Question

GENERAL CHEMISTRY GROUP STUDY SB The first 3 problems are to be submitted today. #1.(30pts, 15 pts each) (c) Consider a 135 meter high waterfall. If the decrease in potential energy of the water (i.e., mgh) is totally converted into heat, calculate the temperature difference between the top and bottom of the waterfall. Data: m-mass, g- acceleration due to gravity 9.81 m/s. The specific heat of water- 4.184 J/(g "C). (d) If a system absorbs 30.0 J of heat from the surroundings, and if it also decreases its internal energy by 33 J, compute the work (w) associated with this process. Interpret the sign of your answer 3(20 pts) Consider the data: AH (kJ/mol) -1226.6 -2967.3 -84.2 -285.7 Pa(s) + 6 Cl(g) >4 PCl,(g) Pds) 5 O2(g) PCls(g)+ Cl2(g) PCl(g) O(g) - P4O10(s) >PCI,(g) >ClaPO(g) Using only the above reaction data, compute the AH of P4O10(s)+ 6 PCl,(g) > 10 CPO (g) The following problems are to be submitted next week, but you should work on them today with your group. #4.(25pts) Consider the reaction AH (kJ/mole) 5 CO:(g) + SigN4(s) ----> 3 SiO(g) + 2 N2O(s) + 5 CO(g) AH 2023.75 kJ/mole (c) Use the above information as well as heat of formation data in the appendix of your textbook to calculate AH (Si,N (s)) in kJ/mole. Additional information: Exn AHO (SiO(g)) 1.94 x 103 kJ/mole (d) How many kJ of heat are absorbed when 0.75 lbs of SigNa(s) react according to the above equation? 1 lb 453.6 g #5.(25pts)) A 25.0 g piece of iron originally at a temperature of 5.2°C and a 35.0 g piece of copper initially at 83.2°C are placed into a calorimeter containing 75.0 g of water at 16.0°C. What is the final temperature of the system (i.e., the two metals + water) in °C assuming that the heat capacity of the calorimeter is negligible? Data: specific heat of Fe 0.450 J/(g°C) and the specific heat of Cu 0.387 (J/g °C)

Answer 1

( 1. (c) wa Potential Energy - heat ton. mgh = in C AT 9.81 m x 135 m = 4.184 AT AT = 3 16.8300 (d) ist law of Thermodynamics Q = utw neutraen en 30J = -33 It we It wt w = 631