University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2011

Transcription

1 Homework Assignment #: Due at 500 pm Wednesday July 6. University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 0 ) he respiratory system uses oxygen to degrade glucose to carbon dioxide and water. he net equation is CH 6 O6( s) + 6O( g) 6CO + 6HO ( ) a) Calculate the enthalpy of reaction at 98. CH 6 O6( s) + 6O( g) 6CO + 6HO ( ) H = 6 H f CO, g + 6 H f HO, H f C6HO6, s H f O, g = 6 ( 393.5kJ ) + 6 ( 85.8kJ ) ( 73.kJ ) = 36.0kJ 74.8kJ + 73.kJ = 80.7kJ b) Assume respiration occurs at constant pressure. Calculate the amount of heat produced by respiration per mole of oxygen absorbed at H 80.7kJ q = = = 467. kj / molo 6 6 c) An average person uses about. moles of oxygen per hour in the course of respiring. Calculate the amount of heat produced by respiration per kilogram per hour by an average person. Assume a weight of 70 kg. ( 467. kj / mole)(. mole / hr) q = = 8.0 kj / hr kg 70kg d) Most of the heat produced by the body is lost. But suppose that a person were clothed in a thermally insulated suit such that all the heat from respiration were retained by the body. After 0 hours of respiration, how much would the body temperature of the person described in part c rise? Assume a heat capacity of 4.8kJ kg - -. q= ( 8.0 kj / kg hr)( 0hr) = 80. kj / kg 80. kj / kg = 4.8 kj / kg = 9.6

2 ) Suppose one mole an ideal monatomic gas contracts adiabatically and reversibly. he initial state of the gas is = atm, =98. he final pressure of the gas is =3 atm. a) Calculate the initial volume, the final temperature and the final volume of the gas. ( R 0.08L atm mol )( 98) = = = 4.5L atm /3 /3 3/ 5/ = = = = /5 /5 ( 98 )( 3) 463 = = = R ( 0.08L atm )( 463) = = = 3atm.7L b) Calculate U and H for the gas. 3R U = nc = n = (.5)( 8.3J )( ) = (.5J )( 65) = 063J 5R H = nc = n = (.5)( 8.3J )( ) = ( 0.8J )( 65) = 348J c) Calculate the work done on the gas. Adiabatic so U = w= 063J 3) Five moles of an ideal monatomic gas contract adiabatically and irreversibly when subjected to a constant external pressure of atm from an initial volume =00L and initial pressure =0.50 atm. Assume the contraction ceases when the system reaches equilibrium. a. Calculate the initial temperature, the final temperature and the final volume of the gas. ( 0.50atm )( 00.0L) = 43 nr = 5mol 0.08L atm mol =

3 Contraction ceases at equilibrium where = = atm ext nr U = wirrev nc ( ) = ext ( ) = ext nr ext nc ( ) = ext nc + nr = ext+ nc 5R 3R n = ext + n ext = + = + 5nR 5 5 5mol 0.08L atm mol 5 ( atm)( 00L) = = 34 b. Calculate U and H for the gas. 3R U = nc = n = (.5)( 8.3J )( ) = (.5J )( 98.0 ) = 5J 5R H = nc = n = (.5)( 8.3J )( 34 43) = ( 0.8J )( 98.0) = 038J c. Calculate the work done on the gas. w= U = 5J 4) In the text, Ch. 3, equation 3.9 it was shown, but not proven that =. Suppose a gas at =98 and =4.0L obeys the equation of state: nb = nr H Using equation 3.9, calculate and for the gas under these conditions. 3 - Assume n= mole. At =98 b = 38.6cm mol. Hint: See section 3.6. = nr nr = = = 0 nb nb From section 3.6

4 H = + + = nr nr ( nb) = nr = + nb = H nr nr = = + nb = nb H = ( mol )( 38.6cm mol ) = 38.6cm 3-3 5) For water at =300 and = atm, the coefficient of thermal expansion 4 0 β = and the isothermal compressibility κ = mn. Calculate for water at =300 and = atm. Will this value of less than for water at =383 and =atm? Explain.. 4 β = = = ( 300) 035Nm 0 κ m N = Nm.0 0 Nm Nm be greater than or Water boils at =373 so at =383 the water is in vapor form. Intermolecular interactions are smaller in the vapor versus the liquid so would be smaller at =383 than at = ) Nitrogen reacts with hydrogen to form ammonia: N ( g) H ( g) NH ( g) +. 3 he table below gives the standard heats of formation at =98 and the heat capacities at constant pressure as a function of temperature : H f,9 C J mol (kj mol - ) NH3 ( g ) N ( g ) H ( g )

5 Note: As in the text, the term / indicates the in terms of degrees elvin () divided by. he ratio / is therefore dimensionless. Calculate the enthalpy of formation of ammonia H f at =000. Note: he original problem had a minor misprint. he enthalpy of formation of ammonia is listed for =9. =98 was intended. If you do the integral from =9 to =000, the answer changes slightly and no credit will be removed as long as you are clear about what you are doing 000 f ( 000 ) f ( 98 ) ( 3) H = H + C d C = C NH C N C H = + C d d = + kj mol = 9.75kJ mol = ( ) 000 f f 98 H 000 = H 98 + C d = 46.kJmol 9.75kJmol = 55.86kJmol 6) Suppose mole of water is initially at =53 and =atm. he temperature is raised to =400 by placing it in contact with hot iron pan that is maintained at =400. he pressure is maintained at =atm. Assume the heat capacity at constant pressure of ice C is constant between 53 and =73 and equals 37 J - mol -. Assume the heat capacity C of liquid water is constant and is equal to 75.3 J - mol -. Assume finally that C of water vapor in the temperature range given is 33 J - mol -. For water H melt =600Jmol - and H vap =4000 Jmol -.Calculate the enthalpy change of the water and the enthalpy change of the iron pan. ( ) ( ) ( ) H = C + H + C + H + C ice liq vap water melt vap - ( 37 J mol )( 0) 600Jmol ( 75.3 J mol )( 00) = + + ( ) Jmol + 33 J mol 7 = 557Jmol - - H = H = 557Jmol pan water -