University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2008

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1 University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 008 Midterm Examination Key July Blue books are required. Answer only the number of questions requested. Chose wisely! If you answer more than the number of problems requested the excess problems will NO be graded. You may keep this exam copy as a souvenir. Good luck Avagadro s Number=N A =6.0x10 Universal gas constant=r=8.1j/mole-k=0.081l-atm/mole-k 1J=1 Nt-m=1kg-m /s 1atm=1.01x10 5 a=1.01x10 5 Nt/m 1000L=1m 101J=1L-atm. 1) (18 points) Define out of the following 6 terms as they are used in the field of thermodynamics and/or mechanics. 1.1) Define the term adiabatic as it is used in thermodynamics. When is the entropy change for an adiabatic process zero and when is it not zero? Answer: A change that proceeds without q exchange i.e. q=0. If the process is δ q reversible and adiabatic then q rev =0. hen S = rev d = 0. But for an irreversible change q irrev = 0 and because qrev > qirrev S 0 1.) Define the thermal expansivity β and the isothermal compressibility κ as derivatives involving state variables V. hese two quantities together with what four other physical properties are used to evaluate thermodynamic U derivatives like etc. Give two examples. V 1 V Answer: thermal expansivity β = and isothermal compressibility V 1 V κ =. he other four are V and the hear capacity at constant V H pressure C =. wo examples: U β CV = = C V κ V H = and V ( 1 β)

2 S Others: = βv on and S C = and U β = the list goes V κ 1.) Define the chemical potential and discuss how we use this quantity to define equilibrium. Give two examples. Answer: Any of the following possibilities will do: U H A G µ i = = = = ni n SV n i n j Sn i n j V n i j n j where i refers to the ith component of a multi-component system. he variable ni is the number of moles of component i. the chemical potential is used as a citerion for equilibrium. Given two phases at equilibrium: α β the criterion for equilibrium is µ α = µ β. Given a chemical reaction equilibrium: ν AA+ νbb νcc+ νdd he equilibrium criterion is ν µ + ν µ = ν µ + ν µ A A B B C C D D 1.4) Define the heat capacity C V. How is the heat capacity related to mechanical; degrees of freedom of a system? When is this relationship valid? Also rank the importance of translational rotational vibrational and electronic motions to the heat capacity for =00K. 1.5) Unless you drive an automobile equipped with a diesel engine your car s engine probably operates according to a thermodynamic cycle called the Otto cycle. Diagram the reversible version of the Otto cycle on a -V axis system and explain each step. Does your car s engine operate reversibly? If it did what would be the advantage(s)? Explain. Answer: Do a sketch of type shown in text figure 5.14a. In-take stroke followed by adiabatic compression (adiabatic because it goes fast) followed by fuel ignition and rapid rise in pressure at essentiallyc onstant volume. hen the power stroke occurs as an adiabatic expansion. henteh exhaust valve opens and the gas escapes followed by a rapid loss of pressure at constant volume. he cycle then repeats itself. he engine cannot run reversibly because it cannot pass through a long sequence of equilibrations. If it did maximum work would be performed and your mileage would be awesome. 1.6) Define an exact differential and explain why state functions are represented by exact differentials.

3 Answer: using a two dimensional differential M ( xydx ) ( ) + N xydyas an example the differential is exact if there exists a function f such that F F df = M ( x y) dx + N ( x y) dy = dx + dy x y y x From this statement Euler s criterion for exactness follows: M N F F = = = y x x y x y y x he integral of an exact differential depends only on the initial state and the final state not on the details of the path: a property which is possessed by state functions their changes depend only on the initial and final equilibrium states. ) (0 points) Answer two out of the four following questions. Your answers should be stated in clear complete sentences and each answer should be supported by a thorough explanation based on the principles of thermodynamics..1) he amplitude of a pendulum consisting of a mass on a long wire is initially adjusted to have a very small value. he amplitude is observed to decrease slowly with time. Is this process reversible? Would the process be reversible if the amplitude did not decrease with time? Solution: he amplitude decreases in time so dissipative forces (e.g. gravity friction) are acting on the system and the motion is not reversible..) he smelting of iron in a blast furnace occurs when carbon derived from coal reacts with oxygen to produce carbon monoxide: C( s) + O ( g) CO( g). At the temperature of the blast furnace CO gas quickly reaches equilibrium with CO gas according to C( s) + CO ( g) CO( g). Under these conditions particles of hematite Fe O are reduced to pure iron in the stack of the furnace above the hot coal: FeO( s) + CO ( g ) Fe( s) + CO( g ). A chimney located in the top of the blast furnace stack allows hot gases to escape. In the 19 th century blast furnaces with enormously tall stacks were constructed with the idea that if hot CO gas could be retained for a longer time in the blast furnace before escaping through the chimney more iron would be produced. Which Law of hermodynamics makes this a bad idea? Explain. Solution: he reaction reaches equilibrium and at the high temperatures within the blast furnace the equilibrium likely is far to the right because the process gives us lots of iron. hat said allowing more time to have hot gases exposed to the hematite yields nothing because once the system reaches equilibrium the second law says the entropy is maximized and further change is not going to happen. Building those big blast furnaces turned out to be a waste of time.

4 .) Consider two identical pieces of spring steel. he first piece of steel is dissolved completely in hydrochloric acid and as a result gives off heat which raises the temperature of the liquid. he second piece of spring steel is coil up and then dissolved in exactly the same amount of hydrochloric acid. he temperature of the liquid also increases. All things being equal other than the fact that one piece of steel is coiled up and the other is not will the temperature increases be the same or different? If they are different which piece of steel gives off more heat when dissolved in acid? Explain. Solution: both springs are of equal weight and composition so they evolve the same amount of heat when they react. But the coiled up spring has addition potential energy which is also converted to heat. So its temperature is higher..4) Is a man who transforms into work through muscular effort the energy of the food he eats and the air he breaths a more or less efficient machine than a reversible steam engine whose boiler temperature is equal to the man s normal body temperature (10K) and whose condenser temperature is the temperature of the surrounding air (00K)? Explain your answer. Solution: he man is more efficient because he is not a heat engine. he efficiency rule h c ε = only applies to heat engines and in this case a heat engine would have a h very low efficiency if it operated between body temperature of 10K and ambient 10K 98K temperature of 98K i.e. ε = = K. (0 points) erform two out of the four calculations given below..1 For water at =00K and =1 bar the thermal expansivity 1 V 4 1 β = = K and the molar volume Vm = 18.1mLmol. V S Calculate the change in entropy with unit change in pressure i.e. at =00K and =1 bar. Calculate the entropy change when the pressure on 1 mole of water increases from 1 bar to 10 bars. Assume β and V m are constants in this pressure range. Hint: Apply Euler s criterion for exactness to the expression for dg written in terms of d and d.

5 Solution: S V dg = Sd + Vd = = βv = ( K )( Lmol ) = LK mol ( )( )( ) = = mol m K mol Nm JK ( )((10 1) 10 ) ds = βvd S = βvd = βv d = LK mol Nm.) Before ethanol can be burned in your car s gas tank it has to be obtained by the chemical conversion of sugar in corn. Assume the sugar in corn is glucose. Assume ethanol is obtained by the fermentation of glucose CH 6 1O 6 via the reaction: CH 6 1O6( aq) CHOH 5 ( aq) + CO( g) 0 At =98K and 1 atm the standard enthalpy change is H98 = kj / mole. he standard absolute entropies are 0 0 S C H O aq = J / K; S C H OH aq = J / K; ( ( )) ( ) ( ) = 1.64 / ( ) ( ) S CO g J K Calculate the entropy change of the system the surroundings and the universe. Solution: ( ) ( ) ( ) S = S CO g + S C H OH aq S glucos e aq 5 ( )( ) ( )( ) = 1.64JK JK 69.45JK = 479.9JK H J Ssurr = = = 4497JK 98K 98K S = S + S = JK = 4977JK univ surr ( ).) Calculate the change in enthalpy when 100. g of liquid methanol initially at 1 bar and 98K undergo changes in temperature and pressure to.50 bars and 50K. he density of liquid methanol is g ml -1 and C m =81.1 JK -1 mol -1. Assume the thermal expansivity is about β=1.00x10-4 K -1. Assume the heat capacity density and thernal expansivity are constants within the tmeperature and pressure ranges given. Justify any approximations that you make. Solution: H H dh = d + d = Cd + ( V V β ) d Cd + Vd Where the approximation results because the thermal expansion of a liquid is small (see text page 57). hen

6 H = nc + nv m ( )( ) m 6 ( 1mL)( 10 m ml ) 100g ( )( ) ( )( ) ( ) = JK mol 50K 98K Nm + gmol 0.791g 1 mol / g =.1mol 417Jmol + 10Jmol = 11J.4) For temperatures between 7K and 000K and at a pressure of 1 bar the heat capacity of carbon monoxide CO is given by Cm 6 = JK mol K K he temperature of 1 mole of CO is increased from 00K to 1000K. Calculate U w and q. Assume CO obeys the ideal gas law between 00K and 1000K. Solution: 1000K 1000K H 6 = C ( ) d d J = + + K K 00K 00K = = ( 8.)( 700) + ( )( ) + ( )( ) = = 114 q = H = 114J ( ) ( ) ( ) ( ) 114 ( 1 )( 8.1 )( 700 ) U = H V = H nr = J mol JK mol K = 114J 5817J = 155J w= V =nr =5817J 4. ( points) Work one of the two multi-step calculations. 4.1) Five moles of an ideal monatomic gas expand adiabatically in two stages: Stage 1: he gas expands adiabatically and reversibly starting at an initial temperature of 98K and an initial pressure of 10 bars and continuing until a pressure of bars is achieved. Stage : Once the gas pressure drops to bars at the end of stage 1 the gas continues to expand adiabatically but now it does so irreversibly against a constant external pressure of 1 bar. he expansion continues until the gas reaches a final pressure of 1 bar. For each expansion stage calculate: the final temperature and final volume. Also for each stage calculate U H and S for the gas and the work done.

7 Solution: Stage 1: Need to get V 1 from ideal gas ( )( 1 nr 5mol 8.1JK mol 1 )( 98K ) 1 law: V1 = = = 0.01m = 1.L 6 10 Nm 1 γ γ hen from V = V V hen also 1/ γ /5 γ 5/ ( 10bar)( 1.L) V = = = 5.5L bar 5 ( 10 Nm )( 0.055m ) ( 5 )( 8.1 ) V = = = 184K nr mol JK mol R U = w = ncvm = n = ( 5mol )( 1.5)( 8.1JK mol )( 184K 98K ) =7105J 5R H = ncm = n = ( 5mol)(.5)( 8.1JK mol )( 184K 98K) =184J S = 0 Stage : ( ) = ( ) nc V V Vm ext nr nr V = ncv m ( ) = ext V ext nr nr + = + extv nr extv + V ext + nr extv = = = + 5 nr ext nr K Nm m 1 = 76 K K = 15 K JK = + ( ) nr ( )( 1 5mol 8.1JK mol 1 )( 15K ) V = = = 0.056m = 56L 5 10 Nm

8 R U = w = ncvm = n = ( 5mol)( 1.5)( 8.1JK mol )( 15K 184K ) = 054J 5R H = ncm = n = ( 5mol)(.5)( 8.1JK mol )( 15K 184K) =5090J 15 1 S = ncm ln nrln = ( 5mol)( 8.1JK mol ) ln ln JK = 6.4JK = ( )( ) 4.) Hydrolysis of the amino acid alanine HNCH ( CH) COOH to ammonia methane and oxygen proceeds according to : ( ) ( ) + 4 ( ) ( ) + ( ) + ( ) H NCH CH COOH s H O NH g CH g O g 4 he following thermodynamic data acquired at =1bar and =98K may be useful H kjmol S JK 1 mol 1 C JK mol f ( ) ( ) ( ) Alanine(s) HO ( ) NH ( g ) CH 4 ( g ) O ( g ) Calculate the enthalpy of reaction and entropy of reaction for the hydrolysis of alanine at 0K. Assume all heat capacities are constant between 98K and 0K. Also calculate the entropy change of the surrounding and the entropy change of the universe. Assume the temperature of the surroundings is stable at =0K. Based on this result will this reaction proceed as written under the conditions described? Confirm your answer by calculating G at =0K. At =98K ( ) ( 4 ) ( ) ( ) 4 ( ) ( )( ) ( )( ) ( ) ( )( ) H = H f NH g + H f CH g + H f O g H f alanine s H f H O = 46.19kJ kJ kJ 56.75kJ = 145kJ ( ) ( ) ( ) ( ) 4 ( ) S = S NH g + S CH g + S O g S alanine s S H O 4 ( )( ) ( )( ) ( ) ( )( ) = 19.51JK JK JK 19.0JK JK = 957.1JK ( ) ( ) ( ) ( ) 4 ( ) C = C NH g + C CH g + C O g C alanine s S H O 4 ( )( ) ( )( ) ( ) ( )( ) = 5.66JK + 5.7JK + 9.6JK 1.6JK 4 75.JK =9.5JK

9 At =0K 0 98 H = H + C = J 19.5JK 0K 98K = 149kJ ( ) 0 S = S + C ln = 957.1JK ( 19.5JK )( 0.10) = 97.96JK 98 0 H J Ssurr = = = 40JK 0K 0K 0 S = S + S = 97.96JK 40JK = 9JK universe surr he reaction cannot occur as written because the entropy of the universe decreases. Another way to put this G = H S = J 0K 97.96JK = 110kJ is: ( )( ) > 0. Same conclusion. G

Chemistry 452 July 23, Enter answers in a Blue Book Examination

Chemistry 452 July 23, Enter answers in a Blue Book Examination Chemistry 45 July 3, 014 Enter answers in a Blue Book Examination Midterm Useful Constants: 1 Newton=1 N= 1 kg m s 1 Joule=1J=1 N m=1 kg m /s 1 Pascal=1Pa=1N m 1atm=10135 Pa 1 bar=10 5 Pa 1L=0.001m 3 Universal