Chapter 6, Problem 6.6P

(a)

Interpretation Introduction

Interpretation:

The following data were collected for the given reaction at $780K$, using these date the order of the reaction has to be determined.

Concept Introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area.

General rate reaction is,

  $aA+bB\to cC+dD$

  $\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Explanation of Solution

Given reaction is,

  $2HI\left(g\right)\underset{}{\to }{H}_2\left(g\right)+I_2\left(g\right)$

Record the given data’s

  $\begin{array}{cccccc}t/s& 0& 1& 2& 3& 4\\ \left[HI\right]/\left(mold{m}^{-3}\right)& 1& 0.43& 0.27& 0.2& 0.16\end{array}$

Let us consider the first order reaction

Rate constant is,

  $\begin{array}{l}\ln {\left[\text{A}\right]}_{\text{t}}=-\text{kt}\text{+}\ln {\left[\text{A}\right]}_0\\ \text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}\\ \text{k}\text{=}\frac{1}{\text{t}}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\end{array}$

Given values are plugging above equation,

  $\begin{array}{c}{k}_1=\frac{1}{1}\left[\frac{1}{0.43}\right]\\ =0.43\times {10}^{-3}{M}^{-1}{s}^{-1}\\ k_2=\frac{1}{2}\left[\frac{1}{0.27}\right]\\ =0.135\times {10}^{-3}{M}^{-1}{s}^{-1}\\ k_3=\frac{1}{3}\left[\frac{1}{0.2}\right]\\ =0.06M^{-1}{s}^{-1}\\ k_4=\frac{1}{4}\left[\frac{1}{0.16}\right]\\ =0.04M^{-1}{s}^{-1}\end{array}$

Hence, the all rate constant are nearly same, so this reaction confirmed to be first order reaction.

(b)

Interpretation Introduction

Interpretation:

Plot the given data in an appropriate fashion, the rate constant has to be determined.

Concept Introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area.

General rate reaction is,

  $aA+bB\to cC+dD$

  $\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

Explanation of Solution

Given reaction is,

  $2HI\left(g\right)\underset{}{\to }{H}_2\left(g\right)+I_2\left(g\right)$

Record the given data’s

  $\begin{array}{cccccc}t/s& 0& 1& 2& 3& 4\\ \left[HI\right]/\left(mold{m}^{-3}\right)& 1& 0.43& 0.27& 0.2& 0.16\end{array}$

In order to determine the rate law, we need to determine the order of the reaction with respect to $\left(HI\right)$.  This can be done graphically.

For a zeroth order reaction $\left[HI\right]$ vs time should result in a straight line, for a first order reaction $\ln \left[HI\right]$ vs time will give a straight line while for a second order reaction $1/\left[HI\right]$ vs time will result in a straight line plot.

In the present case a plot $1/\left[HI\right]$ vs time gives a straight line as shown below (Figure -1), thus the given reaction is first order.

The rate law is thus $r=k{\left[HI\right]}^2$

Figure 1