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3 4 0.20 0.10 0.20 3.2*10^-3 Trial #1 Measured initial rate -/t (mol/L.S) 1 0.10

ID: 677672 • Letter: 3

Question

3

4

0.20

0.10

0.20

3.2*10^-3

Trial #1 Measured initial rate
-/t (mol/L.S) 1 0.10            0.10 0.10 7.9 * 10^-4 2 0.20 0.10 0.10 1.7*10^-3

3

4

0.20 0.10

0.20

0.10

0.10

0.20

3.1*10^-3

3.2*10^-3

The reaction between bromate ions and bromide ions occursaccording to the equation shown below. The following table shows the results of four experiments doneon this system. Use these data to determine the order of reactionwith respect to each reactant, the overall order of reaction, andthe value of the rate constant.

Explanation / Answer

   We Know that :     The Rate of the equation depends on theconcentration of the reactants.                 Rate = K [ BrO3- ]a [ Br- ] b [H+ ] c           Fromthe first two trials :          ( 1 / 2)a    = 0.5                bases are equal so exponents should also be equal.                        a = 1               From the second & third trials :                   ( 1 / 2 )b   =   0.5                     basesare equal so exponents should also be equal.                                          by  trial 4/trial 1  =  3.2*10^-3 mol/L.S   / 7.9 * 10^-4mol/L.S                        =(0.2/0.1)c            (2)c      = (4)                       = (2)2                   Z = 2 The reaction is first order wrt toBrO3- , first order wrt to Br- ,second order wrt H+ . The reaction is a 4th order reaction . rate = k[BrO3-][Br-][H+]2 7.9 * 10^-4 mol/L.S = k ( 0.10 M) ( 0.10 M)(0.10)2                             k   = 7.9 * 10^-4 mol/L.S / ( 0.10 M) ( 0.10 M)(0.10)2                                   =7.9 L3 / mol3 .s                        =(0.2/0.1)c            (2)c      = (4)                       = (2)2                   Z = 2 The reaction is first order wrt toBrO3- , first order wrt to Br- ,second order wrt H+ . The reaction is a 4th order reaction . rate = k[BrO3-][Br-][H+]2 7.9 * 10^-4 mol/L.S = k ( 0.10 M) ( 0.10 M)(0.10)2                             k   = 7.9 * 10^-4 mol/L.S / ( 0.10 M) ( 0.10 M)(0.10)2                                   =7.9 L3 / mol3 .s                             k   = 7.9 * 10^-4 mol/L.S / ( 0.10 M) ( 0.10 M)(0.10)2                                   =7.9 L3 / mol3 .s