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

OCl−+I−→OI−+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

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

Answer :

Answer:

a) The rate law for this reaction:

[tex]R=k[OCl^-]^1\times [I^-]^1[/tex]

b)The rate constant of the reaction is [tex]60.44 M^{-1}s^{-1}[/tex].

c)The rate of the reaction at given concentration will be [tex]6.528\times 10^9 M/s[/tex].

Explanation:

[tex]OCl^-+I^-\rightarrow OI^-+Cl^-[/tex]

a) Rate law of the reaction is given by :

[tex]R=k[OCl^-]^x\times [I^-]^y[/tex]

1) Rate of the reaction when [tex][OCl^-]=1.5\times 10^{-3} M[/tex] and [tex][I^-]=1.5\times 10^{-3} M[/tex].

[tex]1.36\times 10^{-4} M/s=k[1.5\times 10^{-3} M]^x\times [1.5\times 10^{-3} M]^y[/tex]..[1]

2) Rate of the reaction when [tex][OCl^-]=3.0\times 10^{-3} M[/tex] and [tex][I^-]=1.5\times 10^{-3} M[/tex].

[tex]2.72\times 10^{-4}M/s=k[3.0\times 10^{-3} M]^x\times [1.5\times 10^{-3} M]^y[/tex]..[2]

3) Rate of the reaction when [tex][OCl^-]=1.5\times 10^{-3} M[/tex] and [tex][I^-]=3.0\times 10^{-3} M[/tex].

[tex]2.72\times 10^{-4} M/s=k[1.5\times 10^{-3} M]^x\times [3.0\times 10^{-3} M]^y[/tex]..[3]

[1] ÷ [2]

[tex]\frac{1.36\times 10^{-4}M/s}{2.72\times 10^{-4}M/s}=\frac{k[1.5\times 10^{-3} M]^x\times [1.5\times 10^{-3} M]^y}{k[3.0\times 10^{-3} M]^x\times [1.5\times 10^{-3} M]^y}[/tex]

On solving the equation we get value of x :

x = 1

[1] ÷ [3]

[tex]\frac{1.36\times 10^{-4} M/s}{2.72\times 10^{-4} M/s}=\frac{k[1.5\times 10^{-3} M]^x\times [1.5\times 10^{-3} M]^y}{k[1.5\times 10^{-3} M]^x\times [3.0\times 10^{-3} M]^y}[/tex]

On solving the equation we get value of y :

y = 1

The rate law for this reaction:

[tex]R=k[OCl^-]^1\times [I^-]^1[/tex]

b)

Rate of the reaction when [tex][OCl^-]=1.5\times 10^{-3} M[/tex] and [tex][I^-]=1.5\times 10^{-3} M[/tex].

[tex]1.36\times 10^{-4} M/s=k[1.5\times 10^{-3} M]\times [1.5\times 10^{-3} M][/tex]

[tex]k=\frac{1.36\times 10^{-4} M/s}{[1.5\times 10^{-3} M]\times [1.5\times 10^{-3} M]}=60.44 M^{-1}s^{-1}[/tex]

The rate constant of the reaction is [tex]60.44 M^{-1}s^{-1}[/tex].

c)

Let the rate of the reaction when [tex][OCl^-]=1.8\times 10^{3} M[/tex] and [tex][I^-]=6.0\times 10^{4} M[/tex] be R.

[tex]R=60.44 M^{-1}s^{-1}\times [1.8\times 10^{3} M]^1\times [6.0\times 10^{4} M]^1[/tex]

[tex]R=6.528\times 10^9 M/s[/tex]

The rate of the reaction at given concentration will be [tex]6.528\times 10^9 M/s[/tex].

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