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MM:WJB115

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NH3 Molecule

I have investigated the energetics of the haber-process by optimising NH3, N2 and H2 molecules. Then I extracted molecular information: charge, bond length and vibrational properties. When investigating a molecule of Cl2 the same molecular information was found. In addition to this I looked at the molecular orbitals in Cl2 and the relevant atomic orbitals.

Summary

Calculation method:RB3LYP

Basis set: 6-31G(d.p)

Final Energy= -56.55776873 a.u.

RMS Gradient= 0.00000485 a.u.

Point group= C3V

Item

         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

Dynamic Image

Optimised NH3


The optimisation file is liked to here

Display Vibrations

Questions

how many modes do you expect from the 3N-6 rule? 6

which modes are degenerate (ie have the same energy)? 2&3 5&6

which modes are "bending" vibrations and which are "bond stretch" vibrations? bending=1,2,3 bond stretch=4,5,6

which mode is highly symmetric? 4

one mode is known as the "umbrella" mode, which one is this? 1

how many bands would you expect to see in an experimental spectrum of gaseous ammonia? 2 since there are two distinct frequencies for the 6 modes therefore one would expect 2 bands. Two of the frequencies have a small value on the y-axis (relative intensity) which would make them hard to identify. These small intensities are due to a small change in dipole moment.

Charge distribution on ammonia

Charge on N=-1.125

Charge on Hydrogen=0.375

Nitrogen is more electronegative so will have the more negative charge

H-N-H bond angle

Optimized bond angle=105.741

N-H bond distance

Optimized bond distance=1.01798

N2 Molecule

Summary

Calculation method:RB3LYP

Basis set: 6-31G(d.p)

Final Energy= -109.524 a.u.

RMS Gradient= 0.00001902 a.u.

Point group= DinfH

Bond distance= 1.10550 Å 

Charge:0 because both the atoms have the same electronegativity

Item

Item               Value     Threshold  Converged?
 Maximum Force            0.000033     0.000450     YES
 RMS     Force            0.000033     0.000300     YES
 Maximum Displacement     0.000010     0.001800     YES
 RMS     Displacement     0.000015     0.001200     YES

Dynamic Image

Optimised N2

The optimisation file is liked to here

Display vibrations

The frequency was not negative

H2 molecule

Summary

Calculation Method RB3LYP

Basis Set: 6-31G(d,p)

Energy E(RB3LYP) -1.17853929 a.u.

RMS Gradient Norm 0.00012695 a.u.

Point Group D*H

Bond Distance=0.74279 Å 

Charge:0 both atoms have the same electronegativity

Item

Item               Value     Threshold  Converged?
 Maximum Force            0.000220     0.000450     YES
 RMS     Force            0.000220     0.000300     YES
 Maximum Displacement     0.000289     0.001800     YES
 RMS     Displacement     0.000408     0.001200     YES

Dynamic Image

Optimised H2

The optimisation file is liked to here

Display Vibrations

Energy of Haber Process Reaction

Energies in atomic units

E(NH3)= -56.55776873

2*E(NH3)= -113.1155375

E(N2)= -109.524

E(H2)= -1.17853929

3*E(H2)= -3.53561808

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579074

In kJ/mol ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -146.4785879

The product is more stable

Cl2 Molecule

Summary

Calculation Type FREQ

Calculation Method RB3LYP

Basis Set 6-31G(d,p)

Charge 0

Spin Singlet

E(RB3LYP) -920.34987886

RMS Gradient Norm 0.00002510

Point Group D*H

Charge of Cl2 molecule

The two Cl atoms have the same electronegativity so the molecule is neutrally charged

Item

Item               Value     Threshold  Converged?
 Maximum Force            0.000043     0.000450     YES
 RMS     Force            0.000043     0.000300     YES
 Maximum Displacement     0.000121     0.001800     YES
 RMS     Displacement     0.000172     0.001200     YES

Dynamic Image

Optimised Cl2

Link to optimised file here

Display vibrations

The infrared=0 because there is no change in dipole across the molecule

Molecular Orbitals of Cl2

2px π bonding MO
2px orbitals result in π* anti-bonding MO
2pz σ bonding MO, caused by the overlap of 2p orbitals in z orientation
2p orbitals result in σ* Anti-bonding MO
2s σ* Anti-bonding MO

LUMO

The LUMO is the result of two out of phase 3pz orbitals creating a σ* anti-bonding MO

HOMO

The HOMO is a π* bonding MO due to the 3px and 3py orbitals, which give two degenerate molecular orbitals.