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.5.62 Spring 2008 Lecture #34 Page 1 Transition State Theory. II. E‡ vs. Ea. Kinetic Isotope Effect. Want to get kTST into Arrhenius form kTST= !kThK"‡but K"‡= e#G‡RT$ %$ #RTln K‡= &G‡so kTST= !kThe#G‡/RT= !kTheS‡/Re#H‡/RTbecause G‡= H‡# TS‡NOW : H‡= E‡+ &nRTwhere ∆n = (# molecules in TS) – (molecularity of reaction)(molecularity: e.g., unimolecular, bimolecular, etc.) e.g. ∆n = 1 –2 = –1 So: kTST= !kTheS‡Re1e"E‡RTkTST# BTme"E‡RTwhere m = 1Surprisingly, theory predicts a temperature dependence to the pre-exponential factor. This T-dependence is difficult to observe experimentally unless the rate constant ismeasured over a wide temperature range (at least a factor of 5) Now: d lnkTSTdT=d ln!kheS‡/Re1"#$%&'+ lnT + e(E‡)*+,-.dT=1T+E‡RT2(TST) Contrast this to Arrhenius model: d ln kdT=EaRT2revised 4/24/08 11:34 AM5.62 Spring 2008 Lecture #34 Page 2 d ln kdT=d ln kTSTdTEaRT2=1T+E‡RT2! Ea= RT + E‡Again, the experimental Ea is larger than E‡ because Ea is a difference between the average energy of molecules in the pot and the average energy of molecules that react,while E‡ is a microscopic quantity, a threshold energy along the PES. Notice that Ea is not a barrier along PES. COMPARISON OF TRANSITION STATE THEORY WITH COLLISION THEORY Calculate kTST in the limit of the assumptions of collision theory (i.e. simplified TST): 1) collisions of hard spheres2) only translational degrees of freedom Treat H2 as an atom — a hard sphere of mass 2. [No rotation, no vibration] Treat H2F‡ as a diatomic molecule H2+ F ! H2F‡! HF + HWith these assumptions H2FH2F+‡k !kTh(qtrans‡/ N)(qtransH2/ N)(qtransF/ N)"#$%&'qrot‡e(E‡RTNote: no vibrational partition function for H2F‡ is included because the one vibrational mode for the pseudo-diatomic molecule transition state has become the reaction coordinate. Also, no rotational partition function for H2 is included because we are treating H2 as an atom. k !kTh(2"(mH2+ mF)kT)3/ 2h3N(2"mH2kT)3/ 2h3N(2"mFkT)3/ 2h3N#$%%%%&'((((8"2I‡kT)h2e*E‡RTThe reason there is no rotational or vibrational partition function for H2 is not that we are assuming the high-T limit, but rather that we are treating H2 as if it were an atom. revised 4/24/08 11:34 AM5.62 Spring 2008 Lecture #34 Page 3 Now: I‡= µdH2!F2where µ =mH2mFmH2+ mFH2FdH2–Fhard sphere collision diameterk ! N8kT"mH2+ mFmH2mF#$%%&'(()*++,-..1 / 2"dH2/F0e/E‡RTThis looks identical to the collision theory result, and collision theory is not based on thermodynamics. Calculate value for kTST in limit of collision theory assumptions (i.e. what fraction ofkCT=8kT!µ"#$%&'1/ 2!dAB2e(E0RTcollisions are effective because they have sufficient translational energy along the line of centers?): Compare to kTST ! = 1 "dH2#F2= 3 $ 10#19m2k % 3.4 $ 108e#E‡RTm3mol & skTST= 3.9 ! 107e"E‡RTm3mol # skTST is smaller because it reflects the more restrictive co-linear steric requirement. kCT is an upper bound because collision theory treats reactants as spheres with no favored direction of approach (but with an explicit requirement on the effective collision energy). revised 4/24/08 11:34 AM5.62 Spring 2008 Lecture #34 Page 4 TRANSITION STATE THEORY AND KINETIC ISOTOPE EFFECT Consider the H atom (or proton) transfer reaction HX + Y → X + HY (or HX+ + Y → X + HY+) where the original HX bond is broken and a new HY bond is formed HX = A, Y = B, (X + HY) = C A + B → C So the key question is how do we know whether breaking and making a bond to H occurs in the transition state region of the reaction coordinate? The size of the HD kinetic isotope effect tells us whether the H transfer occurs at or before/after the transition state. Kinetic isotope effect — reaction rates are slower if deuterium is substituted for hydrogen and a hydrogen bond is involved in the reaction. Why? Potential energy of interaction the same for HX and DX (Born-Oppenheimer appproximation). 1 2 hνHX 1 2 hνDX Because zero point energy of DX is smaller than for HX Since !"km#$%&'(1/ 2where k = force constantrevised 4/24/08 11:34 AM5.62 Spring 2008 Lecture #34 Page 5 !HX!DX"mDXmHX#$%&'(1/ 2because kHX= kDXIntramolecular potentials are the same. The “shape”of potential curve doesn’t change upon isotopicsubstitution. !DX= !HXmDXmHX"#$%&'1/ 2Since mDX> mHXSo12h!DX<12h!HXSo D0DX> D0HXlarger than that to break H—XKinetic Isotope Effect !kHkDdissociation energy to break D—X bond is Calculate this using transition state theory kHkD=kThqH‡/ N(qHA/ N)(qHB/ N)!"#$%&e'EH‡RTkThqD‡/ N(qDA/ N)(qDB/ N)!"#$%&e'ED‡RTkHkD=qH‡qD‡!"#$%&qDAqDBqHAqHB!"#$%&e'EH‡+ED‡( )RT!EH‡+ ED‡= !V0H!12h "i‡ Hi#+12h "iH+ V0Di#+12h "i‡ Di#!12h "iDi#But V0H= V0DkHkD=qH‡qD‡!"#$%&qDAqDBqHAqHB!"#$%&e'12hi()i‡ H'i()i‡ D'i()iH+i()iD!"#$%&RTMost of the isotope effect is in the difference between the zero point energies of thedeuterated vs. hydrogenated reactants. Can use isotope effect to determine whether ahydrogen bond was involved in the transition state. Standard diagnostic in kinetics! revised 4/24/08 11:34