Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

OCl- + I- → OI-1 +Cl-.
This rapid reaction gives the following rate data:

[OCl-](M) [I]- (M) Rate (M/s)
1.5×10^3 1.5×10^3 1.36×10^4
3.0×10^3 1.5×10^3 2.72×10^4
1.5×10^3 3.0×10^3 2.72×10^4

Write the rate law for this reaction.
Calculate the rate constant with proper units.
Calculate the rate when [OCl-]= 1.8×10^3 M and [I-]= 6.0×10^4 M .

Answer 1

Step-by-step explanation:

(a) As the given chemical reaction equation is as follows.

`OCl^(-) + I^(-) \rightarrow OI^(-1) + Cl^(-1)`

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate =
`K * [OCl^(-)] * [l^(-)]`

(b) Since, the rate equation is as follows.

Rate =
`K [OCl^(-)][l^(-)]`

Let us assume that (
`[OCl^(-)] = [l^(-)]`)

Putting the given values into the above equation as follows.

`1.36 * 10^(-4) = K * (1.5 * 10^(-3))^2`

`1.36 * 10^(-4) = K * (2.25 * 10^(-6))`

K =
`(1.36 * 10^(-4))/(2.25 * 10^(-6))`

=
`60.4 M^(-1)sec^(-1)`

Hence, the value of rate constant for the given reaction is
`60.4 M^(-1)sec^(-1)` .

(c) Now, we will calculate the rate as follows.

Rate =
`K [OCl^(-)][l^(-)]`

=
`60.4 * (1.8 * 10^(3)) * (6.0 * 10^(4))`

=
`6.52 * 10^(5)`

Therefore, rate when
`[OCl^(-)] = 1.8 * 10^(3)` M and
`[I^(-)]= 6.0 * 10^(4)` M is
`6.52 * 10^(5)`.