Chapter 14, Problem 14.16QP

(a)

Interpretation Introduction

Interpretation:

From the given data, the order of the reaction has to be determined.

Concept introduction:

Rate law: An expression relating the rate of a reaction to the rate constant and concentrations of the reactants that are involved in the reaction.

For a given type of reaction:

$\text{aA}\text{+}\text{bB}\stackrel{}{\to }\text{cC+}\text{dD}$

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

$\text{k}$= Rate constant. It is the constant of proportionality between the rate of a reaction and the concentration of the reactants. Its unit is $s^{-1}$.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

It is important to note that x and y are not related to a and b.

Rather, they are determined experimentally.

Order of a reaction:  The sum of exponents of the concentrations in the rate law of chemical reaction is known as order of a reaction.

The order of the reaction for the given rate law is:

$\begin{array}{l}\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}\Rightarrow \text{Rate}\text{law}\\ \text{Oder}\text{=}\text{x}\text{+}\text{y}\end{array}$

If the value of the order is 1, then it is known as the 1^st order reaction.

If the value of the order is 2, then it is known as the 2^ndorder reaction.

If the value of the order is 3, then it is known as the 3^rdorder reaction.

If rate of a reaction is independent of $\left[\text{A}\right]$ and if it is only dependent on $\left[\text{B}\right]$. Then the order of such a reaction is known as zero order and hence the corresponding reaction will be known as zero order reaction.

Answer to Problem 14.16QP

The order of the reaction is zero order.

Explanation of Solution

The given experiments show that when the concentration of

• While holding $\left[\text{X}\right]$ as constant, if $\left[\text{Y}\right]$ is increased or decreased then the initial rate of disappearance of $\text{X}$ changes accordingly. Therefore, the rate of the reaction is directly proportional to the concentration of $\left[\text{Y}\right]$.
• While holding $\left[\text{Y}\right]$ as constant, if $\left[\text{X}\right]$ is increased or decreased then the initial rate of disappearance of $\text{X}$ does not change accordingly. Therefore, the rate of the reaction is not proportional to the concentration of $\left[\text{X}\right]$.

The experimental observations tell that the rate of the reaction is dependent only on the concentrations of $\text{Y}$ and not on the concentration of $\text{X}$.

Hence, form the rate law will be:

${\text{rate = k[X]}}^{\text{o}}\text{[Y]}$

Therefore the order of the reaction is zero order.

Conclusion

The order of the reaction has been found.

(b)

Interpretation Introduction

Interpretation:

From the given data, the initial rate of disappearance of $\text{X}$ has to be determined.

Concept introduction:

Rate law: An expression relating the rate of a reaction to the rate constant and concentrations of the reactants that are involved in the reaction.

For a given type of reaction:

$\text{aA}\text{+}\text{bB}\stackrel{}{\to }\text{cC+}\text{dD}$

The rate law takes the form as:

$\text{Rate}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

$\text{k}$= Rate constant. It is the constant of proportionality between the rate of a reaction and the concentration of the reactants. Its unit is $s^{-1}$.

$\left[\text{A}\right]$-Concentration of the reactant A in molarity.

$\left[\text{B}\right]$-Concentration of the reactant B in molarity.

It is important to note that x and y are not related to a and b.

Rather, they are determined experimentally.

Answer to Problem 14.16QP

The initial rate of disappearance of $\text{X}$ is $\text{0}\text{.1176}\text{M/s}$

Explanation of Solution

To determine the initial rate of disappearance of $\text{X}$, the rate constant has to be calculated.

Among the given experiments in the table, data of any one experiment can be considered to calculate the rate constant for the reaction.

Considering the experiment 1:

$\begin{array}{l}\left[\text{X}\right]\text{=}\text{0}\text{.10M}\\ \left[\text{Y}\right]\text{=0}\text{.50}\text{M}\\ \text{Rate}\text{=}0.147\text{M/s}\end{array}$

$\begin{array}{c}{\text{rate = k[X]}}^{\text{o}}\text{[Y]}\\ \text{k}=\frac{\text{rate}}{{\text{[X]}}^{\text{o}}\text{ [Y]}}\\ =\frac{\text{0}\text{.147}\text{M/s}}{\text{1}\text{×}\text{(0}\text{.50}\text{M)}}\because {\text{[X]}}^{\text{o}}=1\\ =0.294{\text{s}}^{\text{-1}}\end{array}$

Therefore, the rate constant is $\text{k}=0.294{\text{s}}^{\text{-1}}$

Substituting the all the known values in rate law to obtain the rate of the reaction.

Given data:

$\begin{array}{l}\left[\text{X}\right]\text{=}\text{0}\text{.30M}\\ \left[\text{Y}\right]\text{=0}\text{.40}\text{M}\end{array}$

$\begin{array}{c}{\text{rate = k[X]}}^{\text{o}}\text{[Y]}\\ \text{rate = (0}\text{.294}{\text{s}}^{\text{-1}}\text{)(0}{\text{.30M)}}^{\text{o}}\text{(0}\text{.40M)}\\ \text{= (0}\text{.294}{\text{s}}^{\text{-1}}\text{)(1)(0}\text{.40M)}\\ \text{=}\text{0}\text{.1176}\text{M/s}\end{array}$

Therefore, the rate constant is $\text{0}\text{.1176}\text{M/s}$

Conclusion

The initial rate of disappearance of $\text{X}$ has been calculated.