Sample3S - 1 Chem 6 Sample exam 3 (final) brief answers 1....

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1 Chem 6 Sample exam 3 (final) brief answers 1. Because E = hc/ λ , shortest wavelength means biggest energy. This must correspond to a transition from n = infinity to the unknown quantum number n . The energy of this transition is given by E = –2.18x10 –18 J Z 2 (1/n f 2 – 1/n i 2 ) Plug in Z = 1 for hydrogen, n f = n, n i = infinity and solve for n E = hc/ λ = I(–2.18x10 –18 J (1/n f 2 ))I So [(6.626x10 –34 J)(3x10 8 m/s)/(3280 nm x 1m/10 9 nm)]/ 2.18x10 –18 J = 1/n f 2 And n f = 6 . Now we can find the wavelength for line B. The far-right line (largest wavelength) must be the smallest energy transition, from n = 7 to n = 6. So, line B must be due to the transition from n = 8 to n = 6. Plug-n-chug: E = –2.18x10 –18 J (1/6 2 – 1/8 2 ) = –2.65x10 –20 J Convert to wavelength I EI = hc/ λ or λ = hc/ I EI = [(6.626x10 –34 J)(3x10 8 m/s)]/ 2.65x10 –20 J = 7.5x10 –6 m x 10 9 nm/m = 7500 nm 2 . Rate = k 2 [A*] Steady-state: d/dt[A*] = k 1 [A] 2 – k –1 [A*][A] – k 2 [A*] = 0 Solve for [A*] k –1 [A*][A] + k 2 [A*] = k 1 [A] 2 (k –1 [A] + k 2 )[A*] = k 1 [A] 2 [A*] = k 1 [A] 2 /(k –1 [A] + k 2 ) Plug in to the rate expression Rate = k 2 k 1 [A] 2 /(k –1 [A] + k 2 ) There are two possible extremes with the magnitude of the terms in the denominator. If k –1 [A] >> k 2 , then the denominator ~ k [A]. Then the rate = k 2 k 1 [A] 2 /(k –1 [A]) = k 2 k 1 [A]/(k –1 ) or k 2 K eq [A] which is first-order in [A]. Physically, this means that the rate of the second step is very slow, in comparison to the faster first step. Note that if this 2 nd step is slow (rate-determining), then overall rate = rate of 2 nd step = k 2 [A*] but [A*] from the fast 1 st equilibrium step is given by [A*][A]/[A] 2 = K eq or K eq [A] = [A*]. Plug this in to the rate expression to get rate = k 2 K eq [A], same as the result above from the steady-state approach. For the other extreme, at low pressure (i.e. low [A]), then k –1 [A] << k 2 , so that the denominator ~ k 2 . Then the rate = k 2 k 1 [A] 2 /(k 2 ) = k 1 2 which is second-order in [A]. Physical significance: the slow, rate-determining, step is now the first step, bimolecular formation of A*, and its decomposition (step 2) is fast in comparison. Again writing the overall rate as equal to the rate of the first, slow step: Rate = k 1 [A] 2 which, as it should, matches the steady-state result.
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2 3a. Here's the MO energy level scheme for HF. As described in class, the F 2s is too different in energy from the H 1s for these orbitals to form MOs; the F 2p x and 2p y AOs have the wrong symmetry to combine with the H 1s. Therefore these orbitals are nonbonding (same as they started, unchanged F AOs). The F 2p z and the H 1s AOs combine to form sigma-bonding and sigma-antibonding MOs.
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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Sample3S - 1 Chem 6 Sample exam 3 (final) brief answers 1....

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