Lab Test II Review Key

Lab Test II Review Calorimetry 1. Why may the 1st law thermodynamics describe the relationship between the solution and reaction to be equal but opposite sign? The 1 st law of thermodynamic is based on isolated systems where ΔE universe =0. You can relate the expression ΔE system = - ΔE surroundings with heat equilibrium to derive the expression q reaction = -q solution . q reaction can relate to the system while q solution relates to the surroundings. 2. Using a calorimeter, a GTC student dissolved 5.0 g of KNO 3 into 60.0 g of water at 23.5 C. After 12 minutes, the temperature of the solution peaked at 31.3 C. What is the molar heat of solution for dissolving KNO 3 (in kJ / mol KNO 3 )? Heat of solution: (5.0 g + 60.0 g) * 4.184 J/g o C * (31.4 o C – 23.5 o C) = 2121.288 J Heat or reaction: -2121.288 J Molar heat of solution: -2121.288 J / (5.0 g / (1 mol KNO 3 / 101.105 g) = -421894 J/mol  - 43 kJ/mol 3. The temperature for 17.2 g of an unknown metal is raised by 70.0 o C when 580.5 J of heat is added. What would be the specific heat of the metal? 580.5 J / (17.2 g * 70.0 o C) = 0.482 J/g o C

4. 50.0 mL of 0.10 M NaOH was reacted with 50.0 mL of 0.10 M HCl initially at 18.4 o C in a constant-pressure calorimeter. It was observed the temperature rose to a final reading at 19.5 o C. Assume the neutralization is the same as water in terms of density and specific heat. What is the molar heat of neutralization (in kJ/ mol H 2 O)? Heat of solution (50 g + 50 g)* 4.184 J/g o C * (19.5 o C – 18.4 o C) = 460.24 J Heat of neutralization -460.24 J Molar heat of neutralization -460.24 J / (0.1 M * 0.05 L) = -92.048 J/mol H 2 O  -92 kJ/ mol H 2 O 5. Why is the heat of neutralization of a strong acid/strong base larger than that of a weak acid/strong base? A strong acid fully dissociates ensuring all of the acid is neutralized, but weak acids partially dissociates meaning a smaller portion of water will be formed. Spectroscopy 6. Why is preparing a blank solution important before measuring a sample in a spectrophotometer? Preparing a blank or baselining a solution will prevent the instrument from reading the absorbance of other substances. 7. How many grams of solid copper (II) sulfate pentahydrate are needed to prepare a 50 mL of a 0.5 M solution in a volumetric flask? 0.5 M * 0.05 L * 249.696 g / mol CuSO 4 *5 H 2 O = 6.24 g