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Lecture 11 - Chapter 16 Rates& Mechanisms of...

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Unformatted text preview: Chapter 16 Rates & Mechanisms of Chemical Reac6ons Kine6cs: Rates and Mechanisms of Chemical Reac6ons 16.1 Factors That Influence Reac6on Rates 16.2 Expressing the Reac6on Rate 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Concentra6on Changes over Time 16.5 The Effect of Temperature on Reac6on Rate 16.6 Explaining the Effects of Concentra6on and Temperature 16.7 Reac6on Mechanisms: Steps in the Overall Reac6on 16.8 Catalysis: Speeding Up a Chemical Reac6on Figure 16.1 Reac6on rate: the central focus of chemical kine6cs. Concentra6on of the reactant decreases with concomitant increase of the product over 6me Figure 16.4 Plots of [C2H4] and [O2] vs. 6me. Factors That Influence Reac6on Rate Under a specific set of condi6ons, every reac6on has its own characteris6c rate, which depends upon the chemical nature of the reactants. Four factors can be controlled during the reac6on: 1.  2.  3.  4.  Concentra6on  ­ molecules must collide to react Physical state  ­ molecules must mix to collide Temperature  ­ molecules must collide with enough energy to react The use of a catalyst Figure 16.2 Collision energy and reac6on rate. Expressing the Reac6on Rate Reac6on rate  ­ changes in the concentra6ons of reactants or products per unit of 6me. Reactant concentra6ons decrease while product concentra6ons increase. General reac6on, Rate of reac6on =  ­ A B change in concentra6on of A change in 6me  ­ Δ (conc A) Δt =  ­ conc A2  ­ conc A1 t2  ­ t1 Figure 16.3 The concentra6on of O3 vs. 6me during its reac6on with C2H4. rate =  ­  ­ Δ [C2H4] Δt Δ [O3] Δt = Figure 16.4 Plots of [C2H4] and [O2] vs. 6me. In general, for the reac6on, aA + bB cC + dD Rate =  ­ 1 Δ[A] a Δt =  ­ 1 Δ[B] b Δt = + 1 Δ[C] c Δt = + 1 Δ[D] d Δt The numerical value of the rate depends upon the substance that serves as the reference. The changes in concentra6on of the other reac6on components are rela6ve to their coefficients in the balanced chemical equa6on. Sample Problem 16.1 PROBLEM: Expressing Rate in Terms of Changes in Concentra6on with Time Because it has a nonpollu6ng product (water vapor), hydrogen gas is used for fuel aboard the space shuele and in prototype cars with Earth ­bound engines: 2H2(g) + O2(g) 2H2O(g) (a) Express the rate in terms of changes in [H2], [O2], and [H2O] with 6me. (b) When [O2] is decreasing at 0.23 mol/L•s, at what rate is [H2O] increasing? PLAN: Choose [O2] as a point of reference since its coefficient is 1. For every molecule of O2 which disappears, 2 molecules of H2 disappear and 2 molecules of H2O appear, so [O2] is disappearing at half the rate of change of H2 and H2O. SOLUTION: (a) (b)  ­ Δ[O2] Δt 1 rate =  ­ 2 =  ­ 0.23 mol/L•s Δ[H2] Δ[O2] 1 Δ[H2O] =  ­ = + 2 Δt Δt Δt Δ[H2O] Δ[H O] = + 1 ; 2 = 0.46 mol/L•s 2 Δt Δt ...
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