Question

The vibrational peak and the first overtone of CO appear at 2143.31 and 4260.04 cm⁻¹. Assuming a Morse potential, calculate the dissociation energy of the molecule in kJ/mol. Please show all work.

Answer 1

For Morse potential,the energy eigen values are given by:

Gv=(v+1/2)e-(v+1/2)^2 e*Xe

v=vibrational quantum numbers=0,1,2..........Vmax

Xe=anharmonicity constant (dimensionless)=e/4De

De=Dissociation energy from the bottom of the potential well

True Dissociation energy=Do=De-G(0)   as the minimum energy is G(0) for a molecule called the zero point energy

Specific selection rule for tranition :v=1,2............

So,for fundamental peak ,v=01 ,v(cm-1)=e-2eXe

first overtone,v=02,v(cm-1)=2e-6eXe

given,The vibrational peak and the first overtone of CO appear at 2143.31 and 4260.04 cm⁻¹.

So,v(cm-1)=e-2eXe=2143.31 cm⁻¹.

first overtone,v=02,v(cm-1)=2e-6eXe=4260.04 cm⁻¹.

or,e-2eXe=2143.31 cm⁻¹..................(1)

2e-6eXe=4260.04 cm⁻¹.....................(2)

eqn (2)-eqn(1),

2116.73 cm-1=-4eXe

So,eXe=2116.73 cm-1/-4=(-529.182 cm-1)

eXe=529.182 cm-1

putting this value in eqn (1),

,e-2(-529.182 cm-1)=2143.31 cm⁻¹.

or,e=1084.946 cm-1

De=e/4Xe=we^2/4eXe=(1084.946 cm-1)^2/4*(-529.182 cm-1)=-556.098 cm-1

Do=De-G(o)=(-556.098 cm-1)-(2143.31cm-1)=-2699.408 cm-1

Do or dissociation energy=2699.408 cm-1