MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.  2 5.62 Lecture #17: Chemical Equilibrium. II. Examples Readings: Hill, pp. 182-187Maczek, pp. 83-92Metiu, pp. 191-196 Dissociation of a Diatomic Molecule AB →← A + B p°)(pB p°)Kp = (q*A / N)(q*B / N) e+ΔD00RT [unitless] =(pA (q*AB / N) (pAB p°) Kp =(qtrans,B N)(qtrans,A N) g0,Bg0,A qrot,Bqrot,A q*vib,B q*vib,A e+ΔD00RT (qtrans,AB N) g0,AB qrot,AB q*vib,AB Kp = (2πmB)3/2 (kT)5/2 (2πmA)3/2 (kT)5/2 h3ph3p h3p (2πmAB )3/2 (kT)5/2 g0,Bg0,A ⋅1⋅1⋅σθrot,AB(g0,AB T ⋅1⋅1 1− e−θvib,AB T )e+ΔD00 RT × p is in units of bar because the standard state p° = 1 bar = 105 pascal. But all terms in statistical mechanical expression for Kp are evaluated in S. I. units. Be careful! (2πµ)3/2 (kT)5/2 g0,Bg0,A σθrot (1− e−θvib T )e+ΔD00 RT Kp = h3p g0,AB T where µ= mAmB = mAmB reduced mass mA + mB mAB kg/molecule for SI 2 I2 ← 2I Kp =→ pI [p’s in bar]pI2 mI = 0.1269 kg mol–1 µ I2 = 0.06345 kg mol−1 g0, I = 4 g0, I2 = 1 σI2 = 2 ωe = 214.5 cm–1 θvib = 308.6K Be = 0.03737 cm–1 θrot = 0.05377K5.62 Spring 2008 Lecture #17, Page 2 D00, I2 = 12440cm–1 = 17889K (determined by laser spectroscopy!) [Be careful about units here!] ΔD00 = ∑ p(D00 ) −∑ r (D00 )= 0 − (17889K) = −17889K p r σ θrot Kp =(2π0.0634 6 ·1023 )3/ 2 (kT)5 / 2 4·4 2·0.0537 (1− e−308.6 T )e−17889 T h3·105 1 T 1 bar = 105 pascal. 1 pascal = 1N/m2. 1N = 1kg m s–2 )(16) 0.1074 (1− e−308.6/T ) e−17889/T Kp =(13.115 T5/2 T Kp = 22.537 T3/2 (1 − e−308.6/T ) e−17889/T T[K] Kp(calc) Kp(expt) % error 1274 0.1761 0.170±0.001 3% 1173 4.9999⋅10–2 (4.68±0.03) ⋅10–2 2.6% 1073 1.14⋅10–2 1.10⋅10–2 3% 973 1.93⋅10–3 1.82⋅10–3 5.4% 872 2.13⋅10–4 (1.84±0.17) ⋅10–4 –14% probably more accurateM. J. Perlman and G. K. than expt because Kp is Rollefson, J. Chem. so small at low T, thatPhys. 9, 362 (1941)partial pressue ofdissociated I atoms is too small to measure accurately revised 1/10/08 12:55 PM5.62 Spring 2008 Lecture #17, Page 3 3 Isotope Exchange Reaction →H2 + D2 ← 2HD Kp = qHD * N( )2 qH2 * N( ) qD2 * N( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ e+ΔD00 RT Kp = qtrans,HD / N ( )2 qtrans,H2 N( ) qtrans,D2( N) g0,HD 2 g0,H2 g0,D2 qvib,HD *2 qvib,H2 * qvib,D2 * qrot,HD 2 qrot,H2 qrot,D2 e+ΔD00 RT Kp = 2πmHD ( )3 (kT)5 h6 p2 h3p 2πmH2( )3 /2 (kT)5/2 h3p 2πmD2( )3 /2 (kT)5/2 g0,HD 2 g0,H2 g0,D2 D2 σD2(1− e −θvib,H2 T )(1− e−θvib,D2 T ) ⎛ kT ⎞ 2 ⎛ hcBeH2 σH2 ⎞ ⎛ hcBe (1− e−θvib,HD T )2 ⎝⎜ hcBeHDσHD ⎠⎟ ⎜⎝kT ⎠⎟ ⎝⎜⎜ kT mH2 = 2 amu σH2 = 2 g0(H2) = 1 D00(H2) = 36,100 cm–1 mHD = 3 amu σHD = 1 g0(HD) = 1 D00(HD) = 36,394 cm–1 mD2 = 4 amu σD2 = 2 g0(D2) = 1 D00(D2) = 36,742 cm–1 H2ωe(H2) = 4401 cm–1 θvib = 6337 K Be(H2) = 60.8 cm–1 ωe(HD) = 3813 cm–1 HD = 5419 K Be(HD) = 45.7 cm–1θvib (D2) = 3116 cm–1 D2Be(D2) = 30.4 cm–1ωeθvib = 4487 K ΔD00 = ∑ ( −∑r D0 )p D0 ) ( p r 0 ([ ] = −54 cm−1 = −78K ΔD0 = 2 36, 394) − 36,100 + 36, 742 ⎞ ⎟ e+ΔD00 RT ⎟⎠ revised 1/10/08 12:55 PM5.62 Spring 2008 Lecture #17, Page 4 What is Kp at T = 298K? mHDg0,HD2 H2 T D2 T3 (1 − e−θvib )(1 − e−θvib ) BHe 2 BeD2 σH2 σD2 e+ΔD00RT HD T )2 2HD Kp = mH3 22mD3 2 2 g0,H2g0,D2 (1 − e−θvib(Be )σHD Kp = 0.003 / 6 ⋅1023( )3 0.002 6.1023( )3/2 0.004 6.1023( )3/2 12 (1 − e−4487/298 ) 1⋅1 (1 − e−5419/298 )2 × (1 − e−6337/298 )(60.8)(30.4) 2 ⋅ 2e−78/298 (45.7)21 = 3.27 T[K] Kp[CALC] Kp[EXP] 298 3.27 3.28 383 3.47 3.50 741 3.82 3.75 DEPENDENCE OF Kp ON T Kp Kp I2→←2I H2 + D2→←2HD Qualitative difference in behaviors: →→Ι2 ← 2ΙΗ2 + D2 ← 2HD 0Kp ≈ q2trans−Ie+ΔD0RT q2trans,HD qrot, HD 2 qvib,HD 2 e+ΔD 00 /RT qtrans−I2qrotqvib Kp ≈ qtrans,H2qtrans,D2qrot, H2qvib,H2qrot, D2qvib,D2 revised 1/10/08 12:55 PM5.62 Spring 2008 Lecture #17, Page 5 qtrans I ≈ qtransI2(ignore factor of 2 in mass) qtrans,HD  qtrans,H2  qtrans,D2 32 3/2 ⎤ ⎥⎦≈ 1⎡ ⎢⎣ ⎧ ⎪⎪ ⎪ ⎪⎪⎨ignoring mass 2·4 qrot,H2  qrot,D2  qrot,HD ⎡⎢⎢⎣ ⎪ ⎪ ⎪ ⎪⎪⎩functions )2µHD 2 µH2 µD2 ⎤(2 / 3(1 / 2) 1( ) ≈ 1⎥⎥⎦ = qvib,H2  qvib,D2  qvib,HD Kp ∝ qtrans e+∆ D00 RT Kp ∝ σ2e+ ∆ D00RT qrot qvib qtrans  1030, qrot  103, qvib  1,"∆ D00  −18, 000K ∆ D00  −78K Kp ∝ 1027e−18,000 /T Kp ∝ 4e−78/T * large T dependence and large Kp “Small” values of Kp: because of 1027 factor — gain in translational entropy due to* no gain in entropy except for change in number of moles symmetry # (factor of 4) * results in shift of equilibrium* Kp → 4 at modest T because of small toward separated atoms at high T difference in zero point energy. * actually qtrans ∝ T5/2 , 1 1∝ T, the qrot pre-exponential factor is T- dependent * as T increases, both pre- exponential and exponential factors increase and shift equilibrium toward dissociation. Recall from 5.60: ∆ G°(T) = ∆ H°(T) − T∆ S°(T) = −RTln K(T) K(T) = e∆ S°( T)/R  e− ∆ H°(T)/RT  pre-exponentialfactor This gives us an intuitive understanding of the T-dependence of equilibrium constants.Mostly, ∆S°(T) is determined by change in number of moles (strong T-dependence),secondarily in changes in floppiness (approximately T-independent). Mostly ∆H°(T) isdetermined by bond energies (or differences in dissociation energies), but if you want tocompute K(T) from microscopic quantities, use K(T) = e− ∆ G° RT and use statistical mechanics to calculate ∆G°(T) directly, not both ∆H°(T) and ∆S°(T) separately. revised 1/10/08 12:55 PM5.62 Spring 2008 Lecture #17, Page 6 In using statistical mechanics to compute equilibrium constants, it iscomputationally most compact and intuitively most instructive to assemble the relevantfactors in qC * N( )c qD * N( )d qA * N( )a qB * N( )b by assembling all of the relevant information factored according to degee of freedom (translation)(electronic)(vibration)(rotation) Translation Key factors are * does the number of moles change* the only species-specific quantity is mass Electronic Key factor is degeneracy of ground state For CO X1∑+  C(3P) + O(3P) g: 1 3 × 3 3 × 3 The electronic factor is usually