Chemistry 431 Problem Set 1 Fall 2018 Solutions

Transcription

1 Chemistry 43 Problem Set Fall 208 Solutions. It is known that 25.0 ml of liquid benzene (C 6 H 6 ) contains carbon atoms. Determine the density of liquid benzene. ( ) ( ) ( ) (78. ) atoms molecule mol g 25.0 ml 6 atoms molecules mol = g ml 2. (a) A 5.0 kg ball collides with a 0.0 kg ball initially at rest. The collision is elastic and the final velocity of the 5.0 kg ball is m s. Determine the initial velocity of the 5.0 kg ball. m v,i + m 2 v 2,i = m v,f + m 2 v 2,f Given v 2,i = 0, we have 2 m v 2,i + 2 m 2v 2 2,i = 2 m v 2,f + 2 m 2v 2 2,f Let v = v,i x = v 2,f 5.0v = 5.0( 0.333) + 0.0x 2 (5.0)v2 = 2 (5.0)( 0.333) x 2 5.0v +.67 x = 0.0 ( ) 5.0v v 2 = v v 0.42 = 0

2 v = 0.84 ± (0.42)(.25) 2(.25) =.0, 0.35 Only the positive solution is physical so that v =.0 m s (b) The two balls having the same initial velocities as in part a collide and stick together. Use momentum conservation to find the final velocity of the two-ball unit. Show the kinetic energy is not conserved. (5 kg) v f = 5 kg m s v f = 3 m s KE f = 2 (5 kg)(/3 m s ) 2 = J KE i = 2 (5.0 kg)(.0 m s ) 2 = 2.5 J Then KE i KE f and ( ) J =.67 J goes into heat. 3. In this problem we apply the treatment of classical inelastic collisions to the gas-phase reaction H (g) + O-Cl (g) H-O-Cl (g) where the hydrogen atom collides head-on with the OCl molecule to form a bound, linear triatomic molecule. We assume the collision takes place in one dimension, and we assume initially the OCl molecule is at rest and the H atom has an initial velocity of m s (the average speed of a hydrogen atom at 298K). Calculate ) the final velocity of the HOCl molecule after the collision, 2) the final kinetic energy of the HOCl molecule and 3) the change in the total energy in the bonds making up HOCl after the collision. p i = p f m H v H + m OCl v OCl = m HOCl v HOCl (2730 ms ) + 0 = 52.5v HOCl v HOCl = 52.0 ms KE f = 2 m HOClvHOCl 2 = ( ) kg (52.0 ms ) 2 = J 26 KE i = ( ).00 kg (2730. ms ) 2 = J KE i + E bonds,i = KE f + E bonds,f E bonds = E bonds,f E bonds,i = KE i KE f = J 2

3 4. In a study of the elementary, molecular collision process ClO + O ClO 2 the ClO molecule is initially stationary (i.e. v = 0) and aligned along the x axis. The O atom moves in the positive x direction directly along the ClO bond. After the O and ClO species collide, the ClO 2 molecule forms, and the total internal vibrational energy of ClO 2 is found to be greater than the initial ClO vibrational energy by J. Calculate the initial speed of the O atom before the collision. Let KE be the initial kinetic energy of the oxygen atom and KE 2 be the final molecular kinetic energy. Let E v be the initial vibrational energy of ClO and E v2 be the final vibrational energy of ClO2. Then KE + E v = KE 2 + E v2 or KE = KE 2 + E v2 E v = KE J Using momentum conservation m O v 0 = m ClO2 v ClO2 where m O is the mass of the oxygen atom, m ClO2 is the mass of ClO 2, v 0 is the initial velocity of the O atom along the x direction and v ClO2 is the final molecular speed. Then v ClO2 = m O v 0 m ClO2 and 2 m Ov0 2 = 2 m ClO 2 vclo J 2 m Ov0 2 = ( ) 2 2 m mo v 0 ClO J v 2 0 ( v 2 0 = m ClO2 m O v J m ClO2 m O /2 ) = v 0 = 628. ms J 6.0 kg/ A constant force of 500. Newtons is applied upwards to a 20.0 kg block over 2.0 meters. In addition to the force of gravity, a frictional force of -5.0 Newtons also acts on the block. 3

4 (a) What is the change in the total energy of the block? E = (500 N)(2 m) = 6000 J (b) What is the change in the kinetic energy of the block? F net = 500 N (9.8 m s 2 )(20.0 kg) 5.0 N = 299. N KE = (299 N)(2 m) = 3588 J (c) What is the change in the potential energy of the block? P E = (20.0 kg)(9.8 m s 2 )(2.0 m) = 2352 J (d) Calculate the amount of work that is dissipated into heat. q = (2.0 m)(5.0 N) = 60.0 J 6. At what temperature does neon gas have a density of 0.80 g L, when the pressure is.0 bar? p = nrt T = P nr = P (m/m)r = P M ρr where M is the gram molecular mass and ρ is the density. Then T = (.0 bar)(20.8 g mol ) (0.80 g L )( L bar mol K ) = 303K 7. An ideal gas at 2.0 bar pressure occupies a bulb of unknown volume. A syringe is inserted into the bulb and some of the gas is withdrawn at constant temperature. It is found that the volume of the gas in the syringe is.0 cm 3 and the pressure is 4.0 bar while the pressure in the bulb is reduced to.92 bar. Find the volume of the bulb. P = P 2 2 ( ) = (4.0)(.0)cm 3 = 50.cm 3 8. A 0.0 liter bulb is used to confine 5.0 grams of CO 2 gas and 3.0 grams of N 2 gas at 25 C. Find the mole fractions and partial pressures of each gas. ( ) mol (5.0 g) = 0. mol CO g 4

5 P = nrt ( ) mol (3.0 g) = 0. mol N g χ N2 = χ CO2 = 0.5 = (0.22 mol)( L bar mol K )(298 K) 0.0 L = 0.54 bar P N2 = P CO2 =.5(0.54 bar) = 0.27 bar 9. Molecular iodine gas dissociates into iodine atoms according to the reaction I 2(g) 2I (g). When.28 moles of pure gas phase molecular iodine are introduced into a 00. liter flask at a temperature of 497K and allowed to come to equilibrium, the final total pressure is found to be 2.0 bar. Calculate the partial pressure of molecular iodine vapor in the equilibrium mixture. n I2 n I initial.28 mol 0 mol change x mol 2x mol equilibrium (.28 x) mol 2x mol n tot = n I2 + n I = (.28 + x) mol P = n totrt 2.0 bar = ((.28 + x) mol)( L bar mol K )(497. K) 00. L x = 0.33 mol P I2 = (( ) mol)( L bar mol K )(497. K) =.2 bar 00. L 0. A sample of 20.0 grams of N 2 O 4 gas is introduced into a 5.0 liter flask at 298K and allowed to dissociate and come to equilibrium according to the reaction N 2 O 4(g) 2NO 2(g). Given that at equilibrium 85.6% of the initial quantity of N 2 O 4 gas has dissociated, calculate the equilibrium total pressure in the flask g n N2 O 4 = 2(4. g mol ) + 4(6. g mol = 0.27 mol ) 5

6 n N2 O 4 n NO2 initial 0.27 mol 0 mol change (0.856)(0.27) mol 2(0.856)(0.27) mol equilibrium mol mol n tot = ( ) mol = mol P = n totrt = (0.403 mol)( L bar mol K )(298. K) 5.0 L = 2.0 bar. A tank used in scuba diving contains 00. L of compressed air with an initial pressure of 00.0 bar. After a diver inhales three times at full lung capacity at constant temperature and with lung pressure of bar, the final pressure in the tank is 99.8 bar. Calculate the volume of the diver s lungs. Answer P = P 2 2 ( )(00. L) = = 20.5 L lung = 2 /3 = 6.83 L 2. Gas phase PCl 5 dissociates according to the reaction PCl 5(g) PCl 3(g) + Cl 2(g) When pure PCl 5(g) is placed in a flask of fixed volume at 250. C, the initial pressure of the pure PCl 5(g) is bar. After equilibrium is reached, the final total pressure is bar. Calculate the mole fraction of Cl 2(g) in the flask containing the equilibrium mixture. n P Cl5 n P Cl3 n Cl2 initial n change αn 0 αn 0 αn 0 equilibrium ( α)n 0 αn 0 αn 0 n tot = n P Cl3 + n Cl2 + n P Cl5 = n 0 ( + α) P = n totrt = ( + α)n 0RT = ( + α)p = ( + α)43.48 α = χ Cl2 = n Cl 2 n 0 α = n tot n 0 ( + α) = α + α =

7 3. Sulfur trioxide in the gas phase dissociates into sulfur dioxide gas and oxygen gas according to the reaction SO 3(g) SO 2(g) + 2 O 2(g). When pure sulfur trioxide gas is placed in an enclosed container of fixed volume at 00. K, the initial pressure is measured to be bar. The system is allowed to come to equilibrium at a fixed temperature of 00 K, and the final total pressure is bar. Calculate the fraction, α (i.e. the degree of dissociation), of the original sulfur trioxide that dissociates when equilibrium is reached. Let n 0 be the initial number of moles of sulfur trioxide. n SO3 n SO2 n O2 initial n change αn 0 αn 0 αn 0 /2 equilibrium ( α)n 0 αn 0 αn 0 /2 n tot = n SO3 + n SO2 + n O2 = n 0 ( + α/2) P tot = n totrt = n 0RT ( + α/2) = P 0 ( + α/2) = 2.548( + α/2) α = A sample of pure NOCl gas is placed in a container of fixed volume at a temperature of 500. K and allowed to dissociate according to the reaction NOCl (g) NO (g) + 2 Cl 2(g). When equilibrium is reached, the degree of dissociation of NOCl is found to be α = 0.265, and the total pressure is measured to be 2.00 bar. Calculate the initial pressure of the NOCl gas before the dissociate reaction commences. Let n 0 = the initial number of moles of NOCl. n NOCl n NO n Cl2 initial n change αn 0 αn 0 αn 0 /2 equilibrium ( α)n 0 αn 0 αn 0 /2 n tot = n NOCl + n NO + n Cl2 = n 0 ( + α/2) P tot = n totrt = P 0 ( + α/2) 2.00 bar = P 0 ( /2) P 0 =.77 bar 7

8 5. Gas-phase phosgene dissociates into carbon monoxide gas and chlorine gas according to the reaction COCl 2(g) CO (g) + Cl 2(g). When pure phosgene is introduced into a closed flask at 500. C the initial pressure is found to be.82 bar and the total pressure when equilibrium is reached is found to be 3.50 bar. Calculate the mole fraction and partial pressure of phosgene gas in the equilibrium mixture. n COCl2 n CO n Cl2 initial n change -αn 0 αn 0 αn 0 equilibrium n 0 ( α) αn 0 αn 0 n tot = n COCl2 + n CO + n Cl2 = n 0 ( + α) P tot = n totrt = ( + α)n 0RT = ( + α)p 0 α = P tot P 0 = = χ COCl2 = n COCl 2 n tot = α + α = P COCl2 = χ COCl2 P tot = 0.40 bar 8