Question

Suppose that the C=O group in a peptide bond can be regarded as isolated from the rest of the molecule. Given that the force constant of the bond in a carbonyl group is 908 N m−1 , calculate the vibrational frequency of (a) 12C=16O and (b) 13C=16O.

Answer 1

Answer:

a) v = 1497.2 cm^-1

b) v = 1465 cm^-1

Explanation:

In the attached image is the procedure explained to reach the answer.

Answer 2

The vibrational frequency of 12C = 16O is 1498.9 cm^-1 and for 13C = 16O is 1465.6cm^-1

Vibrational Frequency

For a diatomic system, the vibrational frequency is given as

$v = \frac{1}{2\pi c} * (k/u)^\frac{1}{2}$

• v = frequency
• c = velocity of light = $3 * 10^8$
• k = force constant of the bond
• u = reduced mass of the system

The formula of reduced mass given as

$u = \frac{m_1m_2}{m_1 + m_2}$

where m1 and m2 are mass of the atoms.

a)

for 12C = 16O

$v = \frac{1}{2\pi c} (k/u)^\frac{1}{2}$

let's find the reduced mass

$u = \frac{12 * 16}{12 + 16}= 6.85g\\u = \frac{6.85}{6.023*10^2^3} \\u = 1.14 *10 ^ -^2 ^6kg$

substitute the value of reduced mass and solve for the vibrational frequency.

$v = \frac{1}{2\pi c} (\frac{908}{1.14*10^-^2^6})^\frac{1}{2}\\ v = 1.4989 m^-1\\v = 1498.9 cm^-^1$

b)

For 13C = 16O

The reduced mass is

$u = \frac{13 * 16}{13 +16} * \frac{10^-^3}{6.023*10^2^3} \\ u = 1.19*10^-^2^6$

Substitute the value of reduced mass and solve for the vibrational frequency.

$v = \frac{1}{2\pi c} (\frac{908}{1.19*10^-26})^\frac{1}{2}m^-^1\\ v = 1.4656*10^5m^-^1\\ v = 1456.6cm^-^1$

The vibrational frequency of 12C = 16O is 1498.9cm^-1 and the vibrational frequency of 13C = 16O is 1465.6cm^-1

Learn more on vibrational frequency here;

https://brainly.com/question/15531840

https://brainly.com/question/15531840