Chemistry 5350 Advanced Physical Chemistry Fall Semester 2013

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1 Chemistry 5350 Advanced Physical Chemistry Fall Semester 2013 Name: Quiz 2: Chapters 3, 4, and 5 September 26, 2013 Constants and Conversion Factors Gas Constants: J mol 1 K Pa m 3 mol 1 K L orr mol 1 K L atm mol 1 K L bar mol 1 K 1 cubic meters to liters: 1 m 3 = 10 3 L Equations Ideal Gas Law: P = nr van der Waals Equation: P = R a m b m 2 Work: w = Pd First Law of hermodynamics: U ( = q +w U as a function of and constant : ) = ( ) P P Enthalpy: H = U + (P) Enthalpy of Reaction: Hreaction = ν products Hf Entropy: products S = qrev hemochemical Data Substance Hf (kj mol 1 ) CH 4 (g) C 2 H 6 (g) C 2 H 4 (g) 52.4 C(g) H(g) reactants ν reactants H f

2 1. For the internal energy U = U(,), (a) Derive the differential du as a function of and du = d + d (b) Evaluate ( ) for an ideal gas, and modify the equation for du accordingly, showing that, for an ideal gas, U is a function of only. ( ) = 0 for an ideal gas, and ( ) = C. hus, U = C ( 2 1 ) (c) Evaluate ( ) for the van der Waals equation of state. P = R m b a = ( ) P 2 m P = a 2 m (d) Derive an expression for the change in internal energy in compressing a van der Waals gas from an initial molar volume, m,i to a final molar volume, m,f at constant temperature. U,m = m,f m,i m,f U,m = d m m,i [ ] m,f [ a 1 1 d m 2 m = a = a 1 ] m m,i m,i m,f (e) By comparing parts (b) and (d), explain whether or not U,m will be zero for a gas and why? At constant temperature, U = 0 for an ideal gas because U,m = C 0 ( 2 1 ) but that will not be true for the van der Waals gas because U,m is a function of volume.

3 2. Given the thermochemical data in the table on the cover page, calculate the bond enthalpy and bond energy for the following cases. Use the results from part (a) to solve parts (b) and (c). a. he C-H bond in CH 4. CH 4 (g) C(g)+4H(g) H reaction = 4 H f(h, g)+ H f(c, g) H f(ch 4, g) H reaction = kj mol kj mol kj mol 1 = 1663 kj mol 1 he average C-H bond enthalpy in CH 4 = 1663 kj mol 1 4 = 416 kj mol 1 U = H nr = 1663 kj mol 1 4 (8.314 J mol 1 K 1 ) ( K) = 1654 kj mol 1 he average C-H bond energy in CH 4 = b. he C-C single bond in C 2 H 6. C 2 H 6 (g) 2C(g)+6H(g) 1654 kj mol 1 4 = 413 kj mol 1 H reaction = 6 H f(h, g)+2 H f(c, g) H f(c 2 H 6, g) H reaction = kj mol kj mol kj mol 1 = 2825 kj mol 1 H reaction = 6 x C-H bond enthalpy + C-C bond enthalpy C-C bond enthalpy = 2825 kj mol 1-6 x 416 kj mol 1 = 329 kj mol 1 U = H nr = 2825 kj mol 1 7 (8.314 J mol 1 K 1 ) ( K) = 2808 kj mol 1 U = 6 x C-H bond energy + C-C bond energy C-C bond energy = 2808 kj mol 1-6 x 413 kj mol 1 = 329 kj mol 1 c. he C=C double bond in C 2 H 4. C 2 H 4 (g) 2C(g)+4H(g) H reaction = 4 H f(h, g)+2 H f(c, g) H f(c 2 H 4, g) H reaction = kj mol kj mol kj mol 1 = 2253 kj mol 1 H reaction = 4 C-H bond enthalpy+c=c bond enthalpy C=C bond enthalpy = 2253 kj mol kj mol 1 = 589 kj mol 1 U = H nr = 2254 kj mol 1 5 (8.314 J mol 1 K 1 ) ( K) = 2240 kj mol 1 U = 4 C-H bond enthalpy+c=c bond enthalpy C=C bond energy = 2240 kj mol kj mol 1 = 588 kj mol 1

4 3. (a) One mole of an ideal gas at is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because it is very large, the temperature of the thermal water bath resevoir in the surroundings remains essentially constant at during the process. Calculate S, S surroundings, and S total. Because this is an isothermal and reversible process: U = 0, and q reversible = w q reversible = w = nr f i d = nr ln f i q reversible = 1.00 mol J mol 1 K 1 ln 10.0 L 25.0 L = J he entropy change of the system is given by: S = dqreversible = q reversible = J = J K 1 he entropy change of the surroundings is given by: S surroundings = q surroundings = q system = J = 7.62 J K 1 he total change in entropy is given by S total = S + S surroundings = J K J K 1 = 0

5 (b) One mole of an ideal gas at is isothermally compressed at constant external pressure equal to the final pressure in Part (a). At the end of the process, P = P external, and because of this, the process is irreversibe at all but the final state. he initial and final volumes are, respectively, 25.0 L and 10.0 L, and the temperature of the surroundings is. Calculate S, S surroundings, and S total. Determine the direction of spontaneous change, i.e. either compression or expansion, and provide a reason for your answer. First calculate the external pressure P external = nr = 1 mol J mol 1 K 1 10 L 1 m L = Pa and the initial pressure of the system P initial = nr = 1 mol J mol 1 K 1 25 L 1 m L = Pa Because P external > P initial, the direction of spontaneous change will be the compression of the gas to smaller volume. Because this is an isothermal and irreversible process: U = 0, and q = w and q = w = P external ( f i ) = Pa ( m m 3 ) = J he entropy change of the system must be calculated by a reversible path and has the same value as obtained in part (a): S = dqreversible = q reversible = J = J K 1 he entropy change of the surroundings is given by S surroundings = q surroundings = q = J = J K 1 S total = S + S surroundings = J K J K 1 = 4.85 J K 1 It can be seen that S < 0 and S surroundings > 0 confirming that the entropy of an isolated system always increases.

Chemistry 5350 Advanced Physical Chemistry Fall Semester 2013

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