(a)
Interpretation:
The vibrational wavenumber of CO has to be stated.
Concept introduction:
Each vibrational spectrum at high resolution exhibits closely spaced components. There is very less spacing between these components signifying that these components are due to rotational transitions. During simultaneous vibrational and rotational changes, the rotational quantum number changes by a factor of one during vibrational transition of a molecule. On combining the vibration and rotation terms, the vibration-rotation spectra gives the equation as shown below.
˜S(ν,J)=(ν+12)˜ν+˜BJ(J+1)
(a)
Answer to Problem 11C.17P
The vibrational wavenumber of CO is at 2150 cm−1_.
Explanation of Solution
The vibrational wavenumber is given by the position of band centre. In case of carbon monoxide, a low resolution spectrum shows that the band centre occurs at 2150 cm−1. Therefore, the vibrational wavenumber of Carbon monoxide occurs at 2150 cm−1_.
(b)
Interpretation:
The molar zero-point vibrational energy of CO has to be stated.
Concept introduction:
The same concept introduction as discussed in part (a).
(b)
Answer to Problem 11C.17P
The molar zero-point energy is 12869.014 J mol−1_.
Explanation of Solution
The zero point energy is calculated by the formula shown below.
E=(v+12)hc˜ν+˜BJ(J+1) (1)
Where,
- ν is vibrational quantum number
- h is the reduced Planck’s constant
- c is the
speed of light - ˜ν is the wavenumber
- ˜B is the rotational constant.
- J is the rotational quantum number,.
Substitute ˜ν as 2150 cm−1, h as 6.626×10−34 J⋅s , c as 2.998×1010 cm s−1 ν as 0, J as 0 in equation (1).
E=(0+12)(6.626×10−34 J⋅s)(2.998×1010 cm s−1)(2150 cm−1)+˜B×0(0+1)=(12)(6.626×10−34 J⋅s)(2.998×1010 cm s−1)(2150 cm−1)=4.274×10−20 J2=2.137×10−20 J
For molar zero energy, multiplying the above value by Avogadro’s number as shown below.
E(J/mol)=E(J)×NA (2)
Substitute E(J) (2)
E(J/mol)=2.137×10−20 J×6.022×1023 mol−1=12869.014 J mol−1_
Therefore, the molar zero-point energy is 12869.014 J mol−1_.
(c)
Interpretation:
The force constant of CO has to be stated.
Concept introduction:
The strength of a bond in a bond is measured by force constant. The force constant of a molecule is related to the effective mass and vibrational frequency. The vibrational energy of a molecule is given by the equation as shown below.
Eν=(ν+12)ℏω.
(c)
Answer to Problem 11C.17P
The force constant for CO is 1863.01Nm-1_.
Explanation of Solution
The wavenumber of a molecule is given by the equation as shown below.
˜ν=12πc(kfmeff)12 (3)
Where,
- kf is the force constant.
- meff is the effective mass.
- c is the speed of light
Rearrange the above equation for force constant as shown below.
˜ν=12πc(kfmeff)124˜ν2π2c2=kfmeff
Thus the equation reduces to
kf=4π2c2˜ν2meff (4)
The effective mass of CO is calculated as shown below.
meff=mambma+mb (5)
Where,
- ma is the mass of first atom
- mb is the mass of second atom
For CO, substitute the value of ma as 12 g/mol and mb as 16 g/mol and in equation (5).
meff=12 g/mol×16 g/mol12+16 g/mol=19228 g/mol=6.85 g/mol
Convert the effective mass in kg as shown below.
1 g/mol =10−3 kg6.022×1023/mol6.85 g/mol=6.85 g/mol×10−3 kg6.022×1023 /mol=1.138×10−26 kg
Substitute ˜ν as 2150 cm−1, meff as 1.138×10−26 kg and c as 2.998×1010 cm s−1 in equation (4).
kf=4×(3.14)2×(2.998×1010 cm s−1)2×(2150 cm−1)2×(1.138×10−26kg)=4×9.8596×8.98×1020 cm2 s−2×4622500 cm−2×1.138×10−26kg=1863.01Nm-1_
Therefore, the force constant for CO is 1863.01Nm-1_.
(d)
Interpretation:
The rotational constant of CO has to be stated.
Concept introduction:
The energy for a rotational spectrum relates the quantum number and rotational constant to give the expression, E=˜BJ(J+1). The rotational constant further gives a relation with the moment of inertia of the molecule.
(d)
Answer to Problem 11C.17P
The rotational constant is 3.8275 cm−1_.
Explanation of Solution
In a high resolution spectrum, it is observed that the band centre is split into two peaks. As the vibrational spectrum splits to give two peaks with a separation of 7.655 cm−1. This separation between two peaks is given by 2˜B. Therefore, rotational constant is calculated as shown below.
2˜B=7.655 cm−1˜B=7.6552cm−1=3.8275 cm−1_
Therefore, the rotational constant is 3.8275 cm−1_.
(e)
Interpretation:
The bond length of CO has to be stated.
Concept introduction:
The same concept introduction as mentioned in part (d)
(e)
Answer to Problem 11C.17P
The value of bond length of CO is 8.02×10−11m_.
Explanation of Solution
The value of rotational constant helps in determining the moment of inertia of a molecule as shown below.
˜B=ℏ4πIc (6)
Where,
- ℏ is the reduced Planck’s constant.
- I is the moment of inertia
- c is the speed of light.
Substitute the value of ˜B (6)
3.8275 cm−1=1.05457×10−34 J⋅s4×3.14×I×2.998×1010 cm s−1I=1.05457×10−34 J⋅s4×3.14×3.8275 cm−1×2.998×1010 cm s−1I=1.05457×10−34 J⋅s1.44×1012I=7.32×10−47 kg m2
The moment of inertia is given by the equation as shown below.
I=μR2 (7)
Where,
- μ is the reduced mass
- R is the bond length of the molecule.
Substitute μ as 1.138×10−26 kg and I as 7.32×10−47 kg m2 in equation (7) as shown below.
7.32×10−47 kg m2=1.138×10−26 kg×R2R2=6.43×10−21 m2R=8.02×10−11m_
Therefore, the value of bond length of CO is 8.02×10−11m_.
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