Analytical Chemistry Lecture Notes PDF

# Giacomo Luigi Ciamician: Father of Modern Molecular Photochemistry - **Born:** August 25, 1857 (1857-08-25) - Trieste, Austria - **Died:** January 2, 1922 (aged 64) - Bologna, Italy - **Education:** University of Vienna - **Employer:** University of Bologna His first photochemistry experimen...

# Giacomo Luigi Ciamician: Father of Modern Molecular Photochemistry - **Born:** August 25, 1857 (1857-08-25) - Trieste, Austria - **Died:** January 2, 1922 (aged 64) - Bologna, Italy - **Education:** University of Vienna - **Employer:** University of Bologna His first photochemistry experiment was published in 1886 and was titled "On the conversion of quinone into quinol by light". # Characteristics of Light - Light= band of waves of different wavelengths | | | | :----: | :----: | | **v = 2 Hz** | λ | | **v = 4 Hz** | λ | - Increase in frequency (energy) - Time (1 second) - Wavelength (nm) - 400, 500, 600, 700 # (Light) Electromagnetic Spectrum important for photochemistry: - The Electromagnetic Spectrum | Type | Size | | :----: | :----: | | Gamma ray | 10^-12 cm | | X-ray | 10^-10 cm | | Ultraviolet | 10^-8 cm | | Visible | 10^-6 cm | | Infrared | 10^-4 cm | | Microwave | 1 cm | | Radio | 1 km | # Particle Characteristics of Light: - **Light** = flux of discrete units (i.e quanta) called photons. - **Fireworks** # Photochemical Reactions: - Photochemical reaction includes the absorption of visible radiation during photosynthesis. - Without photochemical processes, the Earth would be simply a warm, sterile, rock. # Schematic of photosynthesis in plants. The carbohydrates produced are stored in or used by the plant. | | | | | :----: | :----: | :----: | | 6CO2 - Carbon dioxide | 6H20 - Water | Light - C6H12O6 - Sugar | | | | 6O2 - Oxygen | - Overall equation for the type of photosynthesis that occurs in plants # Difference between photochemical and thermochemical reactions | Characteristic | Photochemical Reactions | Thermochemical Reactions | | :----: | :----: | :----: | | 1. | These involve absorption of light radiations. | These reactions involve absorption or evolution of heat. | | 2. | The presence of light is the primary requirement for reactions to take place. | These reactions can take place in dark as well as in light. | | 3. | Temperature has a very little effect on the rate of photochemical reactions. | Temperature has a significant effect on the rate of a thermochemical reaction. | | 4. | ΔG for photochemical spontaneous reactions may be +ve or -ve. | ΔG for a thermochemical reaction is always negative. | | 5. | Photochemical activation is highly selective. The absorbed photon excites a particular atom or group of atoms which become site for the reaction. | Thermochemical activation is not selective in nature. | # Transmission Transmission, T, is simply defined as the fraction of light that reaches a detector after passing through a sample - $T = \frac{I}{I_0}$ - %T=100 $\frac{I}{I_0}$ # Example What is the absorbance of a sample with a 1.0% transmission? - %T = 1.0=> $I/I_0$= 0.010 or $I_0/I$= 100 - $A= log_{10}(100)= 2.0$ # Example 2 Cytosine has a molar absorption coefficient of 6 x 10³ mol-1 cm¯¹ at 270 nm at pH 7. Calculate absorbance of 1 x 10-3 M cytosine solution in 1mm cell at 270 nm - $A = abc$ - (6 x 10³)x (0.1) x (1 x 10–3) = 6 x_10-1 # Absorbance Spectrometry - Determines concentration of a substance in solution. - Measures light absorbed by solution at a specific wavelength. - **Visible region:** low energy electronic transition due to. - Compounds containing transition metals. - Large aromatic structures & conjugated double bond systems. - **UV region (200-400 nm):** - Small conjugated ring systems. - Carbonyl compounds. # Determination of absorbed intensity A photochemical reaction occurs by the absorption of photons of light by the molecules. Therefore, it is essential to determine the absorbed intensity of light for a study of the rate of reaction. - Schematic diagram of the spectrophotometer used for measurement of light intensity: - **Collimator** (Lens) - **Wavelength Selector** (Slit) - **Detector** (Photocell) - **Light source** - **Monochromator** (Prism or Grating) - **Sample Solution** (in Cuvette) - **Digital Display or Meter** - A sample UV-Vis spectrum. # LAWS OF PHOTOCHEMISTRY - **Grothus-Draper Law** (qualitative aspect) - It is only the absorbed_light_radiations that are effective in producing a chemical reaction. - **Stark-Einstein Law of Photochemical Equivalence** - Stark and Einstein (1905) studied the quantitative_aspect_of photochemical reactions by application of Quantum theory of light. - One photon (hv) - Illustration of Law of Photochemical equivalence; absorption of one photon decomposes one molecule. - Reactant molecule - A - Excited molecule - A - Product molecule - B # Laws of Photochemistry - **1.** Only light that is absorbed can produce photochemical change (Grotthus, Draper) - **2.** A molecule absorbs a single quantum of light is becoming excited (Stark, Einstein) # Quantum yield (or Quantum efficiency) (Φ) $Φ = \frac{No. of molecules reacted or formed}{No. of photons absorbed}$ - For a reaction that obeys strictly the Einstein law, one molecule decomposes per photon, the quantum yield Φ = 1. - When two or more molecules are decomposed per photon, Φ > 1 and the reaction has a high quantum yield. - If the number of molecules decomposed is less than one per photon, Φ < 1 the reaction has a low quantum yield. # Calculation of Quantum Yield - By definition, the quantum yield, Φ, of a photochemical reaction is expressed as: - Φ = (No. of molecules decomposed or formed) / (No. of photons of radiation energy absorbed) - Φ = (No. of moles decomposed or formed) / (No. of moles of radiation energy absorbed) - Thus we can calculate quantum yield from: - (a) The amount of the reactant decomposed in a given time and - (b) The amount of radiation energy absorbed in the same time - The radiation energy is absorbed by a chemical system as photons. Therefore we should know the energy associated with a photon or a mole of photons. # The energy of photons; einstein - We know that the energy of a photon (or quantum), ∈, is given by the equation: - ∈ = hv = hc/λ...(1) - h = Planck's constant (6.624 × 10–27 erg-sec) - v = frequency of radiation - λ = wavelength of radiation - c = velocity of light (3 × 1010 cm - The energy, E, of an Avogadro number (N) of photons is referred to as one einstein: - E = Nhc/λ...(2) - Substituting the values of N (= 6.02 x 1023), h and c, in (2), we have: - E = (1.196 × 108)/λ erg mol-1 the energy is expressed in ergs. # If λ is expressed in Å units (1Å = 10-8 cm), - E = (1.196 x 1016)/λ erg mol-1...(3) # Since 1 cal = 4.184 × 107 erg, energy in calories would be: - E = (1.196 x 1016)/(λ x 4.184 × 107)...(4) - = (2.859 x 10^8)/λ cal mol-1 - E = (2.859 x 10^5)/λ kcal mol-1...(5) - Electron-volt (eV) is another commonly used energy unit, 1 eV = 1.6 x 10-19 J. - It is evident from (3) that the numerical value of einstein varies inversely as the wavelength of radiation. - The higher the wavelength, the smaller will be the energy per einstein # SOLVED PROBLEM 1 Calculate the energy associated with (a) one photon; (b) one einstein of radiation of wavelength 8000 Å. h = 6.62 × 10–27 erg-sec; c = 3 × 1010 cm sec-1. - **SOLUTION** - **(a) Energy of a photon** - ∈ = hv = hc/λ = (6.62 x 10-27 x 3 x 10^10)/(8000 x 10^-8) = (6.62 x 3 x 10^-12)/8.0 = 2.4825 × 10-12 erg - **(b) Energy per einstein** - E = Nhc/λ = (6.02 x 10^23 x 6.62 x 10^-27 x 3 x 10^10)/(8000 x 10^-8) = (6.02 x 6.62 x 3 x 10^11)/8.0 = 1.4945 × 10^12 erg # SOLVED PROBLEM 2 When a substance A was exposed to light, 0.002 mole of it reacted in 20 minutes and 4 seconds. In the same time A absorbed 2.0 × 106 photons of light per second. Calculate the quantum yield of the reaction. (Avogadro number N = 6.02 × 1023) - **SOLUTION** - Number of molecules of A reacting = 0.002 × N = 0.002 × 6.02 × 1023 - Number of photons absorbed per second = 2.0 × 106 - Number of photons absorbed in 20 minutes and 4 seconds = 2.0 × 106 × 1204 - Quantum yield = (No. of molecules reacted) / (No. of photons absorbed) = (0.002 × 6.02 × 1023) / (2.0 x 106 x 1204) = 5.00 × 1011 # SOLVED PROBLEM 3 When irradiated with light of 5000 Å wavelength, 1x10-4 mole of a substance is decomposed. How many photons are absorbed during the reaction if its quantum efficiency is 10.00. (Avogadro number N = 6.02 × 1023) - **SOLUTION** - Quantum efficiency of the reaction = 10.00 - No. of moles decomposed = 1 × 10–4 - No. of molecules decomposed = 1 × 10–4 × 6.02 × 1023 = 6.02 x 1019 - we know that, Φ = (No. of molecules decomposed) / (No. of photons absorbed) = (6.02 x 1019) / (No. of photons absorbed) - No. of photons absorbed = (6.02 x 1019)/10 = 6.02 × 1018