Chapter 15, Problem 27E

Interpretation Introduction

Interpretation:

The rate law should be calculated along with the value of rate constant for the given reaction.

Concept Introduction:

Rate law gives the relationship between the rate of the reaction and the reactant concentrations.

The general reaction is:

  $aA+bB\to cC+dD$

Rate of above reaction is expressed as:

  $rate=k{\left[A\right]}^a{\left[B\right]}^b$

Where, k = rate constant

The rate constant is defined as the proportionality constant which shows the relationship between concentration of reactants and rate of chemical reaction.

Answer to Problem 27E

Rate law expression is:

  $rate=k{\left[I^{-}\right]}^1{\left[OC{l}^{-}\right]}^1{\left[O{H}^{-}\right]}^{-1}$

The rate constant is equal to $60{\text{ s}}^{-1}$ for the given reaction.

Explanation of Solution

Given reaction:

The chemical reaction is:

  ${\text{I}}^{-}(aq)+{\text{OCl}}^{-}(aq)\to {\text{IO}}^{-}(aq)+{\text{Cl}}^{-}(aq)$

The given data is:

  

The given chemical reaction is:

  ${\text{I}}^{-}(aq)+{\text{OCl}}^{-}(aq)\to {\text{IO}}^{-}(aq)+{\text{Cl}}^{-}(aq)$

The general rate law for above reaction is:

  $rate=k{\left[I^{-}\right]}^x{\left[OC{l}^{-}\right]}^y{\left[O{H}^{-}\right]}^z$

Here, concentration of hydroxide affect the rate, thus considered in rate law.

Where, $x,y\text{ and z}$ represents the order of the reaction with respect to each reactant

Now, put the values in above expression for first and second experiment.

  $9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z\text{ (1)}$

  $18.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0026\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z\text{ (2)}$

Divide (1) and (2)

  $\frac{18.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}}{9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}}=\frac{k{\left[0.0026\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z}{k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z}$

  $2.0=\frac{{\left[0.0026\text{ mol/L}\right]}^x}{{\left[0.0013\text{ mol/L}\right]}^x}$

  $2.0={\left(2.0\right)}^x$

Hence, value of x = 1.

Now, put the values in above expression for first and third experiment.

  $9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z\text{ (1)}$

  $4.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.0060\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z\text{ (3)}$

Divide (1) and (3)

  $\frac{4.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}}{9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}}=\frac{k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.0060\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z}{k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z}$

  $0.50=\frac{{\left[0.0060\text{ mol/L}\right]}^y}{{\left[0.012\text{ mol/L}\right]}^y}$

  $0.50={\left(0.50\right)}^y$

Hence, value of y = 1.

Now, put the values in above expression for first and fifth experiment.

  $9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z\text{ (1)}$

  $18.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.050\text{ mol/L}\right]}^z\text{ (4)}$

Divide (1) and (4)

  $\frac{18.7\times {10}^{-3}\text{ mol/L}\cdot \text{s}}{9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}}=\frac{k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.050\text{ mol/L}\right]}^z}{k{\left[0.0013\text{ mol/L}\right]}^x{\left[0.012\text{ mol/L}\right]}^y{\left[0.10\text{ mol/L}\right]}^z}$

  $2.0=\frac{{\left[0.050\text{ mol/L}\right]}^z}{{\left[0.10\text{ mol/L}\right]}^z}$

  $2.0={\left(\frac{1}{2}\right)}^z$

Hence, value of z= $-1$ .

Put the value of x, y and z in rate law expression:

  $rate=k{\left[I^{-}\right]}^1{\left[OC{l}^{-}\right]}^1{\left[O{H}^{-}\right]}^{-1}$

The above expression is rearranged as:

  $rate=k\frac{\left[I^{-}\right]\left[OC{l}^{-}\right]}{\left[O{H}^{-}\right]}$

Put the values in above expression from experiment 1 to determine the value of rate constant (k).

  $9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k\frac{\left[0.0013\text{ mol/L}\right]\left[0.012\text{ mol/L}\right]}{\left[0.10\text{ mol/L}\right]}$

  $9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}=k\left(0.00156\text{ mol/L}\right)$

  $k=\frac{9.4\times {10}^{-3}\text{ mol/L}\cdot \text{s}}{0.000156\text{ mol/L}}$

  $k=60{\text{ s}}^{-1}$

Thus, the rate constant is equal to $60{\text{ s}}^{-1}$ for the given reaction.