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MSU CHM 170 - Exam 1 study guide

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CHM 170 1st EditionExam # 1 Study Guide Lectures: 1 - 8Lecture 1 (January 13)For this exam know how to write rate reactions and the different types of rates there are such as average rates, instantaneous rates, and initial rates. Also know the factors that affect the reaction rates such as collision, energy, catalyst, enzymes and whether it is homogeneous or heterogeneous. Chemical Kinetics: the study of reaction rates, including how reaction rates change with varying conditions and which molecular events occur during the overall reaction.Definition of rate: how much a quantity change in a given period of timeDefining reaction rate: the rate of a chemical reaction is measured by how much the concentration of a reactant deceases in a period of time.The formula for reaction rates is: Rates=∆ concentration∆ time– Another way of understanding the formula is Rate=∆[ product]∆ time=−∆[reactant ]∆ timeRate of reaction is always positive. Therefore, the reactants has a negative sign and the products have a positive sign.–The units for the brackets are Molarity (molesliter). Example for expressing rates.–Consider the following balanced reaction aA+ bB ↔ cC + dDoThe rate expression would be:Rate= - 1a∆[A]∆ t=−1b∆[B]∆t=1c∆[C]∆ t=1d∆[D]∆ tIn most reactions, the coefficients of the balanced equation are not all the same. For example: H2 (g) + I2 (g)  2 HI (g)For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another.For the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made, therefore the rate of change will be different.In order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient. For example: Rate =−Δ[ H2]Δt=−Δ[ I2]Δt=+(12)Δ[HI ]Δt Remember that the product side is always positive and the reactant side is negative.Answer the following question:The correct answer should be B.The average rate is the change in measured concentrations in any particular time period.– Linear approximation of a curve– The larger the time interval then the more the average rate deviates from the instantaneous rate.Examples:For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the [H+].H2O2 (aq) + 3 I(aq) + 2 H+(aq)  I3(aq) + 2 H2O(l)The instantaneous rate is the change in concentration at any one particular time.The initial rate is the tangent to the curve at time zero.Example:The answer is D.First write a rate expression then plug the concentration into the formula. Remember that it is final – initial. Factors affecting reaction ratesThe factors that react faster are small molecules, gases which react faster than liquids which react faster than solids, powdered solids are more reactive than blocks, and ions react faster than molecules. Increasing temperature increases reaction rates. Catalyst are substances which affect the speed of a reaction without being consumed. Most catalyst speeds up reactions they are called positive catalyst. Catalyst that slows down reaction are negative catalyst. Homogeneous are present in the same phase while heterogeneous are present in different phase.Basically the larger the concentration of reactant molecules then the faster the reaction. High pressure equals higher concentration.Rate Laws is the rate of the reaction and the concentration of the reactant. For example: for the reaction aA + bB  products the rate law would have the form given belown and m are called the orders for each reactant.k is called the rate constant.– The exponent on each reactant in the rate law is called the order with respect to that reactant. The sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction2 NO(g) + O2(g) ® 2 NO2(g) is Rate = k[NO]2[O2] The reaction is – second order with respect to [NO], – first order with respect to [O2], – third order overall– Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.– Fractions, zero and negative orders are also possible.– The concentration of a reactant with a zero-order dependence has no effect on the rate of the reaction.Lecture 2 (January 15)Know how to write the appropriate units for orders. Rates unit is MS and k (constant) unit is1s.If the concentration is doubled of the reactant then the rate doubles. If both concentration and rate is doubled then it is first order.First order: Rate=k [A]Rate = (1 S-1) [1]1 =1  Rate = (1 S-1) [2]1 =2 (doubled)Rate = (1 S-1) [1]1 =1 Rate = (1 S-1) [0.5]1 =0.5 (halved)The units would be 1sIf the concentration is doubled then the rate quadruple.Second order: Rate=k [ A]2Rate = (1 M-1s-1) [1]2 =1  MS = (1S)(M2)  1S M1 (M)  1S1M1Third order: Rate=k [ A]3Rate=k [2]3 (8x rate of reaction)MS = (1S) (M3)  1S M2 (M)  1S M2If [A] was doubled what effect would it have on the rate of the reaction? (Assume other reactants are held constant)Rate = k [ A ]1[B]2[C ]3– According to the rate law, since A is a first order then it would double the rate.If [B] was doubled what effect would it have on the rate of the reaction? (Assume other reactants are held constant)– Since B is a second order then the rate will quadruple. – Rate = KM3–MS=1S M2 (M)– The inverse would be M−2s−1Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g)  NO(g) + CO2(g) given the data belowFirst determine by what factor the concentrations and rates changes in these two experiements. Then determine what power the concentration factor must be raised to equal the rate factor.Substitute the exponents into the general rate law to get the rate law for the reaction.Substitute the concentrations and rate for any experiment into the rate law and solve for k. don’t forget the units.Lecture 3 (January 20)Know how to use integrated rate laws for first, second and zero order. Know each graph for eachorder and understand the slopes. Know the half-life equation and how to solve a half-life.First order rate equationLn [A]t = -kt + ln [A]o Second order rate equation1[ A ]t−1[A]=kt∨1[A]t=kt1[ A]Zero order rate


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