RateLaw - NO and substituting the initial rate and concentrations from the first experiment we obtain = Rate = 0.048 M/s[NOHOZ(0.015 M)2(0.015 M k = 1.4

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Unformatted text preview: NO and substituting the initial rate and concentrations from the first experiment, we obtain = Rate = 0.048 M/s [NOHOZ] (0.015 M)2(0.015 M) k = 1.4 x 104/(M2-s) Try repeating the calculation for experiments 2—4, and show that you get the same value of k. Note that the units of k in this example are 1/(M2 ' 5), read as “one over molar squared second.” In general, the units of I: depend on the number of con- centration terms in the rate law and on the values of their exponents. Units for some common cases are given as follows: Rate Law Overall Reaction Order Units for It Rate = k Zeroth order M/Sm-Ms—l Rate = HA] First order 1 / s or 5—1 Rate = k[A][B] Second order 1,1(M - s) or M_1 5—1 Rate = k[A][B]2 Third order 1f{M2-s) or M‘2s_1 Be careful not to confuse the rate of a reaction and its rate constant. The rate depends on concentrations, whereas the rate constant does not (it is a constant). The rate is usually expressed in units of M/s, whereas the units of the rate constant depend on the overall reaction order. Worked Example 12.3 gives another instance of how a rate law can be determined from initial rates. WORKED EXAMPLE 12.3 DETERMINING A RATE LAW FROM INITIAL RATES Initial rate data for the decomposition of gaseous N205 at 55 °C are as follows: Initial Rate of Decomposition Experiment Initial N205 of N205 (Nils) 1 0.020 3.4 x 10‘5 2 0.050 8.5 x 10'5 {a} What is the rate law? lb} What is the value of the rate constant? (cl What is the initial rate of decomposition of N205 at 55 °C when its initial concen- tration is 0.030 M? ...
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