Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Chem Worksheet - Week 8 - Answers Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra Key Questions & Exercises 1. Calculate ΔHo for the reaction, C2H2(g) + H2(g) ÷ C2H4(g) Given: (a) 2 C2H2(g) + 5 O2(g) ÷ 4 CO2(g) + 2 H2O(l) (b) C2H4(g) + 3 O2(g) ÷ 2 CO2(g) + 2 H2O(l) (c) H2(g) + ½ O2(g) ÷ H2O(l) ΔHo = –2599.2 kJ ΔHo = –1410.9 kJ ΔHo = –285.8 kJ The (a) equation has C2H2(g) on the left, where we need it in the target equation, but its coefficient is 2. If we take one half of equation (a) we will have the correct coefficient. The ΔHo contribution will be (½)(–2599.2 kJ) = –1299.6 kJ. Equation (b) has C2H2(g) with the correct coefficient for our target equation, but we need it on the right, not the left as given. Therefore, we will reverse it, making its ΔHo contribution be +1410.9 kJ. Finally, if we take equation (c) as given, it will introduce the H2(g) with the correct coefficient, and it will result in cancellations of the CO2(g), H2O(l), and O2(g) introduced by our manipulations of the first two equations, which are absent in the target equation. The following sum results: (½a) (–b) (c) C2H2(g) + 5/2 O2(g) ÷ 2 CO2(g) + H2O(l) ΔHo = –1299.6 kJ 2 CO2(g) + 2 H2O(l) ÷ C2H4(g) + 3 O2(g) ΔHo = +1410.9 kJ H2(g) + ½ O2(g) ÷ H2O(l) ΔHo = –285.8 kJ _________________________________________________________ C2H2(g) + H2(g) ÷ C2H4(g) ΔHo = –174.5 kJ 2. Write the balanced thermochemical equations that pertain to the given compounds and their standard enthalpies of formation. Compound ΔHof (kJ/mol) CCl4(g) –106.7 C(s) + 2 Cl2(g) ÷ CCl4(l) Fe2O3(s) –822.16 2 Fe(s) + 3/2 O2(g) ÷ Fe2O3(s) HNO3(g) –134.3 1/2 H2(g) + 1/2 N2(g) + 3/2 O2(g) ÷ HNO3(g) NaHCO3(s) –947.7 Na(s) + 1/2 H2(g) + C(s) + 3/2 O2(g) ÷ NaHCO3(s) Thermochemical Equation 3. i. Given: (a) (b) N2O4(g) + 1/2 O2(g) ÷ N2O5(g) HNO3(g) ÷ 1/2 N2O5(g) + 1/2 H2O(l) ΔHo = +1.67 kJ ΔHo = –2.96 kJ Calculate ΔHo for the reaction N2O4(g) + H2O(l) + 1/2 O2(g) ÷ 2 HNO3(g) ΔHo = ? The two given equations, manipulated as shown below, will add to give the target reaction and its enthalpy: (a) N2O4(g) + 1/2 O2(g) ÷ N2O5(g) ΔHo = +1.67 kJ ÷ 2 HNO3(g) ΔHo = +5.92 kJ (–2b) N2O5(g) + H2O(l) __________________________________________________________ N2O4(g) + H2O(l) + 1/2 O2(g) ÷ 2 HNO3(g) ΔHo = 7.59 kJ ii. Given following standard enthalpy of formation data: Compound ΔHof (kJ/mol) N2O4(g) +9.66 HNO3(g) –134.31 H2O(l) –285.83 Calculate ΔHo for the reaction N2O4(g) + H2O(l) + 1/2 O2(g) ÷ 2 HNO3(g) ΔHo = ? Compare your answer to your answer in part i. The enthalpy for the reaction is the sum of the enthalpies of formation for products minus those for reactants, each multiplied by its stoichiometric coefficient in the balanced equation: ΔHo = 2ΔHof (HNO3) – {ΔHof (N2O4) + ΔHof (H2O) + ½ΔHof (O2)} = (2)(–134.31 kJ) – {+9.66 kJ + (–285.83 kJ) + 0 kJ} = 7.55 kJ Note that the value of ΔHof(O2) is taken as 0, because O2(g) is the standard state for elemental oxygen. As expected by the Law of Hess, the answers are virtually the same. In many cases there is a slight difference, due to minor differences in the values from the experimental data. 4. Calculate the heat of combustion of methane, CH4(g), defined by the following thermochemical equation: CH4(g) + 2 O2(g) 6 CO2(g) + 2 H2O(l) ΔHorxn = ? Given the following standard enthalpies of formation: –74.85 kJ CH4(g) CO2(g) –393.5 kJ H2O(l) –285.8 kJ This can be calculated as follows: ΔHorxn = {ΔHof(CO2) + 2ΔHof(H2O)} – {ΔHof(CH4) + 2ΔHof(O2)} = {(–393.5 kJ) + (2)(–285.8 kJ)} – {(–74.85 kJ) + 0} = –890.25 kJ = –890.2 kJ Note that the value of ΔHof(O2) is taken as 0, because O2(g) is the standard state for elemental oxygen. 5. Calculate the enthalpy of formation, ΔHof, for benzene, C6H6(l), given that the heats of formation of CO2(g) and H2O(l) are –393.5 kJ and –285.8 kJ, respectively, and that the heat of combustion of C6H6(l) is –3267.7 kJ. To do this, carry out the following steps. i. Write the balanced thermochemical equation that defines the enthalpy of formation of benzene, C6H6(l). ΔHof(C6H6) = ? 6 C(graphite) + 3 H2(g) ÷ C6H6(l) ii. Write the balanced thermochemical equation for the heat of combustion of benzene, C6H6(l). ΔHocomb = –3267.7 kJ C6H6(l) + 15/2 O2(g) ÷ 6 CO2(g) + 3 H2O(l) iii. Based on your answer to question ii, write an expression for the heat of combustion of benzene, ΔHocomb, in terms of the enthalpies of formation of the reactants and products. Using the data given in the problem, solve this for the unknown value of ΔHof(C6H6), the enthalpy of formation of benzene. We can express the heat of combustion as ΔHocomb = 6ΔHof(CO2) + 3ΔHof(H2O) – ΔHof(C6H6) We have values for everything in this equation except ΔHof(C6H6). Therefore, we can rearrange and solve for it: ΔHof(C6H6) = 6ΔHof(CO2) + 3ΔHof(H2O) – ΔHocomb = (6)(–393.5 kJ) + (3)(–285.8 kJ) – (–3267.7 kJ) = +49.3 kJ 6. For each of the following, indicate which kind of radiation has higher energy. red light or blue light infrared radiation or radio waves x-rays or visible light 7. For each of the following, indicate which has higher frequency light with λ = 490 nm or light with λ = 520 nm (ν = c/λ) light with energy of 3.0 x 10–19 J or light with energy of 4.5 x 10–19 J (E = hν) 8. An argon laser emits green light with a wavelength of 514.5 nm. Calculate the following for this light: (a) the wavelength in Å; (b) the frequency in Hz (s-1); (c) the energy in joules, J. (a) Note that to convert from nanometers to Ångstroms we need only move the decimal point to the right one place. (b) Use ν = c/λ. We need wavelength in meters, because the units on the speed of light are mAs-1. We could convert nm to m first, but with a calculator it is simpler to enter nanometers as 10–9 m; i.e., “514.5E-9". (c) Given the preceding results, we can calculate energy by either E = hν or E = hc/λ. Using the first of these, E = hν = (6.626 x 10–34 JAs)(5.827 x 1014 s–1) = 3.861 x 10–19 J 9. Is energy emitted or absorbed when the following transitions occur in hydrogen: (a) from n = 1 to n = 3 The transition is to a higher energy state, so energy corresponding to the difference between the states must be absorbed. (b) from n = 5 to n = 2 The transition is to a lower energy state, so energy corresponding to the difference between the states must be emitted. (c) an H+ ion acquires an electron into the n = 2 state. An H+ ion is essentially a hydrogen atom with an electron an infinite distance away, the state for which n = 4. By adding an electron to the n = 2 state, it has lower energy (the electron is closer to the nucleus to which it is attracted), so energy will be emitted 10. The four visible lines of the Balmer series in the emission spectrum of hydrogen are violet (410 nm), blue (434 nm), blue-green (486 nm), and red (656 nm). Assign these as state-tostate transitions of the type ni ÷ nf, giving the n values involved for each line. All transitions in the Balmer series are to nf = 2 from ni > 2. The energy of light increases going from red to violet, so the red line is the smallest state-to-state separation, and the violet line is the largest. red (656 nm) 3÷2 blue-green (486 nm) 4÷2 blue (434 nm) 5÷2 violet (410 nm) 6÷2 11. Calculate the energy of the first line in the Lyman series for the hydrogen atom, which arises from a transition from ni = 2 to nf = 1. What is the wavelength of the radiation emitted? In what region of the electromagnetic spectrum does it fall? Ephoton = 1.635 × 10–18 J A wavelength of 121 nm falls well within the ultraviolet region.