1 (289)

Prévia do material em texto

Chapter 14 Chemical Kinetics 289 14.95 (a) There are two elementary steps in the reaction mechanism because there are two peaks in the reaction progress diagram. (b) Intermediates Reactants Products Reaction progress (c) The first step is the rate-limiting step because it has the higher activation energy. (d) The overall reaction is exothermic because the products are at a lower energy than the reactants. 14.97 Given: n-butane desorption from single crystal aluminum oxide, first order; k = 0.128 s⁻¹ at 150 K; initially completely covered Find: (a) (b) t for 25% and for 50% to desorb; (c) fraction remaining after 10 S and 20 S Conceptual Plan: (a) k (b) [C₄H₁₀]₀, k t (c) k = + Solution: (a) (b) = kt + [C₄H₁₀] Rearrange to solve for t. For 25% desorbed, = 0.75 [C₄H₁₀]₀ and t = 1 = 1 = 2.2 For 50% desorbed, = 0.50 and t = 1 = (c) For In = kt + C₄H₁₀] = + = 1.28 = = 0.28 = fraction covered For In = -kt + C₄H₁₀]₀ = + = -2.56 = = 0.077 = fraction covered Check: The units (s, none, and none) are correct. The half-life is reasonable considering the size of the rate constant. The time to 25% desorbed is less than one half-life. The time to 50% desorbed is the half-life. The fraction at 10 S is consistent with about two half-lives. The fraction covered at 20 S is consistent with about four half-lives. 14.99 (a) Given: table of rate constant versus T Find: and A Conceptual Plan: First, convert temperature data into kelvin (°C + 273.15 = K). Because In = R + In A, a plot of In k versus VT will have a slope = and an intercept = In A. Solution: The slope can be determined by measuring on the plot or by using functions such as "add trendline" in Excel. Because the slope = 10759 K = = -(slope)R = Copyright © 2017 Pearson Education, Inc.