Chapter 13, Problem 13.32QP

(a)

Interpretation Introduction

Interpretation:

The given statements have to be answered.

Concept Introduction:

The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.

Half life for first order reactions:

The half life for the first order reaction is constant and it is independent of the reactant concentration.

Half life period of first order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To complete the pictures

Explanation of Solution

The reaction follows first order with presence of half-life of ten seconds.

There are 16 AB particles present in the container,

After one half life (10s) 8 particles will be reacted and 8 remains unreacted.

After two-half lives (20s) 12 particles will be reacted and 4 remains unreacted.

The completed pictures are,

Figure 1

(b)

Interpretation Introduction

Interpretation:

The given statements have to be answered.

Concept Introduction:

The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.

Half life for first order reactions:

The half life for the first order reaction is constant and it is independent of the reactant concentration.

Half life period of first order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To explain the changes in completed figure if the reaction was second-order with same half life

Explanation of Solution

If the half-life is similar for second-order reaction, the container t=20s would have more number of AB and fewer A and B when compared to part a.

(c)

Interpretation Introduction

Interpretation:

The given statements have to be answered.

Concept Introduction:

The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.

Half life for first order reactions:

The half life for the first order reaction is constant and it is independent of the reactant concentration.

Half life period of first order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To give the relative reaction rates for first order reaction at the start of reaction and after 10 seconds elapsed

Explanation of Solution

After 10 seconds, the concentration of the particles is one-half their initial value. Then relative rate of reactions for first-order at the start and after 10 seconds are,

$\frac{{\text{Rate}}_{\text{0}}}{{\text{Rate}}_{\text{10}}}\text{=}\frac{\text{k}{\left[\text{A}\right]}_{\text{o,0}}}{\text{k}{\left[\text{A}\right]}_{\text{o,10}}}\text{=}\frac{{\left[\text{A}\right]}_{\text{o,0}}}{{\left[\text{A}\right]}_{\text{o,10}}}\text{=}\frac{{\left[\text{A}\right]}_{\text{o}}}{\text{1/2}{\left[\text{A}\right]}_{\text{o}}}\text{=2}$

(d)

Interpretation Introduction

Interpretation:

The given statements have to be answered.

Concept Introduction:

The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.

Half life for first order reactions:

The half life for the first order reaction is constant and it is independent of the reactant concentration.

Half life period of first order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

To give the relative reaction rates for second order reaction at the start of reaction and after 10 seconds elapsed

Explanation of Solution

After 10 seconds, the concentration of the particles is one-half their initial value. Then relative rate of reactions for second order at the start and after 10 seconds are,

$\frac{{\text{Rate}}_{\text{0}}}{{\text{Rate}}_{\text{10}}}\text{=}\frac{\text{k}{\left[\text{A}\right]}^{\text{2}}{}_{\text{o,0}}}{\text{k}{\left[\text{A}\right]}^{\text{2}}{}_{\text{o,10}}}\text{=}{\left(\frac{{\left[\text{A}\right]}_{\text{o,0}}}{{\left[\text{A}\right]}_{\text{o,10}}}\right)}^{\text{2}}\text{=}{\left(\frac{{\left[\text{A}\right]}_{\text{o}}}{\text{1/2}{\left[\text{A}\right]}_{\text{o}}}\right)}^{\text{2}}{\text{=2}}^{\text{2}}\text{=4}$