UNIVERSITY OF KWAZULU-NATAL WESTVILLE CAMPUS DEGREE/DIPLOMA EXAMINATIONS: NOVEMBER 006 CHEMISTRY CHEM30W: PHYSICAL CHEMISTRY TIME: 180 MINUTES MARKS: 100 EXAMINER: PROF S.B. JONNALAGADDA ANSWER FIVE QUESTIONS. ALL QUESTIONS CARRY EQUAL MARKS. GENERAL DATA: Gas constant, R = 8.314 J K -1 mol -1 Avogadro constant, N A = 6.0 x 10 3 mol -1 Atmospheric pressure, 1 atm = 101.3 kpa = 760 mm Hg = 760 Torr Planck's constant, h = 6.6 x 10-34 Js = 6.6 x 10-34 kg m s -1 Boltzmann constant, k = 1.381 x 10-3 J K -1 Faraday constant, F = 9.6485 x 10 4 C mol -1 Permittivity = 8.854 x 10-1 J -1 C m -1 Speed of light, c =.998 x 10 8 m s -1 = 6.66 x 10-34 J kg m s -1 Rest mass of proton = 1.673 x 10-7 kg Atomic mass unit, u = 1.66 x 10-7 kg 35 Cl = 34.97 u; 1 H = 1.008 u; Br = 78.9 u; H =.0 u. Graph paper is needed. QUESTION 1 IS ON PAGE /...
QUESTION 1 (0 Marks, 36 Minutes) 1.a Derive an integrated equation of the second order reaction, A Products. (6) 1.b The carbon-14 decay rate of a sample obtained from a young tree is 0.480 disintegrations per second per gram of the sample. Another wood sample prepared from the object recovered at an archaeological excavation gives a decay rate of 0.30 disintegrations per second per gram of the sample. What is the age of the object? (Half life C-14 is 5.73 x 10 3 yr). (6) 1.c. Consider the second-order reaction., A + B C + D With A o = 0.50 mol dm -3 and B o = 0.40 mol dm -3, if it took 380s for 50% consumption of A, calculate the second-order rate coefficient. (5) 1.d Discuss the methods for determining the partial order with respect to each reactant in a chemical reaction. (3) QUESTION : 0 MARKS.a A certain pseudo first-order reaction is 0% complete in 15 minutes at 40 ο C and in 3 minutes at 60 ο C. Calculate (i) the rate constants at 40 ο C and 60 ο C. (3) (ii) the energy of activation for the reaction and (4) (ii) the pre-exponential factor, A. (3).b Derive the Kirchoff equation. (4).c It is often useful to be able to anticipate, without doing a detailed calculation, whether an increase in temperature will result in a raising or a lowering of a reaction enthalpy. The constant-pressure molar heat capacity of a gas of linear molecules is approximately 3.5 R, whereas that of a gas of nonlinear molecules is approximately 4R. Decide whether the standard enthalpies of the following reactions will increase or decrease with increasing temperature: (6) i) H (g) + O (g) H O (g) ii) N (g) + 3 H (g) NH 3 (g) iii) CH 4 (g) + O (g) CO (g) + H O (g) QUESTION 3: 0 MARKS 3.a Calculate the values per mol of w, q, U and S, when 140 g of nitrogen at 0 C and 1 bar is expanded reversibly and isothermally to a pressure of 0.15 bar. (Assume ideal gas behaviour). (6) 3.b 1 mole of Argon gas is compressed adiabatically and reversibly and from 600 dm 3 at 189.4 K to 4 dm 3. Calculate the final temperature. (C p = 9.14 J K -1 mol -1 ) (6)
3.c Given G = G + RT ln p, derive an equation relating the standard free energy change, G and equilibrium constant, K p. (4) 3.d Write concise notes on the constant volume bomb calorimeter. (4) QUESTION 4: 0 MARKS. 4.a Starting from the relation, ( G/ T) P = - S, derive the Gibbs-Helmholtz equation and show that ( ( G o /T)/ T) P = - H o /T (6) 4.b. The vant Hoff equation is expressed as ln (K /K 1 ) = Given the equilibrium constant of the reaction, H 1 R T1 1 T C 3 H 6 (g) C H 4 (g) + C 4 H 8 (g) at 390 K is 0.059 and the standard enthalpy change for the reaction in the temperature range 300 to 500 K is.73 kj mol -1, calculate the values of equilibrium constant, G θ and S θ at 410 K. (8) 4.c The molar heat capacity, C P for a compound is given as follows: C P (s) = 19.0 J K -1 mol -1 10 to 175 K C P (liq) = 6.34 J K -1 mol -1 175 to 500 K Molar entropy, S at 10 K = 10.1 J K -1 mol -1 and at the melting point 175 K. H o = 175.5 J mol -1. Calculate the third-law molar entropy of the compound at 500K. (6) QUESTION 5: 0 MARKS 5.a Write the two Kolraush laws, explaining what the symbols represent and outline their scope. (4) 5.b A conductivity cell filled with 0.1 molar KCl at 5 o C has measured resistance of 4.36 S -1. The conductivity, κ of 0.1 molar KCl is 1.1639 S cm -1. Filled with 0.05 M formic acid, the cell resistance is 45.9 S -1 at 5 o C. Calculate the molar conductance, Λ of 0.05 molar formic acid. (6) 5.c The molar conductance, Λ of aqueous acetic acid solutions were measures at 5 o C in a cell with following results: C/(mol dm -3 ) 0.00 0.0 0.08 0. Λ/(S cm mol -1 ) 3.95 11.4 5.69 3.59 The molar conductivity at infinite dilution was found to be 387.95 cm mol -1. Draw the appropriate graph to show that Oswald s dilution law is obeyed and obtain the molar conductivity at infinite dilution and the values for acid dissociation constant, K a and pk a. (10) 3
QUESTION 6: 0 MARKS 6.a At 5 o C, the molar conductivities of Li + and NO 3 - are 3.87 x 10-3 S m mol -1 and 7.40 x 10-8 S m mol -1 respectively, estimate their mobilities? (3) 6.b Write concise notes on the methods of determination of transport numbers.(3) 6.c. Describe how a concentration cell can be used for determination of solubility product of a sparingly soluble substabce, AgCl in aqueous solution. At 5 o C: AgCl(s) + e- Ag(s) + Cl-(aq) E o = 0. V Ag(s) Ag + (aq) + e - E o = - 0.80 V 6.d State the Beer Lambert law, explain the involving terms. () 6.e Sketch and label the setup for a spectrometer. (4) 6.f Explain how a UV-Visible spectrophometer meter can be used for (i) the determination of the stoichiometric ratio of a chemical reaction () (ii) monitoring the reaction kinetics of a chemical reaction () QUESTION 7: 0 MARKS 7.a The vibrational wave number of HBr is at 650 cm -1. (i) Does this give rise to an absorption in the IR spectrum? Justify your answer using the selection rule for vibrational transitions. (3) (ii) Calculate the reduced mass for HBr and its force constant of the bond. (6) 7.b Describe the various possible degrees of freedom for a tri-atomic linear molecule. (4) 7.c. Sketch the energy curves for the harmonic and anharmonic vibrations showing the quantized energy levels. Indicate are the differences and similarities in the two models? (5) 7.d. Outline the selection rules for the rotational transitions. () END OF PAPER 4
CHEM30: NOVEMBER EXAMIANTIONS 006: SOLUTIONS 1.a Reactants A and B the same. r = - (1/a) d[a]/dt = k[a] 1 - d[a]/dt = k A [A] Rearranging the differential equation, we get d[a]/[a] = - k A dt A t t A t da/ [A] = - k A dt (OR) da/[a] = - k A dt 3 A 0 o A 1 t 1 (1/[A] o ) - (1/[A] t ) = -k A t (OR) (1/[A] 1 ) - (1/[A] ) = - k A (t - t 1 ) 4 [A] t = A o /( 1 + k A t [A o ]) 5 Substituting x = [A] 0 - [A] t, eqn. 4 can be written in the form x/{a o (A o - x)} = - k A t 1.b The carbon-14 decay rate of a sample obtained from a young tree is 0.480 disintegrations per second per gram of the sample. Another wood sample prepared from the object recovered at an archaeological excavation gives a decay rate of 0.30 disintegrations per second per gram of the sample. What is the age of the object? (Half life C-14 is 5.73 x 10 3 yr). t ½ = 5.73 x 10 3 yr, then k =? N o = 0.480, N t = 0.140 and t =? k = 0.693/t 1/ = 0.693/(5.73 x 10 3 ) yr = 1.1 x 10-4 yr -1 t = (1/k) ln (N 0 /N t ) N 0 = 0.480 disintegration per second per gram of the sample. N t = 0.30 disintegration per second per gram of the sample. t = (1/(1.1 x 10-4 yr-1) ln (0.48/0.3) Age of the object = 3.35 x 10 3 yr 1.c. Consider the second-order reaction., A + B C + D With A o = 0.50 mol dm -3 and B o = 0.40 mol dm -3, if it took 380s for 50% consumption of A, calculate the rate coefficient. For second order reaction, A + B k C + D 1 b( a x) k = ln t( a b) a( b x) [A] o = 0.5 mol dm -3 ; [B] o = 0.40 mol dm -3 ; t = 380 sec. x = 0.5 mol dm -3 k = 1 380 x ln 0.10 0.40 x 0.5 = 7.57 x 10-3 mol -1 dm 3 s -1 0.50 x 0.15 1.d Trial/error method/ integrated equations Differential method Oswald s Isolation method Time for definite fraction method.a. At 40 o C, 15 minutes 0 % is completed or 80 % reactant left unreacted. 1 A 1 100 Therefore k 1 = ln o = ln =.479 x 10-4 s -1 t A t 15 x 60 80 Similarly at 60 o C, at 80% reactant felt at 3 min. 5
1 100 4 1 ln = 1.397 x 10 s k = 3 x 60 80 From the Arrhenius equation, k = A e -E/RT k Ea Or log = {(T T ) / T 1 T } k.303r 1 RT1T 8.314 x 313 x 333 E a = ln( k / k1) = ln (1.397/.479) T 0 = 4338 x 1.6096 = 69.74 kj mol -1 One can calculate Ea directly, without estimating the k 1 and k values by using the equation RT T E a = 1 ln (t 1 /t ) as t 1 is inversely proportional to k. T.b Derivation of Hº = Hº 1 + C P dt.c i) Change will be x 4R - 3 x 7/ R = -.5R, decrease with increasing temperature. ii) decrease, 8R - /R - 3 x 7/ R = -6R iii) increase, 8R + 7/R - 4R - x 7/ R 3.a Calculate the values per mol of w, q, U, H and S, when 140 g of nitrogen at 0 C and 1 bar is expanded reversibly and isothermally to a pressure of 0.15 bar. (Assume ideal gas behaviour). (RMM N = 8 g) (6) Temp. = 0 + 73 = 93 K; For an isothermal expansion of an ideal gas, U = H = 0. -w = q = nrt ln V /V 1 = nrt ln P 1 /P = (140/8)8.314 x 93 x ln (1/0.15) = 3106.9 J/mol S = 3106.9 J mol -1 /93 = 78.86 J K -1 mol -1. 3.b 1 mole of Argon gas at 189.4 K is compressed adiabatically and reversibly from 60 dm 3 to 4 dm 3. Calculate the final temperature.c p = 9.14 J K -1 mol -1 3.c T ν γ 1 C C C 1 p p ν = T1 γ = γ = ν 1 and C V = (9.14 8.314) = 0.81 Cν Cθ ( r 1) 60 8.314 T = (189.4 and = 4 (9.14 8.314) and γ -1 = 8.314/0.81 = 0.3995 R C ν Therefore T = 189.4 x (.5) 0.3995 = 189.4 x 1.44 = 73.15 K 3.c Let the reaction be a A + b B c C + d D a, b, c and d are the numbers of moles of respective reagents involved. Consider the gases A, B, C and D at constant temperature behave ideally. The free energy of a moles of A is given by Similarly, b G b = b G B + b RT ln P B c G c = c G C + c RT ln P C d G D = d G D + d RT ln P D a G A = a G A + a RT ln P A 6
Free energy change for the reaction, G = ΣG products - ΣG reactants = c G C + dg D - ag A - bg B Substituting the values of G A... in to the above eqn. we get, G = cg C + dg D - ag A - bg B + RT ln { (P C ) c (P D ) d /(P A ) a (P B ) b } = G + RT ln {(P C ) c ( P D ) d /( P A ) a (P B ) b } (P A, partial pressure of A) At equilibrium, G = 0, i.e. G = - RT ln (P C c P D d / P a a P B b ) = - RT ln K P (K P = Equilibrium Constant) 3.d Calarimetry Calorimeters are used to determine the H (constant pressure) or U (constant volume) (when gases are involved). Reactions should be fast and should go to completion. Example: Heat of combustion, halogenations, neutralization etc. CONSTANT VOLUME BOMB CALORIMETER, q = 0, W = 0 R + K (a) P + K at 5 C U = 0 at 5 C + T (c) (b) U el P + K at + 5 C U a = U b + U c = 0, U c = - U b = - C K+P T R - Reactants, P- Products and C K+P = Heat capacity of the Bomb calorimeter and contents (i.e. of bomb walls, water bath plus products). U b = U el = V I t (Voltage, Current, Time) i.e. H = U + P V For an ideal gas, H = U + ng RT/mol If n = 0, i.e when the stoichiometry of the reactant and product gas molecules is same, then H = U 4.b. The vant Hoff equation is expressed as ln (K /K 1 ) = Given the equilibrium constant of the reaction, H 1 R T1 1 T C 3 H 6 (g) C H 4 (g) + C 4 H 8 (g) at 390 K is 0.059 and the standard enthalpy change for the reaction in the temperature range 300 to 500 K is.73 kj mol -1, calculate the values of equilibrium constant, G θ and S θ at 410 K. H 1 1 The vant Hoff equation is expressed as ln (K /K 1 ) = R T1 T Substitution gives: ln K = -.83 + 730/8.314 {(1/390) (1/410)} = -.83 + 38.36 (0.00564 0.00439) = -.79 Therefore K = 0.0615 G θ = - RT ln K = - 8.134 x 410 x -.79 = + 9510 J mol -1 G θ = H θ - T S θ Therefore, 7
S θ Θ Θ H G.73kJmol 9.51kJmol = = T 400K = -16.9 J K -1 mol -1 1 1 x4.c T T S = Ι S= (C P /T) dt T 1 T 1 175 S50-150K= (19.0/T) dt =19.0(ln175 -ln 10) 10 = 19.0 (5.165 -.303) = 54.95 J K -1 mol -1 S fusion = H fusion/t = 175.5/175 = 9.86 J K -1 mol -1. 500 S150-300K= (6.34/T) dt = 6.34 (ln500 - ln 175) = 6.34 (6.15-5.165) 175 = 7.657 J K -1 mol -1 S at 300K = S 0-10K + S 10-175K + S fusion + S175-500K = 10.1 + 54.95 + 9.86 + 7.657 = 10.589 J K -1 mol -1 5.b κ = Cell constant/r Cell constant = κ x Resistance of KCl solution = 1.1639 S m -1 x 4.36 S -1 = 8.356 m -1. Conductivity of 0.05 M formic acid, κ = 8.356/45.9 S m -1 = 0.1153 S m -1 Therefore, Λ = κ = 0.1153 S m -1 (10-3 m 3 dm -3 ) = 4.61x 10-3 S m mol -1 C 0.05 mol dm -3 5.c c/mol dm-3 0.00 0.0 0.08 0. Lamda/S cm mol-1 3.95 11.4 5.69 3.59 LamdaC/Scm-1 6.59E-05 0.00084 0.000455 0.000718 1/Lamda 0.030349 0.087565674 0.175747 0.7855 Slope 38.7 Intercept = 0.008 1/intercept = 357.149 K = (1/slope) x (Intercept) (1/38.7)*0.008*0.008*10+3 =.05091E-05 6.a. u + = Λ + /ZF = 3.87 x 10-3 S m mol -1 / 96487 C mol -1 = 4.0 x10-8 m V -1 s -1 u- = Λ - /ZF = 7.40 x 10-8 S m mol -1 /96487 C mol -1 = 7.67 x 10-1 m V -1 s -1 (1 C S -1 = 1 A s S -1 = 1 V s) 6.b Brief description of the Hittorf method and moving boundary method. 7.a Reduced mass, µ 8
1 = 1 + 1 = 1 + 1. µ m1 m (1.008 x 1.66x 10-7 kg) (78.9 x 1.66 x 10-7 kg) = (0.59763 + 0.007633) x 10 7 kg -1 µ = 1 kg = 1.65 x 10-7 kg and π = 3.14 6.056 x 10 6 Wave number, ΰ = (1/ π c) (k/µ) Rearranging force constant, k = 4π c ΰ µ = 4x 3.14 x (3.0 x 10 10 cm s -1 ) (650 cm -1 ) x1.65 x 10-7 kg = 4.14 x 10 9 x10-7 kg s - = 41.4 N m -1 (N = kg m s - ) 9