Question

In: Chemistry

Consider this initial-rate data at a certain temperature for the reaction described by :                           &nbsp

Consider this initial-rate data at a certain temperature for the reaction described by :                              OH-(aq)

                                                                                                                                     OCl-(aq) + I- (aq) --------------> OI-(aq) + Cl-(aq)

Trial                    [OCl-]0 (M)               [I-]0 (M)                 [OH-]0 (M)                     Initial rated (M/s)

1                           0.00161                  0.00161                    0.530                             0.000335

2                           0.00161                  0.00301                    0.530                             0.000626

3                           0.00279                  0.00161                    0.710                             0.000433

4                           0.00161                  0.00301                    0.880                             0.000377

Determine the tate law and the value of the rate constant for this reaction.

Solutions

Expert Solution

Answer –Given, reaction – OH-(aq) + OCl-(aq) + I- (aq) -----> OI-(aq) + Cl-(aq)

Assume rate law is - Rate = k [OH-]x [OCl-]y [I-]z

In this rate law there are x,y and z are the order with respect to OH-, OCl- and I-

Rate1 = k [OH-]1x [OCl-]1y [I-]1z

Rate2 = k [OH-]2x [OCl-]2y [I-]2z

Rate3 = k [OH-]3x [OCl-]3y [I-]3z

Rate4 = k [OH-]4x [OCl-]4y [I-]4z

Order with respect to I-

Rate2/ Rate1 = k [OH-]2x [OCl-]2y [I-]2z / k [OH-]1x [OCl-]1y [I-]1z

0.000626 / 0.000335 = (0.530)x /(0.530)x * (0.00161)y /(0.00161)y *(0.00301)z /(0.00161)z

   1.86 = (1.86)z

So, z = 1

Order with respect to OH-

Rate4/ Rat2 = k [OH-]4x [OCl-]4y [I-]4z / k [OH-]2x [OCl-]2y [I-]2z

0.000377 / 0.000626 = (0.880)x /(0.530)x * (0.00161)y /(0.00161)y *(0.00301)z /(0.00301)z

    0.602 = (1.66)x

So, x = -1

Order with respect to OCl-

Rate2/ Rate1 = k [OH-]2x [OCl-]2y [I-]2z / k [OH-]1x [OCl-]1y [I-]1z

0.000626 / 0.000335 = (0.530)-1 /(0.530)-1 * (0.00161)y /(0.00161)y *(0.00301)1 /(0.00161)1

   1.86 = (1)y (1.86)

(1)y = 1.86 /1.86

        =1

y = 0

So the order with respect to OH-, OCl- and I-are -1, 0 and 1 respectively.

Overall order of reaction = -1+ 0 +1 = 0

So, rate law

Rate = k [OH-]-1 [I-]

Now we need to put the values and calculate k

0.000335 Ms-1 = k (0.530 M)-1 *(0.00161)

0.000335 Ms-1 = k*0.00304

k = 0.000335 Ms-1 /0.00304

k = 0.110 M s-1

queen_honey_blossom answered 1 year ago
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