Exam2SW05 - C h e m 5, 9 Section Exam 2 Solutions Winter,...

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Chem 5, 9 Section Winter, 2005 Exam 2 Solutions February 22, Page 1 of 6 1. (4 points each) Here are a few quick problems to get you going. (a) Imagine formic acid, HCOOH( g ), decomposing into CO 2 ( g ) and H 2 ( g ) according to HCOOH( g ) CO 2 ( g ) + H 2 ( g ). The enthalpy change for this reaction is –30.5 kJ mol –1 , and the standard molar enthalpy of formation of CO 2 ( g ) is –393.5 kJ mol –1 . What is the standard molar enthalpy of formation, H f o , of HCOOH( g )? From the general relationship between enthalpies of formation and the reaction enthalpy change, we can write for this reaction H r o = H f o (CO 2 ( g )) + H f o (H 2 ( g )) – H f o (HCOOH( g )) –30.5 kJ mol –1 = –393.5 kJ mol –1 + 0 – H f o (HCOOH( g )) which we solve for H f o (HCOOH( g )) = –363 kJ mol –1 . (b) A saturated solution of lanthanum(III) iodate, La(IO 3 ) 3 , in pure water has an iodate concentration, [IO 3 ], equal to 2.0 × 10 –3 M. What is K sp for La(IO 3 ) 3 ? The net reaction here is La(IO 3 ) 3 ( s ) La 3+ ( aq ) + 3 IO 3 ( aq ). Thus, if [IO 3 ] = 2.0 × 10 –3 M, [La 3+ ] = [IO 3 ]/3 = 6.7 × 10 –4 M because there is no other source for either of these ions except the La(IO 3 ) 3 solid. We now know the equilibrium concentrations that determine the solubility product: K sp = [La 3+ ] [IO 3 ] 3 = 6.7 × 10 –4 2.0 × 10 –3 3 = 5.3 × 10 –12 (c) If 6.02 kJ is transferred as heat to 1 mol of ice, H 2 O( s ) at a constant 1 atm pressure, all the ice melts to H 2 O( l ). Thus, we can see that H for the process H 2 O( s ) H 2 O( l ) is 6.02 kJ because H = q P . Why is E for this process essentially the same number? In general, H and E are related through H = E + ( PV ). So if H E , we need to consider why ( PV ) 0. In this case, the pressure is constant, but that alone isn’t enough to ensure ( PV ) 0. Here, we write ( PV ) = PV (H 2 O( l )) – PV (H 2 O( s )) = P [ V (H 2 O( l )) – V (H 2 O( s ))] and note that the volume of one mole of ice is almost the same as the volume of one more of liquid water, which makes [ V (H 2 O( l )) – V (H 2 O( s ))] 0. 2. (4 + 6 + 6 + 4 points) A sample of a fixed amount of an ideal gas is subjected to a series of state changes as shown in the P , V diagram below. The gas starts in state at a temperature of 400 K, a pressure of 4 atm, and a volume of 1 L. The dashed lines in the diagram show how the external pressure varied as the gas was taken to a series of three intermediate equilibrium states, through , before ending at the final equilibrium state . The P , V coordinates of each state are indicated by a black dot. 5 4 3 2 1 0 012345 P /atm V /L
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Chem 5, 9 Section Winter, 2005 Exam 2 Solutions February 22, Page 2 of 6 (a) Calculate the highest and lowest temperatures experienced by the gas and indicate which equilibrium state (or states) have those temperatures. highest T : 2500 K at state(s) number 4 lowest T : 400 K at state(s) number 1 and 2 We note that state 1 has T 1 = 400 K, and in general, PV / T = nR = a constant. For state 1, P 1 V 1 = 4 L atm, and for state 2, P 2 V 2 also equals 4 L atm. Thus, T at state 2, T 2 , is also 400 K. For state 3, P 3 V 3 = 10 L atm, so T 3 must be ( P 3 V 3 / P 1 V 1 ) T
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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Exam2SW05 - C h e m 5, 9 Section Exam 2 Solutions Winter,...

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