CHEM 108 1st Edition Lecture 11Chapter 14: Chemical Kinetics Continued:Method of determining initial rate:- To find the order in NO, we will first identify two experiments in which the concentrationof O2 remains constant. Then, we will divide the rate laws for those two experiments. Finally, we will solve the equation to find the order in NO.- To find the order in O2, we will repeat this procedure, this time choosing experiments in which the concentration of NO remains constant. - Once the order in NO and the order in O2 are known, data from any experiment can be substituted into the rate law. The rate law can then be solved for the rate constant k.- Ex: Experiment [NO] (M) [O2] (M) Initial reaction rate1 0.1 0.1 1e-62 0.1 0.05 0.5e-63 0.05 0.1 2.5e-7[NO] remains constant in experiments 1 and 2. The generic rate law is rate = k[NO]m [O2]n. We will put the result from experiment 1 in the numerator because its rate is larger than that of experiment 2, and use that to find the order of [O2]- (Rate 1/Rate 2)= (k [NO]n1[O2]m1)/([NO]n2[O2]m2)- (1e-6)/(0.5e-6)= (0.1n *0.1m)/(0.1n *0.05m) - 2= 2m log2=m log2 m=1 [O2] is first order- We can just repeat that process to find the order of [NO]Integrated Rate Law for first order reactions:- Mathematical expression describing the change in concentration of a reactant with time.- Ex: O3(g) O2(g) + O(g)o You can find Rate law through integration of the regular rate lawo Ln([O3]/[O3]0)= -kto initial conditions: [A] at t = 0 is [A]0o OrLn([A]/[A]0)= -kto We can rearrange this equation: ln[A]= -kt + ln[A]0(y = mx + b form of equation)o Also rearrange it into this: [A]=[A]0e-ktIntegrated Rate law for second order reaction:- Reaction: N2 (g) + O2 (g) → 2 NO(g)- Rate law: (-Δ[NO])/Δt = k[NO]2- Integrated Rate law: (1/[NO])= kt + (1/[NO]0)(y = mx + b form of equation)- General law: (1/[A])= kt + (1/[A]0)Integrated Rate law for Third Order of reaction:- Rate law = -k[A]3- Integrated Rate law = (1/[A]2)= kt + (1/[A]20)(y = mx + b form of equation)Other Orders of reaction:- Pseudo-first-order: All reactants but one are present at such high concentrations that they do not change significantly during the course of reaction; reaction rate determined by the concentration of the limiting reactant. - It is considered pseudo-first-order reaction because the reaction appears to obey first order kinetics, and has the same integrated rate law as the first order of reaction, but the rate depends on the concentration of more than one reactanto Rate law: Rate = k[A][B]- Zero-order: Rate of reaction is independent of concentration. o Rate = −Δ[A]/Δt = k[A]0 = k. o Integrated rate law: [A] = −kt + [A]0o Half-life: t1/2 = [A]0 /2kReaction Half-Lives:- The time required for concentration of reactant to decrease by half. Half-life is inversely related to the rate constant of a reaction: the higher the reaction rate, the shorter the half-life.- From integrated rate law, when [A]t = ½ [A]0- Ln([A]/[A]0)= -kt = -0.693- Rearrange the equation: t1/2= 0.693/k- t1/2is always constant, so whenever a substance’s concentration is decreased by half it willhappen again in the same amount of time. This is due to the fact that the rate constant is always constant in a reaction.- Second order of reaction: t1/2=
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