Chapter 14, Problem 14.26QP

(a)

Interpretation Introduction

Interpretation:

The relative rate of the reaction in the given three containers 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,

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

Explanation of Solution

The given first order reaction is,

$X\to Y$

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

X molecules are placed in three equal-volume containers at the same temperature.

Fig (1)

The relative rates of the reaction in the given three containers can be determined as follows,

For convenience, we can use the number of molecules to represent the concentration. Therefore, the relative rates of reaction for the three containers are,

$\begin{array}{l}\left(\text{i}\right)\text{Rate}\text{=}\text{8k}\\ \left(\text{ii}\right)\text{Rate}\text{=}\text{6k}\\ \left(\text{iii}\right)\text{Rate}\text{=}\text{12}\text{k}\end{array}$

To get the relative rates dividing each rate by $2k$

Therefore

Relative rate of the reaction in three containers are $4:3:6$

(b)

Interpretation Introduction

Interpretation:

To determine the relative rates be affected if the volume of the each container were doubled

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,

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

Explanation of Solution

The given first order reaction is,

$X\to Y$

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

X molecules are placed in three equal-volume containers at the same temperature.

Fig (1)

The relative rates of the reaction in the given three containers can be determined as follows,

For convenience, we can use the number of molecules to represent the concentration. Therefore, the relative rates of reaction for the three containers are,

$\begin{array}{l}\left(\text{i}\right)\text{Rate}\text{=}\text{8k}\\ \left(\text{ii}\right)\text{Rate}\text{=}\text{6k}\\ \left(\text{iii}\right)\text{Rate}\text{=}\text{12}\text{k}\end{array}$

To get the relative rates dividing each rate by $2k$

Therefore

Relative rate of the reaction in three containers are $4:3:6$

The relative rates would be unaffected if the volume of the each container were doubled. Therefore, the relative rates between the three containers would remain same and so the actual rate would decrease by $50\%$  .

(c)

Interpretation Introduction

Interpretation:

The relative half-life of the reactions in $\left(i\right)to\left(iii\right)$ 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.

For first order reaction $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ Is the concentration of reactant A at time t ${\left[A\right]}_0$ is the initial concentration of reactant, k is the rate constant.

$t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

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,

$X\to Y$

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

X molecules are placed in three equal-volume containers at the same temperature.

Fig (1)

We know that, the half-life of a first order reaction is independent on substrate (reactant) concentration; it does not depend on substrate concentration.

Therefore, the relative half-life of the reactions in $\left(i\right)to\left(iii\right)$ will be same; 1:1:1