Problem 34 - Chemistry Central Science

Question

(a) A complex absorbs light with wavelength of \(530 \mathrm{~nm}\). Do you expect it to have color? (b) A solution of a compound appears green. Does this observation necessarily mean that all colors of visible light other than green are absorbed by the solution? Explain. (c) What information is usually presented in a visible absorption spectrum of a compound? (d) What energy is associated with the absorption at \(530 \mathrm{~nm}\) in \(\mathrm{kJ} / \mathrm{mol}\) ?

Step-by-Step Solution

Answer

(a) Yes, the complex will have color since it absorbs light with a wavelength of \(530 \mathrm{~nm}\), which falls within the visible light spectrum. (b) No, the solution appearing green does not necessarily mean that all other colors are absorbed; it primarily reflects or transmits green light while absorbing other colors. Some other colors may still be reflected or transmitted, but green is the most predominant. (c) A visible absorption spectrum presents information about the wavelengths of light absorbed by the compound within the visible light range, shown as a graph with peaks indicating wavelengths absorbed. (d) The energy associated with the absorption at \(530 \mathrm{~nm}\) is approximately \(225.5 \, \mathrm{kJ/mol}\).

Step 1: a) Determining the Color Presence

Visible light has wavelengths ranging from approximately 400 nm (violet) to 700 nm (red). Given that the wavelength of absorbed light for the complex is 530 nm, it falls within the visible light spectrum. Therefore, the complex will have a color since it absorbs light in the visible range.

Step 2: b) Green Compound and Absorption of Other Colors

A solution appears green because it primarily reflects or transmits green light while absorbing other colors of the visible light spectrum. It does not necessarily mean that all other colors are absorbed by the compound. In some cases, a small portion of other colors may still be reflected or transmitted, but the green light is the most predominant, making the solution appear green.

Step 3: c) Information in a Visible Absorption Spectrum

A visible absorption spectrum of a compound presents information about the wavelengths of light absorbed by the compound within the visible light range (approx. 400-700 nm). This data is typically presented in the form of a graph, where the x-axis represents the wavelength of light and the y-axis represents the absorbance or extinction coefficient. Peaks in the graph indicate the specific wavelengths of light that are absorbed by the compound and can give insights into the electronic transitions and molecular structure of the compound.

Step 4: d) Calculating Energy Associated with Absorption

In order to calculate the energy associated with the absorption at 530 nm in kJ/mol, we can use the Planck's equation: E = \(h\nu\), where E is the energy, h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{~Js}\)), and \(\nu\) is the frequency of light. However, we are given the wavelength, not the frequency. To find the frequency, we can use the relationship between wavelength (λ), frequency (ν), and the speed of light (c): \(\mathrm{c = \lambda\nu}\). Rearranging the equation, we get: \(\nu = \frac{c}{\lambda}\). First, convert the wavelength to meters: \(\lambda = 530\, \mathrm{nm} \times \frac{1 \, \mathrm{m}}{10^9 \, \mathrm{nm}} = 5.3 \times 10^{-7} \, \mathrm{m}\). Now, calculate the frequency: \(\nu = \frac{3.0 \times 10^8 \, \mathrm{m/s}}{5.3 \times 10^{-7} \, \mathrm{m}} = 5.66 \times 10^{14} \, \mathrm{s}^{-1}\). Next, determine the energy E for one photon: \(E = (h)(\nu) = (6.626 \times 10^{-34} \mathrm{~Js})(5.66 \times 10^{14} \, \mathrm{s}^{-1}) = 3.75 \times 10^{-19} \, \mathrm{J}\). We now convert this energy to kJ/mol: \(E = 3.75 \times 10^{-19} \, \mathrm{J} \times \frac{1\, \mathrm{kJ}}{10^3 \, \mathrm{J}} \times \frac{6.022 \times 10^{23} \, \mathrm{photons/mol}}{1\, \mathrm{photon}}\) \(E \approx 225.5 \, \mathrm{kJ/mol}\) The energy associated with the absorption at 530 nm is approximately \(225.5 \, \mathrm{kJ/mol}\).