1 (284)

Prévia do material em texto

284 Chapter 14 Chemical Kinetics 14.67 Given: table of rate constant versus T Find: Ea and A Conceptual Plan: Because In k = + In A, a plot of In k versus 1/T will have a slope = and an intercept = In A. 4 Solution: The slope can be determined by 2 measuring on the plot or by using functions such as "add trendline" in Excel. 0 Because the slope = -2767.2 K = -2 = (slope)R -4 y = 2767.2x + 25.112 = -6 kJ -8 = 23.0 and intercept = 25.112 = In A then 0.008 0.009 0.01 0.011 0.012 1/Temperature (1/K) A = = = 8.05 X 10¹⁰ s⁻¹. Check: The units (kJ/mol and s⁻¹) are correct. The plot was extremely linear, confirming Arrhenius behavior. The activation energy and frequency factor are typical for many reactions. 14.69 Given: rate constant = 0.0117/s at 400.0 K and 0.689/s at 450.0 K Find: (a) and (b) rate constant at 425 K Conceptual Plan: (a) k₁, k₂, T₂, Ea then J/mol kJ/mol 1kJ In = R (b) Eₐ, k₁, T₂ k₂ In Solution: (a) In = R Rearrange to solve for = R In k₁ k₂ = 8.314 400.0 1 J mol In 450.0 0.689 1 = (b) In k₁ = R 1 with k₁, = T₁ = 400.0 K, T₂ = 425 K Rearrange to solve for k₂. J 1.22 X mol 1 1 In = + In 0.0117 = -2.2902 J 400.0 K 425 8.314 K Check: The units (kJ/mol and s⁻¹) are correct. The activation energy is typical for a reaction. The rate constant at 425 K is between the values given at 400 K and 450 K. 14.71 Given: rate constant doubles from 10.0 °C to 20.0 °C Find: Conceptual Plan: °C K then k₁, k₂, T₂ then J/mol kJ/mol 1kJ K = °C + 273.15 In Solution: T₁ = 10.0 °C + 273.15 = 283.2 K and T₂ = 20.0 °C + 273.15 = 293.2 K and k₂ = then In Rearrange to solve for Copyright © 2017 Pearson Education, Inc.