Week 8 Assignment F23

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University of Toledo *

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Chemistry

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Jan 9, 2024

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Week 8 Peer-Lead Team Learning CLASSROOM TOPICS: A little more generally on Thermochemistry 5.6 Hess’s Law 5.7 Standard Enthalpies of Formation 6.1 The Nature of Light 6.2 Quantum Theory and the Photoelectric effect PRE-CLASS ASSIGNMENT - IN ADVANCE of meeting with Peer-Leader (0-2 points): 1. What can you measure with a calorimeter? To measure the heat transfer 2. Rather than measuring the heat transfer (enthalpy change) in a chemical reaction with a calorimeter, what “thermodynamic values” can be used to calculate the heat transfer (enthalpy change)? 3. Complete the table below Quantity Symbol Typical Units OR Value and Units if a Constant Wavelength Frequency Energy of one Photon “Speed of light” Planks constant 4. An equation to relate frequency and wavelength: c = λν 5. An equation to relate energy of a photon and frequency: E = hν

6. An equation to relate energy of a photon and wavelength: E = h * c / λ WARM UP Compare answers to the problems above. DURING THE WORKSHOP First, more thermochemistry and a little other review…… 1. A student dissolves 12.2 g of ammonium chloride in 200.g of water in a well-insulated cup. She then observes the temperature of the solution to fall from 20.0 to 15.9 °C over the course of 5 minutes. a) Is ammonium chloride an electrolyte? Explain. Would you have predicted that it easily dissolves in water? Explain. Ammonium chloride (NH4Cl) is an electrolyte because it dissociates into ions (NH4+ and Cl-) when dissolved in water, allowing it to conduct electricity. You would predict that it easily dissolves in water because it contains ionic bonds, and water is a polar solvent that can separate and solvate the ions in the compound. b) Is the reaction exothermic, endothermic or neither? The reaction is endothermic because the temperature of the solution fell from 20.0°C to 15.9°C. In an endothermic reaction, heat is absorbed from the surroundings, resulting in a decrease in temperature. c) If the reaction is either exothermic or endo thermic, calculate the heat that is released or absorbed in the reaction: To calculate the heat (q) absorbed or released in the reaction, you can use the formula: q = m * c * ΔT Where: q = heat (in joules) m = mass of the solution (in grams) c = specific heat capacity of water (approximately 4.18 J/g°C) ΔT = change in temperature (final temperature - initial temperature) m = 200 g (mass of water) + 12.2 g (mass of NH4Cl) = 212.2 g ΔT = 15.9°C - 20.0°C = - 4.1°C (note the negative sign for the decrease in temperature) q = 212.2 g * 4.18 J/g°C * (-4.1°C) = -3,764.37 J Since 1 kJ = 1,000 J, the heat absorbed or released is approximately -3.76 kJ.

d) Determine D H Rxn in kJ/mol then complete the thermochemical equation below. NH 4 Cl (aq) → NH 4 + (aq) + Cl - (aq) The change in enthalpy (ΔH) for the reaction can be determined by dividing the heat (q) by the number of moles of NH4Cl used. To find the number of moles, use the molar mass of NH4Cl: Molar mass of NH4Cl = 14.01 g/mol (N) + 4.02 g/mol (H) + 35.45 g/mol (Cl) = 53.48 g/mol Number of moles of NH4Cl = 12.2 g / 53.48 g/mol ≈ 0.228 moles ΔH = q / moles = (-3.76 kJ) / 0.228 moles ≈ -16.49 kJ/mol So, ΔHrxn for the reaction NH4Cl (aq) → NH4+ (aq) + Cl- (aq) is approximately -16.49 kJ/mol. e) Is any energy exchanged as work in this reaction? ________ Explain your reasoning: No, there is no work involved in this reaction. The reaction occurs in a well-insulated cup, which means the system is isolated from its surroundings. In an isolated system, no energy is exchanged as work, and all the energy changes occur as heat. f) Compare the change in internal energy and the change in enthalpy for this reaction. In an isolated system, where no work is done, the change in internal energy (ΔU) is equal to the heat exchanged (q). Therefore, ΔU ≈ -3.76 kJ, which is the same as the heat. The change in enthalpy (ΔH) is also approximately -16.49 kJ/mol. g) Write an equation that relates change in internal energy, heat, and work for a reaction: The equation that relates change in internal energy (ΔU), heat (q), and work (w) for a reaction in an isolated system is: ΔU = q - w In this case, since it's an isolated system with no work done, the equation simplifies to: ΔU = q

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2. Write a thermochemical equation corresponding to the standard enthalpy of formation for glucose (C 6 H 12 O 6 ). D H f for the process is -1273.7 kJ/mol. The thermochemical equation for the standard enthalpy of formation of glucose (C6H12O6) is as follows: C6H12O6 (s) → 6C (s) + 6H2 (g) + 3O2 (g) The standard enthalpy of formation (ΔHf) for this process is -1273.7 kJ/mol. This means that when one mole of glucose is formed from its elements in their standard states (carbon as graphite, hydrogen as dihydrogen gas, and oxygen as diatomic oxygen gas), it releases 1273.7 kJ of heat energy. The coefficients in the equation are chosen such that the molar quantities on both sides of the equation match the molar quantities in one mole of glucose. This allows the ΔHf value to be directly associated with the formation of one mole of glucose. 3. Calculate the enthalpy for the oxidation of CO to CO 2 : 2CO(g) + O 2 ( g ) → 2CO 2 (g ) using the enthalpy of reaction for the combustion of C to CO and the enthalpy for the combustion of C to CO 2 : C(s) + O 2 ( g ) → CO 2 (g) Δ H = -393.5 kJ 2C(s) + O 2 ( g ) → 2CO(g) Δ H = -221.0 kJ To calculate the enthalpy for the oxidation of CO to CO2: 2CO(g) + O2(g) → 2CO2(g) you can use the given enthalpies of reaction for the combustion of C to CO and C to CO2: C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ 2C(s) + O2(g) → 2CO(g) ΔH = -221.0 kJ First, notice that the oxidation of CO (carbon monoxide) to CO2 (carbon dioxide) can be represented as the reverse of the combustion of CO: CO(g) + 1/2O2(g) → CO2(g) Now, let's consider the desired reaction:

2CO(g) + O2(g) → 2CO2(g) We can break down this reaction into the following steps: 1. Convert 2 moles of CO to 2 moles of C: 2CO(g) → 2C(s) The enthalpy change for this step would be twice the enthalpy of the combustion of C to CO: 2 * ΔH1 = 2 * (-221.0 kJ) = -442.0 kJ 2. Convert 2 moles of C to 2 moles of CO2: 2C(s) + O2(g) → 2CO2(g) The enthalpy change for this step would be twice the enthalpy of the combustion of C to CO2: 2 * ΔH2 = 2 * (-393.5 kJ) = -787.0 kJ Now, you can add the enthalpy changes for both steps to find the overall enthalpy change for the desired reaction: ΔH = ΔH1 + ΔH2 = -442.0 kJ + (-787.0 kJ) = -1229.0 kJ So, the enthalpy change for the oxidation of CO to CO2 is -1229.0 kJ. 4. Calculate the standard heat of combustion for glucose from standard enthalpies of formation. a) Write a balanced reaction. b) What data do you need to look up? Do that and record it below:

c) Show the calculation for the standard heat of combustion for glucose. a) To calculate the standard heat of combustion for glucose (C6H12O6), you need to write a balanced combustion reaction for glucose. The combustion of glucose involves reacting it with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced reaction is as follows: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g) b) To calculate the standard heat of combustion for glucose, you need to look up the standard enthalpies of formation (ΔHf) for each of the compounds involved in the reaction. Specifically, you'll need the ΔHf values for glucose (C6H12O6), carbon dioxide (CO2), and water (H2O). Here are the standard enthalpies of formation for the compounds: - ΔHf for glucose (C6H12O6) = -1273.7 kJ/mol - ΔHf for carbon dioxide (CO2) = -393.5 kJ/mol - ΔHf for water (H2O) = -285.8 kJ/mol c) To calculate the standard heat of combustion (ΔHcomb) for glucose, you can use the formula: ΔHcomb = ΣΔHf(products) - ΣΔHf(reactants) First, identify the moles of each substance in the balanced equation: - Moles of C6H12O6 = 1 (coefficient in front of glucose) - Moles of O2 = 6 (coefficient in front of O2) - Moles of CO2 = 6 (coefficient in front of CO2) - Moles of H2O = 6 (coefficient in front of H2O) Now, apply the formula: ΔHcomb = [6ΔHf(CO2) + 6ΔHf(H2O)] - [ΔHf(C6H12O6) + 6ΔHf(O2)] ΔHcomb = [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] - [-1273.7 kJ/mol + 6(0 kJ/mol)] ΔHcomb = [-2361 kJ + (-1715.4 kJ)] - [-1273.7 kJ] = -2361 kJ - 1715.4 kJ + 1273.7 kJ = -2803.7 kJ The standard heat of combustion for glucose is approximately -2803.7 kJ/mol. This means that when one mole of glucose is completely burned, it releases 2803.7 kJ of heat energy.

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Now light and other forms of electromagnetic energy…. 5. Sketch two electromagnetic waves A and B with the same amplitude but different wavelength: A has a wavelength of “x” and B has a wavelength of “2x”. Label the wavelength on your sketches. a) If wave A has a frequency of “y”, what is the frequency of wave B? b) If a photon of A has an energy of “z”, what is the energy of a photon of B? c) Choose “directly proportional” or “inversely proportional” to describe each of the following pairs: wavelength and frequency energy and frequency energy and wavelength a) If wave A has a wavelength of "x," and wave B has a wavelength of "2x," then the frequency of wave B can be calculated using the wave equation: c = λν Where: c = speed of light (constant, approximately 3.00 × 10^8 m/s) λ = wavelength ν = frequency For wave A: c = x * y y = c / x For wave B: c = 2x * ν_B ν_B = c / (2x) So, the frequency of wave B (ν_B) is half the frequency of wave A (ν), or ν_B = 0.5y. b) The energy (E) of a photon is directly proportional to its frequency (ν) and can be calculated using the equation: E = hν Where: E = energy of the photon h = Planck's constant (constant, approximately 6.626 × 10^(-34) J·s)

ν = frequency If a photon of wave A has an energy of "z," then a photon of wave B will have the same energy because their frequencies are directly proportional. Therefore, the energy of a photon of wave B is also "z." c) Describing the relationships: - Wavelength and frequency: These two are inversely proportional. As wavelength increases, frequency decreases, and vice versa. - Energy and frequency: These two are directly proportional. As frequency increases, energy increases, and vice versa. - Energy and wavelength: These two are inversely proportional. As wavelength increases, energy decreases, and vice versa.

6. Carbon dioxide is a greenhouse gas because it absorbs infrared radiation emitted from the earth after the earth has absorbed energy from the sun. One area of the electromagnetic spectrum in which carbon dioxide absorbs well occurs at a wavelength around 15 m m. a) Identify from the diagram to the right, the region of the electromagnetic spectrum in which this absorption lies. b) Calculate the frequency (in Hz) of radiation at 15 m m. a) The region of the electromagnetic spectrum in which carbon dioxide (CO2) absorbs well, specifically at a wavelength around 15 μm (micrometers), falls within the **infrared region**. Infrared radiation has longer wavelengths than visible light but shorter wavelengths than microwaves. b) To calculate the frequency (ν) of radiation at 15 μm, you can use the speed of light (c) and the wavelength (λ) with the following formula: ν = c / λ Where: ν = Frequency (in hertz, Hz) c = Speed of light in a vacuum (approximately 3.00 × 10^8 meters per second) λ = Wavelength (in meters) First, you need to convert the wavelength from micrometers to meters. 1 μm (micrometer) is equal to 1 × 10^(-6) meters. λ = 15 μm * (1 × 10^(-6) m/μm) λ = 15 × 10^(-6) meters

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Now, calculate the frequency: ν = (3.00 × 10^8 m/s) / (15 × 10^(-6) meters) ν ≈ 2.00 × 10^13 Hz So, the frequency of radiation at 15 μm is approximately 2.00 × 10^13 Hz. 7. (Similar to ALEKS) The ozone hole has been created by reactions involving chlorine atoms and ozone. The chlorine atoms are largely released from a group of man-made compounds known as chlorofluorocarbons or CFCs. It takes about 168 kJ/mol to break a carbon-chlorine bond and release a chlorine atom from a CFC. Calculate the maximum wavelength of electromagnetic radiation (in nm) for which a carbon-chlorine bond could be broken by absorbing a single photon. E = 168 kJ/mol To calculate the maximum wavelength of electromagnetic radiation for which a carbon-chlorine bond could be broken by absorbing a single photon, you can use the energy of the photon (E) and the following formula: E = hc / λ Where: E = Energy of the photon (in joules) h = Planck's constant (approximately 6.626 × 10^(-34) J·s) c = Speed of light (approximately 3.00 × 10^8 m/s) λ = Wavelength of the photon (in meters) Given that E is 168 kJ/mol, we need to convert it to joules per photon and then rearrange the formula to solve for λ. First, convert kJ to joules: 1 kJ = 1000 J 168 kJ = 168,000 J Now, we have: E = 168,000 J/mol We need to find the energy of a single photon by dividing this energy by Avogadro's number (the number of molecules or atoms in a mole): E_photon = (168,000 J/mol) / (6.022 × 10^23 molecules/mol) E_photon ≈ 2.793 × 10^(-19) J/photon Now, you can rearrange the formula to solve for λ:

λ = hc / E_photon λ = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (2.793 × 10^(-19) J) λ ≈ 2.25 × 10^(-7) meters To express the wavelength in nanometers (nm), you can convert from meters: 1 meter = 1 × 10^9 nanometers λ ≈ 2.25 × 10^(-7) meters * 1 × 10^9 nm/meter λ ≈ 225 nm So, the maximum wavelength of electromagnetic radiation for which a carbon-chlorine bond could be broken by absorbing a single photon is approximately 225 nanometers. 8. What is the frequency in s -1 of the radiation described in the problem above? To find the frequency (ν) in s^(-1) of the radiation with a wavelength of 225 nm, you can use the speed of light (c) and the wavelength (λ) with the following formula: ν = c / λ Where: ν = Frequency (in s^(-1)) c = Speed of light (approximately 3.00 × 10^8 m/s) λ = Wavelength (in meters) Given that the wavelength is 225 nm, you should convert it to meters: 1 meter = 1 × 10^9 nanometers λ = 225 nm * (1 × 10^(-9) m/nm) λ = 2.25 × 10^(-7) meters Now, you can calculate the frequency: ν = (3.00 × 10^8 m/s) / (2.25 × 10^(-7) meters) ν ≈ 1.33 × 10^15 s^(-1) So, the frequency of the radiation with a wavelength of 225 nm is approximately 1.33 × 10^15 s^(-1).

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The following problems may be solved by the group this week or next depending on whether it has been covered in class. 9. In relation to the diagram at the right: a) What do the lines labeled n=1 through n=7 represent? b) What would be the relative energy of the nucleus if it were added to the diagram? c) List the obits shown in order from lowest to highest energy. d) In theory, do any other orbits exist? If so, how many? e) At n=∞, what is the energy of the orbit? f) Which orbit shown would have the most negative value for energy? g) Describe how the distance between and energy between successive orbits changes as orbits get further from the nucleus. ( e.g., Compare distance and energy between 2 & 3 to distance and energy between 3 & 4) h) The lines with the arrows represent the changes in energy for an electron transitioning between different orbits. Would photons of light be absorbed or emitted in these transitions? i) Based on the diagram, and what you know about the energy of UV, visible and IR radiation, suggest ideas for the table below. Share how you came to your conclusion. a) The lines labeled n = 1 through n = 7 represent electron energy levels or electron orbits in an atom. These correspond to the principal quantum numbers (n) in the atomic model. Relative Energy Series Name Type of radiation (UV, visible or IR) Relative energy (high, medium, low) Relative wavelength (high, medium, low) Lyman Balmer Paschen

b) The energy of the nucleus is extremely high compared to the energy levels of electrons. It would be located far below the diagram, well outside the visible range. c) The orbits are typically listed in order from lowest to highest energy, starting with n = 1 (the ground state) and increasing with n. d) In theory, an infinite number of orbits exist, but only a finite number are shown in these diagrams. The higher the value of n, the further the orbit is from the nucleus. e) At n = ∞, the energy of the orbit is zero, which represents the energy of an electron that is completely removed from the influence of the nucleus. f) The orbit with the most negative value for energy is the one closest to the nucleus, which corresponds to n = 1. g) As you move further from the nucleus (e.g., from orbit 2 to orbit 3 or from orbit 3 to orbit 4), the distance between orbits increases, and the energy levels become less negative or closer to zero. Electrons in outer orbits have higher energy than those in inner orbits. h) When electrons transition between orbits (as shown by the arrows in the diagram), they either emit or absorb photons of light. When an electron falls to a lower energy level (n → n-1), it emits a photon, and this corresponds to light being emitted. When an electron moves to a higher energy level (n → n+1), it absorbs a photon, and this corresponds to light being absorbed. i) The terms like Lyman, Balmer, and Paschen refer to specific series of spectral lines in the emission spectra of hydrogen or other similar atoms. Each series corresponds to a different transition of electrons between energy levels. Here's how you can relate these to UV, visible, or IR radiation:  Lyman series: These transitions involve very high energy differences, resulting in the emission of ultraviolet (UV) radiation. The relative energy is high, and the relative wavelength is low (short wavelengths).  Balmer series: Balmer transitions result in the emission of visible light (usually in the range of 400-700 nm), so it corresponds to visible radiation. The relative energy is medium, and the relative wavelength is medium.  Paschen series: These transitions result in the emission of infrared (IR) radiation. The relative energy is lower than the other series, and the relative wavelength is higher (longer wavelengths).

10. According to the Bohr model, the energy of an orbit with n = 2 is -5.45 × 10 -19 J. The energy of an orbit with n = 5 is -8.72 × 10 -20 J. Calculate the wavelength (in nm) and frequency (in hertz) of the photon emitted when an electron falls from n = 5 to n = 2. Be sure to include units in your answer. Steps/strategy To calculate the wavelength and frequency of the photon emitted when an electron falls from n = 5 to n = 2 in the Bohr model, you can use the following formulas: 1. The energy change (ΔE) is given by: ΔE = E_initial - E_final Where E_initial is the energy of the initial orbit (n = 5), and E_final is the energy of the final orbit (n = 2). ΔE = (-8.72 × 10^(-20) J) - (-5.45 × 10^(-19) J) ΔE = 4.58 × 10^(-19) J 2. The energy change is also equal to the energy of the emitted photon, E_photon: E_photon = ΔE 3. Now, use the energy of the photon to calculate its wavelength (λ) and frequency (ν) using the following equations: E_photon = hν E_photon = (hc) / λ Where: E_photon = Energy of the photon (in joules) h = Planck's constant (6.626 × 10^(-34) J·s) c = Speed of light (3.00 × 10^8 m/s) λ = Wavelength of the photon (in meters) ν = Frequency of the photon (in hertz) First, calculate the frequency (ν): E_photon = hν ν = E_photon / h ν = (4.58 × 10^(-19) J) / (6.626 × 10^(-34) J·s) ν ≈ 6.91 × 10^14 Hz Now, calculate the wavelength (λ): E_photon = (hc) / λ λ = (hc) / E_photon λ = [(6.626 × 10^(-34) J·s) * (3.00 × 10^8 m/s)] / (4.58 × 10^(-19) J) λ ≈ 1.29 × 10^(-7) meters To express the wavelength in nanometers (nm), you can convert from meters: 1 meter = 1 × 10^9 nanometers λ ≈ 1.29 × 10^(-7) meters * 1 × 10^9 nm/meter

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λ ≈ 129 nm So, the wavelength of the photon emitted when an electron falls from n = 5 to n = 2 is approximately 129 nanometers, and the frequency is approximately 6.91 × 10^14 hertz. 11. Determine the energy of a mol of photons with a wavelength of 2.33 × 10 4 nm. To determine the energy of a mole of photons with a given wavelength, you can use the following formula: E = (hc) / λ Where: E = Energy (in joules) h = Planck's constant, which is approximately 6.626 × 10^(-34) J·s c = Speed of light in a vacuum, which is approximately 3.00 × 10^8 m/s λ = Wavelength of the photons (in meters) First, you need to convert the given wavelength from nanometers (nm) to meters: 1 nm = 1 × 10^(-9) meters So, 2.33 × 10^4 nm = 2.33 × 10^(-9) meters Now, you can calculate the energy: E = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (2.33 × 10^(-9) meters) E ≈ 8.52 × 10^(-19) joules per photon Now, you want to find the energy of a mole of photons. Avogadro's number (6.022 × 10^23) represents the number of entities (in this case, photons) in one mole. So, multiply the energy per photon by Avogadro's number to find the energy of a mole of photons: Energy of a mole of photons = (8.52 × 10^(-19) J/photon) * (6.022 × 10^23 photons/mol) Energy of a mole of photons ≈ 5.14 × 10^5 J/mol So, the energy of a mole of photons with a wavelength of 2.33 × 10^4 nm is approximately 5.14 × 10^5 joules per mole.

Week 8 Assignment F23

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