Homework Problem Set 5 Solutions. E e + H corr (a.u.) a.) Determine the bond dissociation enthalpy of ethane in kcal/mol (1 a.u. = kcal/mol).

Transcription

1 Chemistry Dr. Jean M. Standard Homework Problem Set 5 Solutions 1. Given below are the sum of electronic and thermal enthalpies, E e + H corr, from Hartree-Fock calculations using a 6-31G(d) basis set. Species E e + H corr (a.u.) C 2 H CH a.) Determine the bond dissociation enthalpy of ethane in kcal/mol (1 a.u. = kcal/mol). The bond dissociation reaction for ethane (for breaking the C-C bond) is given by C 2 H 6 2CH 3. To calculate the enthalpy of the bond dissociation reaction, the following equation is employed, ΔH rx = products ( ) E e + H corr ( E e + H corr ) + ΔνRT. reactants For this reaction, the enthalpy of reaction becomes ΔH rx = 2( E e + H corr,ch 3 ) ( E e + H corr,c 2 H 6 ) + RT. The final term in the expression equals RT because Δν = moles products moles reactants = 1. So, ΔνRT = 0.59 kcal/mol. Substituting the sums of electronic and thermal enthalpy terms, we get the bond dissociation enthalpy, ( ) ( E e + H corr,c 2 H 6 ) + RT ( ) ( kcal/mol) + ( 0.59 kcal/mol) ΔH rx = 2 E e + H corr,ch 3 = kcal/mol ΔH rx = 70.11kcal/mol. The values of E e + H corr used in the calculation of the bond dissociation enthalpy were converted to kcal/mol using the conversion factor 1 a.u. = kcal/mol. b.) The experimental bond dissociation enthalpy of ethane is 89 kcal/mol. Explain why the calculation from part (a) is so far off. The reaction as written has 9 electron pairs on the reactant side (two 1s core pairs from the C atoms, six pairs from C-H bonds, and one pair from the C-C bond) and only 8 electron pairs (two 1s core pairs from the C atoms and six pairs from C-H bonds) and 2 unpaired electrons on the product side. The calculation in part (a) performed at the Hartree-Fock level of theory neglects electron correlation. Since the number of electron pairs is different on the reactant and product sides of the reaction, the electron correlation effects on the reactant and product side will be different. Since the amount of electron correlation that is neglected is different on either side of the reaction, this throws off the determination of the reaction enthalpy.

2 2. Given below are the sum of electronic and thermal enthalpies, E e + H corr, from Hartree-Fock calculations using a 6-31+G(d) basis set. Determine the enthalpy of reaction in kcal/mol (1 a.u. = kcal/mol) for the hydrogenation of ethylene. 2 Species E e + H corr (a.u.) C 2 H H C 2 H The reaction for the hydrogenation of ethylene is given by C 2 H 4 + H 2 C 2 H 6. To calculate the enthalpy change of the reaction, the following equation is employed, ΔH rx = products ( ) E e + H corr ( E e + H corr ) + ΔνRT. reactants For this reaction, the enthalpy of reaction becomes ΔH rx = ( E e + H corr,c 2 H 6 ) ( E e + H corr,c 2 H 4 ) ( E e + H corr, H 2 ) RT. The final term in the expression equals RT because Δν = moles products moles reactants = 1. So, ΔνRT = 0.59 kcal/mol. Substituting the sums of electronic and thermal enthalpy terms, we get the hydrogenation enthalpy, ( ) ( E e + H corr,c 2 H 4 ) ( E e + H corr, H 2 ) RT. ( ) ( kcal/mol) ( kcal/mol) 0.59kcal/mol ΔH rx = E e + H corr,c 2 H 6 = kcal/mol ΔH rx = kcal/mol. The values of E e + H corr used in the calculation of the hydrogenation enthalpy were converted to kcal/mol using the conversion factor 1 a.u. = kcal/mol.

3 3. Indicate for which reactions below electron correlation effects are likely to be important in determining the reaction energies. 3 (a.) 2 CH 4 + F 2 2 CH 3 F + H 2 (b.) C 3 H 8 CH 3 + CH 3 CH 2 NO YES (c.) H + + Cl HCl NO (d.) CH O 2 CO H 2 O YES (e.) H + Cl HCl YES (f.) F + CH 4 CH 3 F + H NO Reactions that will be most significantly impacted by electron correlation effects are those in which the number of electron pairs is different on the reactant and product sides. Electron correlation effects may have an impact in determining more accurate reaction energies in reactions in which the number of electron pairs is the same in the reactants and products; however, the impacts are usually lower than for reactions in which the electron pair number changes. In this problem, the number of electron pairs is different on the reactant and product sides for reactions (b), (d), and (e); hence, electron correlation effects will be most important for those reactions. Note that reaction (b) involves the breaking of a C-C bond to produce two radicals, so the number of electron pairs is reduced by one. Reaction (d) has different numbers of electron pairs on the reactant and product sides because the O 2 molecule exists as a triplet state and therefore there are four unpaired electrons on the reactant side (two from each O 2 molecule) but there are no unpaired electrons on the product side. Reaction (e) is a bond formation reaction between hydrogen atom (a radical) and chlorine atom (a radical); the number of electron pairs increases by one when the H-Cl bond is formed.

4 4 4. Given below are results from Hartree-Fock calculations for the formaldehyde molecule. Basis Set Energy (a.u.) r CO (Å) r CH (Å) HCO ( ) STO-3G G G G(d) G(d,p) a.) Are these results in agreement with the Variation Principle? Explain. The results are in agreement with the Variation Principle. The number of basis functions increase in order in the basis sets STO-3G (minimal basis), 3-21G, 6-31G, 6-31G(d), and 6-311G(d,p), respectively. [Actually, 3-21G and 6-31G have the same number of basis functions. However, in the 6-31G basis set, the core and valence orbitals are more accurately represented using more gaussian primitives, so the energy is lower.] The energy decreases monotonically in the series in accord with the Variation Principle. Since the representation of the approximate wavefunction improves as basis functions are added, the approximate energy drops. We do not know the exact energy, but we can be sure that it is lower than any of the approximate values. On the other hand, there is no Variation Principle for geometries or other properties. Therefore, although we would expect increasing the basis set size to generally lead to an improvement in geometry, the convergence may not be monotonic. Thus, a calculation performed using a smaller basis set may by chance give a geometry that is closer to experiment than a calculation using a larger basis set. b.) For each basis set listed in the table above, give the basis set type (for example, minimal basis, double zeta basis, triple zeta basis, split valence basis, etc.). STO-3G - minimal basis set 3-21G - split valence double zeta basis set 6-31G - split valence double zeta basis set 6-31G(d) - split valence double zeta plus polarization basis set (polarization on non-hydrogen atoms only) 6-311G(d,p) - split valence triple zeta plus polarization basis set (polarization on all atoms)

5 5. Consider two gaussian type atomic orbitals, G 1 (r) and G 2 (r), given by 5 G 1 (r) = G 2 (r) = $ & % $ & % 2α 1 π 2α 2 π ' ) ( ' ) ( 3 / 4 3 / 4 e α 1(r r 1 ) 2, e α 2 (r r 2 ) 2, where r 1 and r 2 are the positions of the centers of the two gaussian type orbitals, and α 1 and α 2 are the orbital exponents, which define the size and extent of the orbital. a.) Suppose that the two gaussian type orbitals are centered at the origin (i.e., both centered on the same atom), so that r 1 = r 2 = 0. Note that the radial coordinate r is the distance from the origin, and in terms of cartesian coordinates is defined as r = α 1 =1.0 Å 2 x 2 + y 2 + z 2. Plot the shapes of the two orbitals if and α 2 = 2.0 Å 2. In making such a plot, it is the convention to consider a cut through the three-dimensional function in order that it is more easily plotted. To do this, for example, we might set y=z=0 and then graph the function with x ranging from 2.0 to 2.0 Å. Describe what effect the orbital exponent has on the shape of the orbital. A plot of the gaussian type orbitals is shown below. Notice that the function G 1 (r ), which has an orbital exponent of 1.0 Å 2 is wider than the function G 2 (r ), which has an orbital exponent of 2.0 Å 2. In addition, the smaller exponent leads to a function with a smaller maximum.

6 5. continued 6 b.) Now assume that the two gaussian type orbitals are centered on two different nuclei in a molecule, such that r 1 = 0.0 Å and r 2 = 3.0 Å. Plot the shapes of the two orbitals in this case (your range in r should now be from about 2.0 to 4.0 Å. Do the orbitals overlap significantly? What happens if r 2 = 1.5 Å instead? A plot of the gaussian type orbitals for the case when G 2 (r ) is centered at r 2 = 3.0 Å is given below. Notice that the overall shape of G 2 (r ) does not change; the entire function is shifted to the right so that the maximum occurs at r 2 = 3.0 Å. There is very little overlap between the two functions in this case. A plot of the gaussian type orbitals for the case when G 2 (r ) is centered at r 2 = 1.5 Å is given below. Again, the overall shape of G 2 (r ) does not change; the entire function is just shifted to the right. In this case, the two orbitals are close enough together for substantial overlap to occur.

7 6. Consider performing electronic structure calculations on the ethanol molecule. 7 a.) Describe a minimal basis set for ethanol, including the total number of basis functions. For ethanol, CH 3 CH 2 OH, a minimal basis set consists of the following basis functions: C -- 1s 2s 2p(3) O -- 1s 2s 2p(3) H -- 1s TOTAL NO. BASIS FUNCTIONS = 21 b.) Describe the 6-31G basis set for ethanol, including the total number of basis functions. For ethanol, CH 3 CH 2 OH, the 6-31G basis set consists of: C -- 1s 2s 2s' 2p(3) 2p'(3) O -- 1s 2s 2s' 2p(3) 2p'(3) H -- 1s 1s' TOTAL NO. BASIS FUNCTIONS = 39 c.) Describe the 6-311G(d,p) basis set for ethanol, including the total number of basis functions. For ethanol, CH 3 CH 2 OH, the 6-311G(d,p) basis set consists of: C -- 1s 2s 2s' 2s" 2p(3) 2p'(3) 2p"(3) d(5) O -- 1s 2s 2s' 2s" 2p(3) 2p'(3) 2p"(3) d(5) H -- 1s 1s' 1s" p(3) TOTAL NO. BASIS FUNCTIONS = 90 Note that here we are assuming that 5 d-type functions are used for polarization; that will be the case in any homework or exam questions. d.) Approximately how would the CPU time for single point energy calculations on ethanol scale using the basis sets in parts (a) - (c)? From lecture, the CPU time required for a Hartree-Fock single point energy calculation scales as approximately K 3.5, where K is the number of basis functions. For ethanol, the relative CPU times are shown in the table below. The times are reported relative to the smallest basis set calculation. Basis Set No. Basis Fxs Relative CPU Time minimal G G(d,p)

8 8 7. Consider performing electronic structure calculations on anthracene. a.) Describe a minimal basis set for anthracene, including the total number of basis functions. For anthracene, C 14 H 10, a minimal basis set consists of C -- 1s 2s 2p(3) H -- 1s TOTAL NO. BASIS FUNCTIONS = 80 b.) Describe the 3-21G basis set for anthracene, including the total number of basis functions. For anthracene, C 14 H 10, the 3-21G basis set consists of C -- 1s 2s 2s' 2p(3) 2p'(3) H -- 1s 1s' TOTAL NO. BASIS FUNCTIONS = 146 c.) Describe the 6-31+G(d) basis set for anthracene, including the total number of basis functions. For anthracene, C 14 H 10, the 6-31+G(d) basis set consists of C - 1s 2s 2s' 2p(3) 2p'(3) d(5) sp(4) H - 1s 1s' TOTAL NO. BASIS FUNCTIONS = 272 d.) Approximately how would the CPU time for single point energy calculations on anthracene scale using the basis sets in parts (a) - (c)? For anthracene, the relative CPU times are shown in the table below. The times are reported relative to the smallest basis set calculation. Basis Set No. Basis Fxs Relative CPU Time minimal G G(d)