Chapter 19, Problem 19.89QP

The integrated rate law for the zeroth-order reaction A → B is [A]_t = [A]₀ − kt. (a) Sketch the following plots: (i) rate versus [A]_t and (ii) [A]_t versus t. (b) Derive an expression for the half-life of the reaction, (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when [A]_t = 0.

Interpretation Introduction

Interpretation:

1. (a) (i) The plots for rate versus ${\left[\text{A}\right]}_{\text{t}}$ and (ii) ${\left[\text{A}\right]}_{\text{t}}$ versus time have to be sketched.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Answer to Problem 19.89QP

(a)

(i) The plots for rate versus ${\left[\text{A}\right]}_{\text{t}}$ and (ii) ${\left[\text{A}\right]}_{\text{t}}$ versus time is sketched.

Explanation of Solution

From a rate law, for zero order reaction the exponent is zero.  In other way, the rate of the reaction is just equal to a constant and it doesn’t change with time passes.

1. (i) The rate law can be ${\text{rate = k[A]}}^{\text{0}}\text{ = k}$

Fig (1)

1. (ii) The integrated zero order rate law can be ${\left[\text{A}\right]}_{\text{t}}=-\text{kt}+{\left[\text{A}\right]}_{\text{t}}$

Therefore, a plot for ${\left[\text{A}\right]}_{\text{t}}$ versus time should be a straight line with a slope of –k.

Fig (2)

Interpretation Introduction

Interpretation:

(b) The half-life for the given reaction has to be derived.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Answer to Problem 19.89QP

(b)

The half-life for the given reaction is derived.

Explanation of Solution

In order to calculate the half-lives with integrated rate law is as follows.

${\left[\text{A}\right]}_{\text{t}}={\left[\text{A}\right]}_0-\text{kt}$

At $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ , ${\left[\text{A}\right]}_{\text{t}}=\frac{{\left[\text{A}\right]}_{\text{0}}}{\text{2}}$

Substitute it into the above equation,

$\frac{{\left[\text{A}\right]}_{\text{0}}}{\text{2}}={\left[\text{A}\right]}_0-k{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{{\left[\text{A}\right]}_{\text{0}}}{2k}$

$k=\frac{{\left[\text{A}\right]}_{\text{0}}}{2t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Interpretation Introduction

Interpretation:

(c) The time in half-lives when the integrated rate law is no longer valid i.e., ${\left[\text{A}\right]}_{\text{t}}=0$ has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Answer to Problem 19.89QP

(c)

The time in half-lives when the integrated rate law is no longer valid i.e., ${\left[\text{A}\right]}_{\text{t}}=0$ is calculated.

Explanation of Solution

When, ${\left[\text{A}\right]}_{\text{t}}=0$

${\left[\text{A}\right]}_0=-kt$

$t=\frac{{\left[\text{A}\right]}_0}{k}$

Substitute the above rate constant in the obtained formula.

$t=\frac{{\left[\text{A}\right]}_0}{{\left[\text{A}\right]}_0/2t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$t=2t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$