Chapter 13, Problem 13.130CP

(a)

Interpretation Introduction

To Write:

Balanced equation for the reaction of iodide ion with hypochlorite ion

b)

Interpretation Introduction

To determine:

The rate law and Rate constant for all 4 experiments

Interpretation Introduction

To find the rate Constant: (for I^-& OCl^-)

  $k=\frac{2.303}{10}\log \frac{2.40\times {10}^{-4}}{2.17\times {10}^{-4}}=1.0\times {10}^{-2}$

  $k=\frac{2.303}{10}\log \frac{1.60\times {10}^{-4}}{1.37\times {10}^{-4}}=1.5\times {10}^{-2}$

Experiment − 2: (for I^-& OCl^-)

Given Data:

Initial time = 0 &Final time = 10s
Initial conc. = 1.20 ( 10-4 , Final conc. = 1.08 ( 10-4 & 1.60 ( 10-4 , 1.48 ( 10-4

Formula Used:$k=\frac{2.303}{t}\log \frac{{[I^{-}]}_0}{{[I^{-}]}_{10}}$

$k=\frac{2.303}{t}\log \frac{{[OC{l}^{-}]}_0}{{[OC{l}^{-}]}_{10}}$

Calculation to find the rate Constants:
  $k=\frac{2.303}{10}\log \frac{1.20\times {10}^{-4}}{1.08\times {10}^{-4}}=1.0\times {10}^{-2}$

  $k=\frac{2.303}{10}\log \frac{1.60\times {10}^{-4}}{1.48\times {10}^{-4}}=7.7\times {10}^{-3}$

Experiment − 3: (for I^-& OCl^-)

Given Data:

Initial time = 0 &Final time = 10s
Initial conc. = 2.40 ( 10-4 , Final conc. = 2.34 ( 10-4 & 4.00 ( 10-5 , 3.40 ( 10-5

Formula Used:$k=\frac{2.303}{t}\log \frac{{[I^{-}]}_0}{{[I^{-}]}_{10}}$

$k=\frac{2.303}{t}\log \frac{{[OC{l}^{-}]}_0}{{[OC{l}^{-}]}_{10}}$

Calculation to find the rate Constants:
  $k=\frac{2.303}{10}\log \frac{2.40\times {10}^{-4}}{2.34\times {10}^{-4}}=2.5\times {10}^{-3}$

  $k=\frac{2.303}{10}\log \frac{4.00\times {10}^{-5}}{3.40\times {10}^{-5}}=1.6\times {10}^{-2}$

Experiment − 4: (for I^-& OCl^-)

Given Data:

Initial time = 0 &Final time = 10s
Initial conc. = 1.20 ( 10-4 , Final conc. = 1.14 ( 10-4 & 1.60 ( 10-4 , 1.54 ( 10-4

Formula Used:$k=\frac{2.303}{t}\log \frac{{[I^{-}]}_0}{{[I^{-}]}_{10}}$

$k=\frac{2.303}{t}\log \frac{{[OC{l}^{-}]}_0}{{[OC{l}^{-}]}_{10}}$

Calculation to find the rate Constants:
  $k=\frac{2.303}{10}\log \frac{1.20\times {10}^{-4}}{1.14\times {10}^{-4}}=5.1\times {10}^{-3}$

  $k=\frac{2.303}{10}\log \frac{1.60\times {10}^{-4}}{1.54\times {10}^{-4}}=3.8\times {10}^{-3}$

Interpretation Introduction

(c)

To verify:

If the reaction occurs by a single step mechanism

(d)

Interpretation Introduction

To determine:

A mechanism that is consistent with the rate law, and express the rate constant in terms of the rate constants for the elementary steps in the mechanism