Rep:Mod:AFS17
NH3 Molecule
Calculation Method = RB3LYP
Basis Set = 6-13G(d,p)
Final Energy = -56.55776873 au
RMS Gradient = 0.000000485 au
Point Group = C3V
N-H Bond length = 1.01798 Å
H-N-H Bond Angle = 105.741 degrees
Item Value Threshold Converged? Maximum Force 0.000004 0.000450 YES RMS Force 0.000004 0.000300 YES Maximum Displacement 0.000072 0.001800 YES RMS Displacement 0.000035 0.001200 YES Predicted change in Energy=-5.986287D-10 Optimization completed. -- Stationary point found.
Optimised NH Molecule |
Expect 6 modes from the 3N - 6 rule for non-linear systems, as N = 4 in NH3 as shown in Figure 1.
3 (4) - 6 = 6
Modes 2 and 3, 5 and 6 are degenerate as they have the same energy (same frequency).
Mode 4 is highly symmetric with exactly the same displacement in all directions and mode 1 is the umbrella mode. Modes 1, 2 and 3 are bending modes with 4, 5 and 6 are stretching modes.
Therefore, 4 bands would be expected in an experimental spectrum of gaseous ammonia as there are 4 distinct frequencies.
The charge on N = -1.125 and the charge on H = 0.375 (Figure 2)
This is as expected as nitrogen is more electronegative than hydrogen, so would expect nitrogen to be more negative which is shown in Figure 2. Nitrogen is more electron withdrawing than hydrogen, so nitrogen has a negative charge relative to hydrogen.
H2 Molecule
Calculation Method = RB3LYP
Basis Set = 6-31G(d,p)
Final Energy = -1.17853936 au
RMS Gradient = 0.00000017 au
Point Group = D*H
H-H Bond length = 0.74279 Å
H-H Bond Angle = 180 degrees (linear)
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164080D-13
Optimization completed.
-- Stationary point found.
Optimised HMolecule |
There is only one IR frequency for H2 as there is only one stretching mode at 4465.68 cm-1 as shown in Figure 3. As H2 is linear N = 2, 3N -5
3(2) - 5 = 1
So theres only 1 mode predicted and seen (Figure 3).
N2 Molecule
Calculation Method = RB3LYP
Basis Set = 6-31G(d,p)
Final Energy = -109.52412868 au
RMS Gradient = 0.02473091 au
Point Group = D*H
N-N Bond length = 1.092000 Å
N-N Bond Angle = 180 degrees (linear)
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401095D-13
Optimization completed.
-- Stationary point found.
Optimised N Molecule |
Only one vibration as expected (Figure 4) as its linear with N=2, for linear molecules the number of modes = 3N-5.
3(2) - 5 = 1
Haber-Bosch Process
N2 + 3H2 -> 2NH3
E(NH3) = -56.55776873 au
2*E(NH3) = -113.1155375 au
E(N2) = -109.52412868 au
E(H2) = -1.17853936 au
3*E(H2) = -3.53561808 au
ΔE=2*E(NH3)-[E(N2)+3*E(H2)] = -0.05579 au
ΔE = -0.05579 au x 2625.5 = -146.478 kJ mol-1 (3 d.p.)
N2 is the most stable molecule as it has a strong triple bond and the most negative energy, therefore it requires the most energy to break its bonds and react. The negative enthaply change of the reaction shows that the reaction is exothermic, meaning ammonia is more stable overall than the reactants.
The literature value of this reaction is - 92 kJ mol-1. [1], this is much less negative however as it's still negative it shows our calculations are correct in showing that it's an exothermic reaction. However, our value is probably too low as its derived from perfect optimised molecules using a Gaussview, whereas in real life reactions may not proceed perfectly.
Furthermore, in real life reactants must exceed the activation energy of a reaction, this can be lowered by adding a catalyst[1] which will increase the rate by providing an alternative reaction pathway with a lower activation energy.
Project Molecule CH4
Calculations
Calculation Method = RB3LYP
Basis Set = 6-13G(d,p)
Final Energy = -40.52275298 au
RMS Gradient = 0.00803796 au
Point Group = TD
C-H Bond length = 1.07000 Å
H-C-H Bond Angle = 109.471 degrees
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.256043D-08
Optimization completed.
-- Stationary point found.
Optimised CH4 Molecule
Optimised CH Molecule |
Vibrational Modes
CH4 is non-linear so uses the 3N - 6 rule to determine the number of vibrational modes, with N=5
Therefore the number of expected modes: 3(5) - 6 = 9
Which shown in Figure 5 with the 9 frequencies.
There will be 4 bands on the IR spectra, as there are 4 distinct IR frequencies (Figure 5).
Modes 1, 2 and 3; 4 and 5; 7, 8 and 9 are all degenerate.
Mode 6 is highly symmetric, 6, 7, 8 and 9 are all stretching modes with all the others being bending modes.
Charges on CH4
Charge on Carbon = -0.470 Charge on Hydrogen = 0.118
This is as expected (Figure 6) as carbon is slightly more electronegative than hydrogen, as its further to the right of the periodic table than hydrogen, this means that carbon is more electron withdrawing than hydrogen. Leading to a slight negative charge on carbon and a slight positive charge on each hydrogen.
Molecular Orbitals of CH4
Electronic configuration of Carbon 1s2 2s2 2p2
Electronic configuration of Hydrogen 1s1
When carbon and hydrogen combine to make CH4, atomic orbitals (AOs) combine to form molecular orbitals (MOs).
The first MO (Figure 7) is deep in energy at -10.16707 au as a result of 1s orbitals overlapping on both carbon and hydrogens. This MO is very deep in energy, deeper than those formed from valence shell AOs. There is little overlap and each AO is held tightly to respective nuclei and not involved much in bonding. It is the result of 1s orbitals overlapping, which are close to the nuclei and hence deep in energy.
The next MO at -0.69041 au (Figure 8) is the occupied bonding combination of empty 2s orbitals on hydrogen and filled 2s of carbon. This MO is not as deep in energy as the first MO, so its more involved in bonding and has less reliance on the respective nuclei.
This MO (Figure 9) is identical to 3, 4 and 5 with the same energy of -0.38831 au, they are degenerate. They are perpendicular to the bonds as a result of the 2p AOs of carbon mixing with the empty low lying 2p AOs of hydrogen. These are pi MOs, there are three because each atom has 3 2p orbitals that can combine perpendicular to the bond. Electron density is above and below bond due to AO mixing.
This is the Highest Occupied Molecular Orbital (HOMO).
Figure 10 shows the first antibonding MO of CH4, which is unoccupied and its the result of the combination of 2s AOs, it is so much higher in energy at 0.11824 au and its positive.
This is the LUMO - Lowest Unoccupied Molecular Orbital.
The difference in energy between the HOMO and the LUMO is 0.50655 au.
Figure 11 shows the unoccupied anti-bonding piMO from the overlap of 2p AOs, as there are three degenerate MOs at 0.17677 au, which are perpendicular to the bonds. These MOs are so high in energy that they're positive.
One of the reasons CH4 is so stable is due to the 3 2p AO based pi MOs perpendicular to each bond, which are all occupied across the molecule along with another 2 MOs forming 4 strong bonds with 4 hydrogen atoms. This means that the energy required to break a apart CH4 molecule is very high so it's very stable.
- ↑ 1.0 1.1 This is the literature value for the Haber Process, taken from https://nile.northampton.ac.uk/courses/1/WasteWelcome/content/_362372_1/09_equil.pdf