Chemistry 5350 Advanced Physical Chemistry Fall Semester 2013 Mierm Examination: Thermodynamics and Kinetics Name: October 10, 2013 Constants and Conversion Factors Gas Constants: 8.314 J mol 1 K 1 ; 8.314 Pa m 3 mol 1 K 1 ; 62.36 L Torr mol 1 K 1 8.206 10 2 L atm mol 1 K 1 ; 8.314 10 2 L bar mol 1 K 1 Thermodynamic Equations Ideal Gas Law: PV = nrt van der Waals Equation: P = RT a V m b Vm 2 Wor: w = PdV First Law of Thermodynamics: U = q +w U as a function of V and constant T: U ) = T ) P P V T T V Enthalpy: H = U + PV) Entropy: S = qrev T Enthalpy of Reaction: Hreaction = ν Hf ν Hf Standard Gibbs Energy of Reaction: G reaction = ν G f ν G f Themochemical Data Substance Hf J mol 1 ) G f J mol 1 ) SO 2 g) -298.6 H 2 Ol) -285.8 H 2 Sg) -20.6 CO 2 g) -393.5 NO 2 g) 51.31 N 2 O 4 g) 97.89 Chemical Equilibrium At equilibrium: r G = RT lnk a A g)+b B g) c C g)+d D g) K = yc C yd D P/P ) [c+d) a+b)] ya a yb B Rates of Chemical Reactions dn Rate of Conversion: J dξ = ν J Rate of Reaction: v = 1 d[conc J ] ν J First Order Reaction: ln [A] 0 = t [A] Second Order Reaction: t = 1 1 [A] [A] 0 Arrhenius Equation: ln 2 1 = Ea R ) T 2 T 1 T 2 T 1
1. a) A system consists of a mole of ideal gas that undergoes the following change in state: Xg, 298 K, 10 bar) Xg, 298 K, 1 bar) What is the S m if the expansion is reversible? U = q +w = 0; q = w and w = RT ln 10 bar 1 bar = 8.314 J mol 1 K 1 298 K ln10 = 5705 J mol 1 = q S m = 5705 J mol 1 298 K = 19.14 J K 1 mol 1 What is the value of S m, if the gas expands into a larger container, so that the final pressure is 1 bar? This process becomes irreversible, but the value of S m is the same whether the expansion is reversible or irreversible. b) The same change in state taes place, Xg, 298 K, 10 bar) Xg, 298 K, 1 bar) but we now consider the gas plus a heat resevoir at 298 K to be our system. What is the S m if the expansion is reversible? If the expansion is reversible, then S m = 0, as it must be for any process in an isolated system. What is the value of S m, if the gas expands into a larger container, so that the final pressure is 1 bar? If the expansion is irreversible, then S m = 19.14 J K 1 mol 1, a positive value which it must be for any irreversible process in an isolated system.
2. a. From the data in the following table at 298 K, Reaction H reactionj mol 1 ) Fes)+2H 2 Sg) FeS 2 s)+2h 2 g) -137.0 H 2 Sg)+ 3 2 O 2g) H 2 Ol)+SO 2 g) -562.0 as well as data from the table on the front page, calculate the standard enthalpy of formation of H 2 Sg) and FeS 2 s). Reaction H 2 Ol)+SO 2 g) H 2 Sg)+ 3 O 2 2g) HreactionJ mol 1 ) 562.0 Ss)+O 2 g) SO 2 g) -298.6 H 2 g)+ 1 O 2 2g) H 2 Ol) -285.8 H 2 g)+ss) H 2 Sg) Hf = 20.6 Reaction HreactionJ mol 1 ) Fes)+2H 2 Sg) FeS 2 s)+2h 2 g) -137.0 2H 2 g)+2ss) 2H 2 Sg) 2 x -20.6 Fes)+2Ss) 2H 2 Sg) Hf = 178.2 b. Using the enthalpy of combustion of benzene liquid H combustion C 6 H 6 l)) = 3268 J mol 1 ), as well as data from the table on the front page, determine the standard enthalpy of formation for benzene. 3H 2 Ol)+6CO 2 g) 15O 2 2g)+C 6 H 6 l) H combustion C 6 H 6,l) 6Cs)+6O 2 g) 6CO 2 g) 6 H f CO 2,g) 3H 2 g)+ 3 O 2 2g) 3H 2 Ol) 3 H f H 2 O,l) 3H 2 g)+6cs) C 6 H 6 l) H combustion C 6 H 6,l)+6 H f CO 2,g)+3 H f H 2 O,l) H f C 6 H 6,l) = 3268 J mol 1 6 393.5 J mol 1 3 285.8 J mol 1 = 49.6 J mol 1
3. Suppose that a system initially contains 0.500 mol of N 2 O 4 g), and the equilibrium is attained at 25 and 2.00 bar. a. Obtain r G for the reaction above. G reaction = N 2 O 4 g) 2NO 2 g) ν G f ν G f G reaction = 2 51.31) J mol 1 97.89 J mol 1 = 4.73 J mol 1 b. Calculate K P G reaction = RT lnk P 4730 J mol 1 = 8.314 J mol 1 K 1 298 K lnk P ; K P = 0.148 c. Find the equilibrium composition. The quadratic formula x = b± b 2 4ac 2a form ax 2 +bx+c = 0. can be used for solutions of quadratic equations of the N 2 O 4 NO 2 Initial Amount 0.50 0.50 Equilibrium Amount 0.50 ξ 2ξ Equilibrium Mole Fraction 0.50 ξ a A g)+b B g) c C g)+d D g) 2ξ K = yc C yd D y a A yb B P/P ) [c+d) a+b)] N 2 O 4 g) 2NO 2 g) K P = y2 NO 2 y N2 O 4 ) 2 2ξ 0.50 ξ ) 1 2 = 0.148 1 ) 2 = 0.148; 8ξ 2 = 0.148; ξ = 0.067 0.25 ξ2 n NO2 = 0.134; n N2 O 4 = 0.433
4. a. A certain chemical reaction is first order, and 540 s after initiation of the reaction, 32.5% of the reactant remains. What is the rate constant for this reaction? ln [A] 0 [A] = t ln [A] 0 0.325 [A] 0 = 540 s; = 2.08 x 10 3 /s At what time, after initiation of the reaction, will 90% of the reactant be consumed? ln [A] 0 0.10 [A] 0 = 2.08 x 10 3 /s t; = 1100 s b. Derive the rate law for decomposition of 2N 2 O 5 g), according to the following reaction, and on the basis of the following mechanism: 2N 2 O 5 g) 4NO 2 g)+o 2 g) Rate of Reaction Intermediates d[n 2 O 5 ] d[no] N 2 O a 5 NO2 + NO 3 a NO 2 + NO 3 N 2 O 5 NO 2 + NO b 3 NO2 +O 2 + NO NO + N 2 O c 5 NO2 +NO 2 + NO 2 = a [N 2 O 5 ]+ a[no 2 ][NO 3 ] c [NO][N 2 O 5 ] = b [NO 2 ][NO 3 ] c [NO][N 2 O 5 ] = 0 d[no 3 ] = a [N 2 O 5 ] a[no 2 ][NO 3 ] b [NO 2 ][NO 3 ] = 0 Combine above equations to give the Rate Law for Decomposition ) d[n 2 O 5 ] a b = [N a + 2 O 5 ] b