If the following data was observed for the reaction OCl- + I- à OI- + Cl- ; what will the rate be when [OCl- ] = 2.0×10-3 M and [I- ] = 5.0×10-4 M ?
Exp. 1 ([OCl- ] = 1.5×10-3 M; [I- ] = 1.5×10-3 M; Initial Rate = 1.36×10-4M/s
Exp. 2 ([OCl- ] = 3.0×10-3 M; [I- ] = 1.5×10-3 M; Initial Rate = 2.72×10-4 M/s
Exp. 3 ([OCl- ] = 1.5×10-3 M; [I- ] = 3.0×10-3 M; Initial Rate = 2.72×10-4 M/s
( ) 6.0×10-5 M/s
( ) 2.4×10-4 M/s
( ) 5.7×10-6 M/s
( ) 4.2×10-4 M/s 5.
The reaction A + B à C is found to be zeroth order with respect to A and second ord
Rate will be written as :
r = k*[OCl-]m [I-]n
where k is rate constant of reaction.
From exp 2 and 3 it can be seen that
2.72×10-4 M/s = k * (3.0×10-3 M)m * (1.5×10-3 M)n and,
2.72×10-4 M/s= k* ( 1.5×10-3 M)m * (3.0×10-3 M)n
Dividing these 2 we get,
1 = 2m * 2-n
So, 2m-n = 1 or m-n =0 or m=n
From exp 2 and 1 it can be seen that
2.72×10-4 M/s = k * (3.0×10-3 M)m * (1.5×10-3 M)n and,
1.36×10-4 M/s= k* ( 1.5×10-3 M)m * (1.5×10-3 M)n
Dividing these 2 we get,
2 = 2m
So, m= 1
Consequently Put this in any of the rate equations to obtain k
So, 2.72×10-4 M/s = k * (3.0×10-3 M) * (1.5×10-3 M)
So, k = 60.44 M-1s-1
So, final rate equation is
Rate = 60.44 M-1s-1 * [OCl-] [I-]
In this put [OCl- ] = 2.0×10-3 M and [I- ] = 5.0×10-4 M, you obtain
Rate = 6.0×10-5 M/s (Answer)
If the following data was observed for the reaction OCl- + I- à OI- + Cl-...
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