Question

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

Answer 1

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 = 2^m * 2^-n

So, 2^m-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 = 2^m

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)