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Chapter 12: Chemical Kinetics  Focus on the question – How fast does it happen? o More closely we’ll look at the speeds, or rates at which chemical reactions occur  The area of chemistry concerned with reaction rates and sequence of steps by which reactions occur is called chemical kinetics o Subject of crucial environmental, biological, and economic importance 12.1 Reaction Rates  Chemical reactions differ greatly o Examples:  Combination of Na and Br, occur instantly  Rusting of Fe, imperceptibly slow  To describe a reaction rate quantitatively o Need to specify  How fast the concentration of a reactant or product changes per unit time o Rate =  The reaction rate can be defined either as the increase in the concentration of a product per unit time or as the decrease in the concentration of a reactant per unit time o Example:  2 N O254NO + O 2 2  If we want to find the rate formation of O 2n the decomposition of N O2 5 o 1. Use this equation:  Rate of formation of O =2 = o 2. Plug in the values necessary o 3. Solve o The most common units :  M/s  mol/(L*s) o Defined in terms of concentration because we want the rate to be independent of the scale of the reaction o Don’t forget a minus sign when calculating the rate of disappearance of reactant because reaction rate is defined as a positive quantity  Example:  Rate of decomposition of N O 2 5 = -5 = 1.9 x 10 M/s o Depend on the concentration of at some of the reactants and therefore decrease as the reaction mixture runs out of reactants, as indicated by the decreasing slopes of the curves  The slope of the tangent to a concentration-versus-time curve at a time t is called the instantaneous rate at that particular time  The instantaneous rate at the beginning of a reaction (t = 0) is called the initial rate 12.2 Rate Laws and Reaction Order Ddexxxxdxxedx9drrx9wzxz  The dependence of the reaction rate on the concentration of each reactant is given by an equation called the rate law o Rate = - = k[A] [B] n  k is the proportionality constant called the rate constant  exponents m and n in the rate law indicate how sensitive the rate is to changes in [A] and [B]  Small positive numbers  If m is negative, the rate decreases as [A] increases  If m is zero, the rate is independent of the concentration of A  If m = 1 and [A] is doubled, the rate doubles  If m = 2 and [A] is doubled, [A] quadruples and the rate increases by a factor of 4  The values of the exponents m and determine the reaction order with respect to A and B  The sum of the exponents (m + n) defines the overall reaction order o Example: 2  Rate = k[A] [B]  m = 2, n = 1 so m + n = 2 + 1 = 3 12.3 Experimental Determination of a Rate Law  Methods to determine the values of the exponents in a rate law o The method of initial rates  Carry out a series of experiments in which the initial rate of a reaction is measured as a function of different sets of initial concentrations  Ways to determine rate law: o Establish the reaction order o Evaluate the numerical value of the rate constant k  Rate = k  zeroth order  Rate = k[A]  first order  Rate = k[A][B]  second order 2  Rate = k[A][B]  third order  The rate depends on the concentrations, whereas the rate constant does not  The rate is usually expressed in units of M/s, units of the rate constant depend on the overall reaction order  Example: o Initial rate date for the decomposition of gaseous N O a2 55 degrees are as follows: Experiment Initial N2O 5 Initial rate of Decomposition -5 1 .020 3.4 x 10 2 .050 8.5 x 10 -5 a) What is the rate? Compare experiments 1 and 2 and when we do find that both the initial N O and initial rate of decomposition come out to be 2.5 2 5 and with this we can say the rate is proportional to the concentration of N O2a5d therefore the rate law is of first order Rate = - = k [N 2 ]5m b) What is the value of the rate constant? -3 -1 Solve by substituting k = = = 1.7 x 10 s c) What is the initial rate of decomposition of N O2at555 degrees when its initial concentration is .030M? Substitute initial concentration of .030 M and rate constant from part b -5 Rate = - = k [N 2 ]5= (.030 M) = 5.1 x 10 M/s 12.4 Integrated Rate Law  A first order reaction is one whose rate depends on the concentration of a single reactant raised to the first power o Rate = - = k[A]  With calculus it’s possible to convert the rate law to another form, integrated rate law o ln = - kt  ln = natural log  [A] = concentration of A at some initial time 0  [A] =tconcentration of A at any time t thereafter  = fraction of A that remains at time t o Concentration-time equation that lets us calculate the concentration of A or the fraction of A that remains at any time t o Used to calculate the time required for the initial concentration of A to drop to any particular value or to any
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