Question

odine atoms will combine to form I2 in liquid hexane solvent with a rate constant of 1.5×1010L/mol⋅s. The reaction is second order in I . Since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instantaneously, usually by photochemical decomposition of I2. Suppose a flash of light creates an initial [I] concentration of 2.00×10−2 M .
How long will it take for 94% of the newly created iodine atoms to recombine to form I2? Express your answer using two significant figures.

Answer 1

Answer:

The time taken is  $t = 1.11 *10^{-9} \ s$

Explanation:

From the question we are told that

    The  rate constant is  $k = 1.50 *10^{10} \ L /mol \cdot s$

    The initial concentration of iodine atom  is  $[I] = 2.0*10^{-2} \ M$

Generally the integrated  rate law for a second order reaction is  mathematically represented as

        $\frac{1 }{[I_r]} = \frac{1}{[I]} * k * t$

Where $[I_r]$ is the concentration of the remaining iodine atom  after the recombination which is mathematically evaluated as

      $[I_r ] = [I_o ] *$[100%  - 94%]

The reason for the 94%  is that we are told from the question that only 94% of the iodine atom recombined

=>  $[I_r ] = [I_o ] *$[6%]

=>  $[I_r ] = [I_o ] *0.06$

substituting values

    $[I_r ] = 2.0 *10^{-2}*0.06$

    $[I_r ] = 1.2 *10^{-3}$

So

    $\frac{1 }{1.2 *10^{-3}} = \frac{1}{2.0 *10^{-2}} * 1.50*10^{10} * t$

  $t = 1.11 *10^{-9} \ s$

         

Answer 2

It will take "1.11 × 10⁻⁹ s".

Chemical reaction:

A process wherein the two or even more compounds collide with both the proper orientation as well as enough effort to generate a new substance or the outcomes, is considered as Chemical reaction. This process involves the breaking as well as formation of atom connections.

According to the question,

Rate constant, k = 1.5 × 10¹⁰ L/mol.s

Initial concentration, [I] = 2.0 × 10⁻² M

By using the integrated rate law,

→ $\frac{1}{[I]_r} = \frac{1}{[I]}$ × k × t ...(Equation 1)

Now,

The concentration of remaining Iodine atom,

[$I_r$] = [$I_o$] × [100% - 94%]  

     = [$I_o$] × 6%

     =  [$I_o$] × 0.06

By substituting the values is above equation,

[$I_r$] = 2.0 × 10⁻² × 0.06

     = 1.2 × 10⁻³

hence,

The time taken will be:

$\frac{1}{[I]_r} = \frac{1}{[I]}$ × k × t

By substituting the values,

$\frac{1}{1.2\times 10^{-3}} = \frac{1}{2.0\times 10^{-2}}$ × 1.50 × 10 × t

           t = 1.11 × 10⁻⁹ s

Thus the above answer is correct.

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