1. Three runs were used to study the reaction: A+B ---> 3C +D The following data

Question

1. Three runs were used to study the reaction:
A+B ---> 3C +D

The following data were obtained (initial concentrations of C and D= 0)

          InitialConcs(M)                               Concs after 150s
Run   [A]                 [B]               [A]            [B]              [C]
1       3.000             2.000           2.970M       ________      ________
2       6.000             2.000           5.940M       ________      ________
3       3.000             4.000           2.880M       ________      ________

(a)Fill in the missing concentrations in the table above.

(b)Calculate the rate of the reaction in M/s for each of the threeruns.

(c) Determine the order with respect to A and B

(d)Calculate the rate constant for the reaction, includingunits

Explanation / Answer

a)                     Initial Concs(M)                               Concs after 150s
      Run   [A]                 [B]               [A]            [B]              [C]
    1       3.000             2.000           2.970M         1.970M    0.030M
     2       6.000             2.000           5.940M        1.940M      0.060M
    3       3.000             4.000           2.880M       3.880M       0.120M b) The rate of reaction in run 1is    - d[A]/ dt   =  - d[B] /dt = (2.970-3.000) M / 150 s                                                                  rate= 2.0*10-4 M / s                 ratein run2  is                  -d[A]/dt   =  - d[B] / dt   = (1.940-2.000)M / 150 s                                                                  rate  = 4.0*10-4 M /s           rate in run 3is                  d[A]/ dt   =  - d[B] / dt   =(3.880-4.000) M / 150 s                                                                                                                 rate   = 8.0*10-4 M / s   c) Order with respect to A is 1 and order with respect to B is 2 .     since from the above calculation we know ,when we double the concentration of A by keeping B as fixed weget doulble the rate in A ,hence it is first order reaction.similarly when we double the concentration of B by keepingA as fixed we get quadraple rate in B , so it is a second orderreaction. d) we have formula for rate constant k =rate / [A][B]                                                                  here we calculate the rate constant k byconsidering any one of the above run for the reaction .    so let we take run concentrations of A , Band thier rate in run 1 .                  concentration of A , B are [A] = 3.000M , [B] = 2.000M                        rate = 2.0*10-4 M / s            Now we substitute all these in equation of rate constant                         k = (2.0*10-4 M / s ) / (3.000M)*(2.000M)                           = 3.3*10-5 M-1 s-1             k  = 3.3*10-5 M-1s-1

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