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Question

The table given below gives kinetic data for the following reaction at 298 K:
OCl+IOI+Cl
Expt.
No.
[OCl]
(in mol dm3)
[I]
(in mol dm3)
[OH]
(in mol dm3)
d[IO]/dt
(in 104 mol dm3 s1)
1.0.00170.0017 1.0 1.75
2.0.00340.0017 1.0 3.50
3.0.00170.0034 1.0 3.50
4.0.00170.0017 0.5 3.50
What is the rate law and what is the value of rate constant?

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Solution

Let the rate law be =k[OCl]x[I]y[OH]z
From expt. (1), 1.75×104=k[0.0017]x[0.0017]y[1.0]z ...(i)
From expt. (2), 3.50×104=k[0.0034]x[0.0017]y[1.0]z ...(ii)
Dividing equation (ii) by equation (i),
3.50×1041.75×104=[0.0034]x[0.0017]x
2=2x
x=1, i.e., first order with respect to OCl
From expt. (1),
1.75×104=k[0.0017]x[0.0017]y[1.0]z ...(i)
From expt. (3),
3.50×104=k[0.0017]x[0.0034]y[1.0]z ...(iii)
Dividing equation (iii) by equation (i),
3.50×1041.75×104=[0.0034]y[0.0017]y
or 2=2y
or y=1, i.e., first order with respect to I
From expt. (1), 1.75×104=k[0.0017]x[0.0017]y[1.0]z ...(i)
From expt. (4),
3.50×104=k[0.0017]x[0.0017]y[0.5]z ...(iv)
Dividing equation (i) by equation (iv),
1.75×1043.50×104=[1.0]z[0.5]z
or 12=2z
or 21=2z
z=1, i.e., first order with respect to OH is 1.
Rate law =k[OCl][I][OH]
From expt. (1) k=1.75×104[OH][OCl][I]=1.75×104×1.00.0017×0.0017
=60.55s1

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