HELPPPP! I have a quiz on this tomorrow. I-(aq) + CIO-(aq) rightarrow IO-(aq) +

Question

HELPPPP! I have a quiz on this tomorrow.

I-(aq) + CIO-(aq) rightarrow IO-(aq) + CI-(aq) Iodide ion, I-, is oxidized to hypoiodite ion, IO-, by hypochlorite, CIO-, in basic solution according to the equation above. Three initial-rate experiments were conducted; the results shown in the following table. Experiment [I-] (mol L-1) [CIO] (mol L-1) Initial Rate of Formation of IO-(mol L -1 s-1) 1 0.017 0.015 0.156 2 0.052 0.015 0.476 3 0.016 0.061 0.596 Determine the order of the reaction with respect to each reactant listed below. Show your work. I-(aq) CIO-(aq) For the reaction, write the rate law that is consistent with the calculations in part (a); calculate the value of the specific rate constant, k, and specify units.

Explanation / Answer

let the rate equation be

r = K[I-]^a[ClO-]^b

sp from table

0.156 = K*0.017^a*0.015^b......................(1)

0.476 = K*0.052^a*0.015^b......................(2)

0.596 = k*0.016^a*0.061^b......................(3)

dividing (2) by (1)

(0.052/0.017)^a = 0.476/0.156

a = 0.9977 = 1.................order with respect to [I-]

dividing (3) by (1)

b = 0.9986= 1.................order with respect to [CLO-]

so rate = r = K[I-]*[ClO-]......................order = 1

putting values in (1)

K = 0.156/(0.017^a*0.015^b)

K = 611.765 M^-1 s^-1

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