} + 1 3 × {2 As(s) + 3 Cl 2 ( l ) 2 AsCl 3 ( l )} So Δ H vap ° = – 2 3 × (–305.0 kJ) + 1 3 × (–548.8) = 20.4 kJ/mol Another (maybe easier?) way is to realize that the vaporization reaction is the reverse of the formation reaction for Cl 2 ( l ). So you can do: Δ H ° = –548.8 kJ = 2 Δ H f ° (AsCl 3 ) – 2 Δ H f ° (As) – 3 Δ H f ° (Cl 2 ( l )) Δ H f ° (Cl 2 ( l )) = 1 3 { 2 Δ H f ° (AsCl 3 ) – 2 Δ H f ° (As) + 548.8 } = 1 3 { 2 (–305.0) – 2 (0) + 548.8 } = –20.4 kJ/mol so Δ H vap ° = 20.4 kJ/mol
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B8 © 2015 L.S. Brown (10 pts) 7. An insulated container holds 525 g of molten ( i.e., liquid) lead at 411 ° C. If 269 g of solid lead are heated just to the melting point (328 ° C) and then dropped into the molten lead, the system comes to thermal equilibrium at 328 ° C with only liquid present. (In other words, the solid melts and the liquid cools to a final temperature of 328 ° C.) Find the molar heat of fusion ( H ° fusion ) for lead. Some of the data below may be useful. Heat capacity of solid lead = 26.65 J mol –1 K –1 Melting point of lead = 328 ° C Heat capacity of liquid lead = 29.42 J mol –1 K –1 Boiling point of lead = 1749 ° C I’m going to convert both masses to moles first. Lead is 207.2 g/mol, so: n liq = 525 g/207.2 g/mol = 2.534 mol and n solid = 269 g/207.2 g/mol = 1.298 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
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