For the reaction ClO - + I - - -> OI - + Cl - in aqueous solution at 25°C, initi
ID: 993530 • Letter: F
Question
For the reaction ClO- + I- --> OI- + Cl- in aqueous solution at 25°C, initial rates, r0 as a function of initial concentrations (where c° = 1 mol/dm3) are as follows:
103[ClO-]/c° 4.00 2.00 2.00 2.00
103 [I-]/c° 2.00 4.00 2.00 2.00
103 [OH-]/c° 1000 1000 1000 250
103r0/(c° s-1) 0.48 0.50 0.24 0.94
(a) Find the rate law and the rate constant and (b) Devise a mechanism consistent with the observed rate law.
Explanation / Answer
Suppose the order of the reaction w.r.to ClO-(aq), I-(aq) and OH-(aq) are respectively 'x', 'y' and 'z'
Hence rate law of the reaction can be written as
rate = k * [ClO-(aq)]x *[I-(aq)]y * [OH-(aq)]z
For Data - 1:
0.48 = k * [4.00]x *[2.00]y * [1000]z --------- (1)
0.50 = k * [2.00]x *[4.00]y * [1000]z --------- (2)
0.24 = k * [2.00]x *[2.00]y * [1000]z --------- (3)
0.94 = k * [2.00]x *[2.00]y * [250]z --------- (4)
Dividing equation (3) by equation(4)
=> (0.24 / 0.94) = (1000/250)z
=> 4z = 0.255 = 1/4 = (4)-1
=> z = -1
Dividing equation (3) by equation(4)
=> (0.24 / 0.94) = (1000/250)z
=> 4z = 0.255 = 1/4 = (4)-1
=> z = -1
Dividing equation (2) by equation (3) we get
=> (0.50/0.24) = (4.00/2.00)y * (1000/1000)-1
=> 2y = 2.083
=> y = 1 (nearly equals to 1)
Dividing equation (1) by equation (3) we get
=> (0.48/0.24) = (4.00/2.00)x
=> 2x = 2.0
=> x = 1
Hence rate law of the reaction is
rate = k * [ClO-(aq)] *[I-(aq)] / [OH-(aq)] (answer)
where rate constant 'k' can be calculated by putting the values of x, y, and z in any of the equations
=> 0.48 Cos-1 = k * [4.00 Co] *[2.00 Co] / [1000 Co]
=> k = 60 s-1 (answer)
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