CHM 202 F-19 1. Use the data in Table 13.3, M&F 7th edn, p.500, and the rate law for the reaction: 2 NO(g) + O(g) → NO(g) to calculate a value for the rate constant, k, for each expt and an average value. You must give the units for k. The rate law is: Rate [NO][O₂]. Ans. 2. NO ONO to calculate a value for the mte constant, k. for each ex pt and an average value. You must give the units for A The rate law Ans. 2. On what factor ick dependent? Ans. Rate Ol 13.3 METHOD OF INITIAL RATES: EXPERIMENTA DETERMINATION OF A RATE LAW One method of determining the values of the exponents in a rate low-the mak rates-is to carry out a series of experiments in which the initial rate of way as a function of different sets of initial concentrations. Take, for example, he us nitric oxide in air, one of the reactions that contributes to the formation of wh F - 250/2) 2 NO(g) + O(z) Some initial rate data are collected in TABLE 13.3 TABLE 13.3 Initial Concentration and Rate Date for the are 2 NO(g) + O(g) 2 NO(g) Experiment Initial [NO] Initial (O₂) (4/9) Initial Reaction 0.015 0.015 0.026 1 0.030 0.015 0096 2 0.015 0.030 0.048 3 0.030 0.030 0.192 4 Note that pairs of experiments are designed to investigate the change in initiate occurs when the initial concentration of a single reactant is changed. In the fira says ments, for example, the concentration of NO is doubled from 0.015 M to 0.039 M velet concentration of O, is held constant. The initial rate increases by a factor of 4, from 9 to 0.096 M/s, indicating that the rate depends on the concentration of NO squa When [NO] is held constant and [02] is doubled (experiments 1 and 3), the initial bles from 0.024 M/s to 0.048 M/s, indicating that the rate depends on the concent O, to the first power, [O]'. Therefore, the rate law for the reaction is Rate k[NO][O₂] In accord with this rate law, which is second order in NO, first order in O, and the overall, the initial rate increases by a factor of 8 when the concentrations of both NO are doubled (experiments 1 and 4). The preceding method uses initial rates rather than rates at a later stage of the ma because chemical reactions are reversible and we want to avoid complications from fe verse reaction: reactants products. As the product concentrations build up, the rate t reverse reaction increases and the measured rate is affected by the concentrations of b reactants and products. At the beginning of the reaction, however, the product co tions are zero, and therefore the products can't affect the measured rate. When we an initial rate, we are measuring the rate of only the forward reaction, so only reactam a catalysts; see Section 13.12) can appear in the rate law. One step in determining a rate law, as we've just seen, is to establish the react Another is to evaluate the numerical value of the rate constant k. Each reaction has characteristic value of the rate constant, which depends on temperature but not a trations. To evaluate k for the reaction 2 NO(g) + O2(g) 2 NO2(g) we can use the data from any one of the experiments in Table 13.3. Solving the rate law fir and substituting the initial rate and concentrations from the first experiment, w Rate k [NO][O] we obtain 0.024 M/s (0.015 M)²(0.015 M) = 7.1 X 10/(M²-s) within the range of experimental error. Note that the units of k in this example are Try repeating the calculation for experiments 2-4, and show that you get the same val read as "one over molar squared overall order f Dete Eape STRA T 59 Chemical Kic 13.3 METHOD OF INITIAL RATES: ERIMENTAL DETERMINATION OF A RATE LA One method of determining the values of the exponents in e law-the method of rate-is to carry out a series of experiments in which the initial site of a reaction is me as a function of different sets of initial concentrations. Take, example, the e nitric oxide in air, one of the reactions that contributes to the formation of acid rain 2 NO(g) + O(g) Some initial rate data are collected in TABLE 13.3. - 2 NO(g) TABLE 13.3 Initial Concentration and Rate Data for the Reaction 2 NO(g) + O(g)-2 NO₂(9) oxidation Experiment Initial [NO] Initial [0,] (M/s) Initial Reaction Rate 0.015 0.024 0.015 0.015 0.096 0.030 2 0.030 0.048 3 0.015 0.030 0.192 4 0.030 Note that pairs of experiments are designed to investigate the change in initial rate th ments, for example, the concentration of NO is doubled from 0.015 M to 0.030 M while the occurs when the initial concentration of a single reactant is changed. In the first two exper concentration of O, is held constant. The initial rate increases by a factor of 4, from 0.024 M When NO is held constant and [O,] is doubled (experiments 1 and 3), the initial rate dou to 0.096 M/s, indicating that the rate depends on the concentration of NO squared, NO bles from 0.024 M/s to 0.048 M/s, indicating that the rate depends on the concentration of O, to the first power, [O]'. Therefore, the rate law for the reaction is Rate = k[NO][O] In accord with this rate law, which is second order in NO, first order in O2, and third order overall, the initial rate increases by a factor of 8 when the concentrations of both NO and 0, are doubled (experiments 1 and 4). The preceding method uses initial rates rather than rates at a later stage of the reaction because chemical reactions are reversible and we want to avoid complications from the re- verse reaction: reactants products. As the product concentrations build up, the rate of the reverse reaction increases and the measured rate is affected by the concentrations of both reactants and products. At the beginning of the reaction, however, the product concentra tions are zero, and therefore the products can't affect the measured rate. When we measure an initial rate, we are measuring the rate of only the forward reaction, so only reactants (and catalysts; see Section 13.12) can appear in the rate law. One step in determining a rate law, as we've just seen, is to establish the reaction order. Another is to evaluate the numerical value of the rate constant k. Each reaction has its own characteristic value of the rate constant, which depends on temperature but not on concen trations. To evaluate k for the reaction 2 NO(g) + O2(g) 2 NO2(g) we can use the data from any one of the experiments in Table 13.3. Solving the rate law for k and substituting the initial rate and concentrations from the first experiment, we obtain km Rate [NO][O] 0.024 M/s (0.015 M) (0.015 M) = 7.1 x 10³/(M².s) within the range of experimental error. Note that the units of k in this example are 1/(Ms) Try repeating the calculation for experiments 2-4, and show that you get the same value of k read as "one over molar squared second." The units of k depend on the overall order of the reaction can general sed Units Rate Law RateA Rate A Rate-A Rate K[A Be carefu concentratio, The rate is u on the overal Worked initial rates. w Deterr Initial ra Initial ra Exper (a) V (b) V (c) STR (a) Ha