Hint - Start by focusing on sets of trials in which only one concentration chang

Question

Hint - Start by focusing on sets of trials in which only one concentration changes and the others are held constant. Rate laws take the form Where A, B, and C are reactants raised to their respective orders (x, y, or z) and k is a rate constant unique to the reaction.

Consider this initial-rate data at a certain temperature for the reaction described by OH (aq Trial Initial rates (M/s) 0.000404 0.000747 0.000559 0.000415 Determine the rate law and the value of the rate constant for this reaction. 0.00191 0.00191 0.00357 0.00191 0.00191 0.00353 0.510 0.00191 0.00353 0.918 0.510 2 rate -k 0.689 OH] [OH- Number Units Select answer

Explanation / Answer

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

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

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

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

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

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

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

Order with respect to I-

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

0.000747 / 0.000404 = (0.00191)x /(0.00191)x * (0.00353)y /(0.00191)y *(0.00510)z /(0.00510)z

   1.84 = (1.84)y

So, y = 1

Order with respect to OH-

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

0.000415 / 0.000747 = (0.00191)x /(0.00191)x * (0.00353)y /(0.00353)y *(0.918)z /(0.510)z

    0.56 = (1.8)z

So, z = -1

Order with respect to OCl-

Rate3/ Rate2 = k [OCl -]3x [I-]3y [OH -]3z / k [OCl -]2x [I-]2y [OH -]2z

0.000559 / 0.000747 = (0.00357)x /(0.00191)x * (0.00191) /(0.00353) *(0.689)-1 /(0.510)-1

   0.75 = (1.86)x (0.54)(1.35)

(1.86)x = 1

x = 0

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

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

So, rate law

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

Now we need to put the values and calculate k

0.000335 Ms-1 = k (0.00191 M) / (0.510M)

0.000404 Ms-1 = k*0.00304

k = 0.000404 Ms-1 /0.00374

k = 0.108 M s-1

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