Chemical Reaction Rates: Calculations and Interpretations, Assignments of Chemistry

Download Chemical Reaction Rates: Calculations and Interpretations and more Assignments Chemistry in PDF only on Docsity! CHAPTER 14 HOMEWORK SOLUTIONS 1/31/05 6. 0 .5 0 .7 0 .9 1 .1 1 .3 1 .5 1 .7 1 .9 2 .1 0 1 0 0 2 0 0 3 0 0 4 0 0 t im e ( m in ) [H C l] for rate at t = 0 min – 54.0 min: rate = - ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ s 60 min 1 min 0 -min 4.05 1.85 - 58.1 MM = 8.33 x 10-5 M/s = slope The other four rates are calculated similarly. t = 54 min – 107 min: 6.92 x 10-5 M/s t = 107 min – 215 min: 5.25 x 10-5 M/s t = 215 min – 430 min: 3.4x 10-5 M/s 10. a) t ]Br[ t ]H[ t HBr][ 2 1 22 ∆ ∆ = ∆ ∆ = ∆ ∆ − b) t ]SO[ 2 1 t ]O[ t ]SO[ 2 1 322 ∆ ∆ = ∆ ∆ −= ∆ ∆ − c) t O]H[ 2 1 t ]N[ t ]H[ 2 1 t NO][ 2 1 222 ∆ ∆ = ∆ ∆ = ∆ ∆ −= ∆ ∆ − 16. a) rate = k[H2][NO]2 b) rate = (6.0 x 104 M-2s-1)(0.050 M)2(0.010 M) = 1.5 Ms-1 c) rate = (6.0 x 104 M-2s-1)(0.10 M)2(0.010 M) = 6.0 Ms-1 18. a) 1.7 x 10-7 Ms-1 = k(0.0477 M)(0.100 M) k = 3.6 x 10-5 M-1s-1 b) M-1s-1 c) It would be quartered. Doubling the solvent volume would halve the concentrations of each reactant. (Math shown below. d = dilute, c = concentrated) [C6H5OH]d = ½ [C6H5OH]c and [OH-]d = ½ [OH-]c ratec = k[C6H5OH]c[OH-]c rated = k[C6H5OH]d[OH-]d = k(½[C6H5OH]c)(½ [OH-]c) = ¼ k[C6H5OH]c[OH-]c = ¼ ratec 24. a) yx yx k k 2323 1313 2 1 ]NH[][BF ]NH[][BF rate rate = /s 0.1065 /s .21300 ) (0.125) (0.250 ) (0.250) (0.250 rate rate yx yx 2 1 M M MMk MMk == y ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= M0.125 M0.250 rate rate 2 1 = 2 y = 1 Choosing 2 experiments from 3, 4, and 5 and performing similar math yields x = 1, so: rate = k[BF3][NH3]. b) reaction order = 1 + 1 = 2 ⇒ second order c) use experiment No. 1 0.2130 Ms-1 = k(0.250 M)(0.250 M) k = 3.41 M-1s-1 30. a) t½ = 1-4- s 10 x .07 693.0 = 940 s b) k = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ s 60 min 1 min3.56 693.0 = 2.0 x 10-4 s-1 32. a) ln[N2O5]2.5min = -(6.82 x 10-3 s-1) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ min s 60 (2.5 min) + ln ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ L 0.2 mol 0250.0 [N2O5]2.5min = 4.5 x 10-3 M mol2.5min = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ L mol 0045.0 (2.0 L) = 0.0090 molN2O5 b) ln ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ L 0.2 mol 010.0 = -(6.82 x 10-3 s-1)t + ln ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ L 0.2 mol 0250.0 t = 130 s 34. ln(PCH3NC) t ln(PCH3NC) t 6.219 0 4.559 8,000 5.814 2,000 3.731 12,000 5.193 5,000 3.109 15,000 slope = s 0 - s 15,000 6.219 - 3.109 = 2.07 x 10-4 s-1 k = -slope = 2.07 x 10-4 s-1 t½ = 1-4- s 10x 2.07 693.0 = 3.35 x 103 s