Page 144 - Materials Chemistry, Second Edition
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Mass-Balance Concept and Reactor Design                          127



                 1000



                 Concentration (mg/kg)  100





                   10



                    1
                     0        1        2        3       4        5        6
                                             Time (hr)

           FIGURE 4.1
           Concentration versus time.


           Use these data to determine the reaction order and the value of the rate
           constant.

              Strategy:
              To determine the reaction order, a trial-and-error approach is often
                taken. From Table 4.1, if it is a zeroth-order reaction, the plot of con-
                centration versus time should be a straight line. The plot of ln(C)
                versus time should be a straight line if it is first-order kinetics. If it is
                second-order, the plot of (1/C) versus time will be a straight line. The
                value of k is then obtained from the slope of the straight line.
              Solution:
              Since many reactions of environmental concern are first-order reac-
                tions, first assume that it is first-order and plot the concentration-
                time data on a semilog scale (Figure 4.1).
              A straight line fits the data very well, so the assumption of first-order
                kinetics is valid. The slope of the straight line can be determined as
                0.263/h. It should be noted that the rate constant in Equation (4.11)
                is based on exponential with base  e, and the plot in the figure is
                on log . Consequently, the value of k to be used in Equation (4.11)
                      10
                should be the product of the slope from the semilog  plot and 2.303
                                                              10
                (which is the natural log of 10). That is
                                  k = (0.263)(2.303) = 0.606/h

                   See Figure 4.1.
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