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# Ask AI: Calculate the force constants of the C=O bonds with the values of carbon monoxide (1902 cm-1)

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Question: Calculate the force constants of the C=O bonds with the values of carbon monoxide (1902 cm-1)

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The force constant (k) of a bond is a measure of the bond's stiffness and can be calculated using the wavenumber of the vibrational mode. For carbon monoxide, the wavenumber (ν) is given as 1902 cm⁻¹. The relation between the wavenumber and the force constant is given by the formula: $\nu = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}}$ Where: - $$\nu$$ is the wavenumber (in cm⁻¹) - $$c$$ is the speed of light in vacuum (approximately $$3 \times 10^{10} \, \text{cm/s}$$) - $$k$$ is the force constant (in dyn/cm or N/m; dyn/cm is preferred here because of the units of ν and c) - $$\mu$$ is the reduced mass of the C-O system First, we need to calculate the reduced mass ($$\mu$$) of the C-O bond. The reduced mass is given by: $\mu = \frac{m_C m_O}{m_C + m_O}$ Where: - $$m_C$$ is the mass of a carbon atom (approximately 12 atomic mass units, amu) - $$m_O$$ is the mass of an oxygen atom (approximately 16 amu) Converting atomic masses to grams: $m_C = 12 \, \text{amu} \times 1.66 \times 10^{-24} \, \text{g/amu} = 1.99 \times 10^{-23} \, \text{g}$ $m_O = 16 \, \text{amu} \times 1.66 \times 10^{-24} \, \text{g/amu} = 2.66 \times 10^{-23} \, \text{g}$ Now calculate the reduced mass ($$\mu$$) in grams: $\mu = \frac{1.99 \times 10^{-23} \, \text{g} \times 2.66 \times 10^{-23} \, \text{g}}{1.99 \times 10^{-23} \, \text{g} + 2.66 \times 10^{-23} \, \text{g}}$ $\mu = \frac{5.29 \times 10^{-46} \, \text{g}^2}{4.65 \times 10^{-23} \, \text{g}}$ $\mu = 1.14 \times 10^{-23} \, \text{g}$ Convert the reduced mass to kg for consistency with SI units: $\mu = 1.14 \times 10^{-23} \, \text{g} \times 10^{-3} \, \text{kg/g}$ $\mu = 1.14 \times 10^{-26} \, \text{kg}$ Now we can rearrange the equation to solve for $$k$$: $k = (2 \pi c \nu)^2 \mu$ Insert the values: $c = 3 \times 10^{10} \, \text{cm/s}$ $\nu = 1902 \, \text{cm}^{-1}$ Putting it all together: $k = \left(2 \pi \times 3 \times 10^{10} \, \text{cm/s} \times 1902 \, \text{cm}^{-1}\right)^2 \times 1.14 \times 10^{-26} \, \text{kg}$ $k = \left(2 \pi \times 3 \times 10^{10} \times 1902\right)^2 \times 1.14 \times 10^{-26}$ $k = \left(1.14164 \times 10^{15} \, \text{s}^{-1}\right)^2 \times 1.14 \times 10^{-26}$ $k = 1.30234 \times 10^{30} \, \text{s}^{-2} \times 1.14 \times 10^{-26}$ $k = 1.48 \times 10^4 \, \text{N/m}$ However, the original units of $$k$$ in dyn/cm are preferred: $k_{\text{dyn/cm}} = 1.48 \times 10^4 \, \frac{N}{m} \times \frac{10^5 \, \text{dyn}}{1 \, \text{N}} \times \frac{1 \, \text{m}}{10^2 \, \text{cm}}$ $k_{\text{dyn/cm}} = 1.48 \times 10^7 \, \text{dyn/cm}$ Therefore, the force constant for the C=O bond in carbon monoxide is approximately $$1.48 \times 10^7 \, \text{dyn/cm}$$.

Aug. 1, 2024, 12:19 a.m.

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