Chapter 13, Problem 13.29QP

Consider the first-order reaction A → B shown here. (a) What is the rate constant of the reaction? (b) How many A (yellow) and B (blue) molecules are present at t = 20 s and 30 s?

Interpretation Introduction

Interpretation:

Half-life and the rate constant for the given reaction have to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Half-life is the time required for one half of a reactant to react.

Half-life for a first order reaction is $t_{\frac{1}{2}}=\frac{0.693}{k}$ ; half-life for a first order reaction is independent of initial concentration of reactant

Explanation of Solution

The given first order reaction is,

$A\to B$

For first order reaction, $Rate=k\left[A\right]$

In the given diagram A molecules (yellow spheres) are converted to B molecules (blue spheres)

     

Fig (1)

There are 16 A molecules at $\text{t}\text{=}\text{0s}$ and at $\text{t}\text{=10s}$ there are 8 A molecules; the time of $t=10s$ represents the first half-life of the reaction.

Using the following equation the rate constant of this reaction can be calculated.

Half-life for a first order reaction is

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

$t_{\frac{1}{2}}$ for this reaction is $10s$ and so rate constant k is,

                                             $\begin{array}{c}\text{10}\text{s}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{k}\text{=}\frac{\text{0}\text{.693}}{\text{10}}\\ \text{=}0.0693s^{-1}\end{array}$

Interpretation Introduction

Interpretation:

The number of molecules of A and B present at $\text{t}\text{=}\text{20}\text{s}\text{and}\text{t}\text{=}\text{30s}$ has to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Half-life is the time required for one half of a reactant to react.

Half-life for a first order reaction is $t_{\frac{1}{2}}=\frac{0.693}{k}$ ; half-life for a first order reaction is independent of initial concentration of reactant

Explanation of Solution

The given first order reaction is,

$A\to B$

For first order reaction, $Rate=k\left[A\right]$

In the given diagram A molecules (yellow spheres) are converted to B molecules (blue spheres)

         

   Fig (1)

There are 16 A molecules at $\text{t}\text{=}\text{0s}$ and at $\text{t}\text{=10s}$ there are 8 A molecules; the time of $t=10s$ represents the first half-life of the reaction.

Therefore,

At $\text{t}=\text{20}\text{s}$ the concentration of molecule A will decrease to half of its concentration at $t=10s$, and so there will be 4 A molecules at $\text{t}=\text{20}\text{s}$. The mole ratio between A and B molecule is $1:1$ therefore, four more molecules of B will be produced and there will be 12 B molecules present at $\text{t}=\text{20}\text{s}$

At $\text{t}=3\text{0}\text{s}$ the concentration of molecule A will decrease to half of its concentration at $\text{t}=\text{20}\text{s}$, and so there will be 2 A molecules at $\text{t}=3\text{0}\text{s}$. The mole ratio between A and B molecule is $1:1$ therefore, two more molecules of B will be produced and there will be 14 B molecules present at $\text{t}=3\text{0}\text{s}$