CHAPTER 141. Factors that affect rate of reactions:a. Concentration – more particles are present, meaning more collisions to form productsb. Surface area (physical state) – the more room there is to move, the more particle collisions there will be to form productsc. Catalyst – lowers the activation energy by making collisions that form products more likelyd. Temperature – the higher the temperature, the more the particles move, and the more likely it is for particles to collide and form products.2. Rate expressions:a. Rate of disappearance = −∆ [ A]∆t = −[A]final−[A]initialt final−t initial (change in concentration of reactant divided by the change in time).b. Rate of appearance = ∆[ B]∆ t = [B]final−[B]initialt final−t initial (change in concentration of product divided by the change in time).c. Instantaneous rate = the rate of a reaction at a specific point in time.Initial rate = the rate of a reaction when t=0d. Rate = k[chemical]k = rate constant = rate[chemical] (rate divided by the concentration of the chemical observed)***if there is a graph provided, the rate constant = the slope of the line***e. Example of reaction rates and stoichiometry:2ABtwo moles of A disappear for every one mole of B.Therefore…− ∆[A]∆ t=2∆[B]∆ t… because the rate at which [A] decreases is twice that of the rate at which [B] increases.For any reaction, aA + bB  cC +dD, the rate expression israte = −1a∆[A]∆ t=− 1b∆[B]∆ t=1c∆[C]∆ t=1d∆[ D]∆ tf. Rate expression is used to determine how the reaction of a reactant orproduct varies over time.3. Rate Laws:a. For the generic equation aA + bB  cC + dDthe rate law is: rate = k[A]x[B]ywhere x and y are the reaction orders (can be 0, 1, or 2). The whole reaction will be (x+y)th overall.b. Rate law is used to determine how the rate of a reaction changes as we change the concentration of either the products or reactants.c. *** RATE EXPRESSION ≠ RATE LAW ***d. Varying unites depending on order of the overall reaction:i. 1st order: rate = k[A] k has units of 1/s or s-1ii. 2nd order: rate = k[A]2 or k[A][B] k has units of 1/M*se. Determining “order”:For the reaction: aA + bB  (products)Pick 2 rows where [B] is constantEvidence of first order:If [A] is…DOUBLED and the rate increases by 2, the order for the rectant A is 1st.TRIPLED and the rate increases by 3, the order for reactant A is1st.QUADRUPLED and the rate increases by 4, the order for reactant A is 1st.Evidence of second order:If [A] is…DOUBLED and the rate increases by 4, the order for reactant A is 2nd.TRIPLED and the rate increases by 9, the order for reactant A is2nd.QUADRUPLED and the rate increases by 16, the order for reactant A is 2nd.4. Concentration-Time Calculations:a. 1st Order Integrated Rate Law:ln[A]t = -kt + ln[A]0 where [A]t is the concentration of A at any time t y = mx + b where [A]0 is the concentration of A at t=0b. If you plot reaction times t versus thenatural log of reactant concentration, and a straight line with a negative slope is obtained, then the reaction is 1st order.c. 2nd Order Integrated Rate Law: 1[A]t=kt+1[A]0 y = mx + bd. Zero Order Reactions:[A]t = -kt+[A]05. Half-life:Definition – the time it takes for the concentration of a reactant to decrease by half of its initial concentration.t1/2 = t when [A]t = [A]0/2t1/2 for a first-order reaction = .693kNote that the half-life for a first order reaction is concentration independent.Also:[A] = [A]0/2nwhere n = # of half-lives.t1/2 for a second-order reaction = 1k [ A ]06. Rate-Determining Steps (and the misc. things related)a. Rate-determining step is the slowest step in the sequence of stepsb. The rate-determining step SHOULD predict the rate law of the overall reaction, but only if the FIRST elementary step is the slow step.Example:Step 1: A + B  F SlowStep 2: B + F  2C Fast A + 2B2CThe rate law is: rate = k[A][B]Note that the rate law for the overall reaction conforms to the molecularity of Step 1.When the slow step is not the first step in the reaction mechanism, therate law for the overall reaction is not written directly from molecularity.Example:Step 1: NO(g) + Cl2(g)  NOCl2(g) fastStep 2: NOCl2 (g) + NO(g)  2NOCl(g) slowOverall: 2NO(g) + Cl2(g)  2NOCl(g)If we did it based on the slow step, then rate = k[NOCl2][NO]. But that’s not possible because NOCl2 is an intermediate.Therefore the rate = k[NO]2[Cl2]c. Intermediate – a chemical that is cancelled out in the elementary stepswhen calculating the overall reactiond. Catalyst – a chemical that is in both the reactants and products in the net equation.7. Chemical Relationships:a. Arrhenius Equation: k = A e−EaRTk is the rate constantEa is the activation energyR is the gas constant (8.314 J/K-mol)T is temperature in Kelvin (K)A is a frequency factor, meaning the probability that a reaction will occurIMPORTANT: As Ea (the hill on the graph) increases, k decreases, meaning the reaction goes slower. As temperature T increases, k increases and the reaction goes