Chapter 20, Problem 20.23E

Interpretation Introduction

(a)

Interpretation:

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is to be calculated. The answer is to be compared with that of first order kinetics.

Concept introduction:

The rate law for second order reaction is represented as,

$-\frac{d\text{[A]}}{dt}=k{\text{[A]}}^2$

This represents the case in which identical reactants are present. In the second order kinetics, rate of the reaction is proportional to the square of concentration of the reactant. The integrated rate law for second order reaction is represented as,

$\frac{1}{{\text{[A]}}_t}-\frac{1}{{\text{[A]}}_0}=k\cdot t$

Where,

• ${[\text{A}]}_0$ is the initial concentration.

• ${[\text{A}]}_t$ is the concentration at time $t$.

• $k$ is the rate constant.

Answer to Problem 20.23E

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $33.2\text{s}$.

The amount of time taken in second order kinetics to produce the same amount of gas is higher than that of the first order kinetics.

Explanation of Solution

It is given that the thermal decomposition of mercuric oxide follows second order kinetics and the rate constant is $6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}$. The given reaction is,

$2\text{HgO}\left(\text{s}\right)\to 2\text{Hg}\left(l\right)+{\text{O}}_2\left(\text{g}\right)$

The initial amount of $\text{HgO}$ is $1.00\text{gram}$.

The number of moles of $1.00\text{mL}$ of ${\text{O}}_2$ at STP ($1\text{atm}$, $273\text{K}$) is calculated by the formula,

$PV=nRT$

Where,

• $P$ is the pressure.

• $V$ is the volume.

• $R$ is the gas constant and its value is $0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T$ is the temperature.

• $n$ is the number of moles.

Substitute the values of pressure, volume, gas constant and temperature in the above formula.

$\begin{array}{rll}n& =& \frac{PV}{RT}\\ n& =& \frac{1\text{atm}\left(\frac{1.00\text{mL}}{1000\text{mL}}\right)}{0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 273\text{K}}\\ n& =& 4.46\times {10}^{-5}\text{moles}\end{array}$

Thus, the number of moles of ${\text{O}}_2$ is $4.46\times {10}^{-5}\text{moles}$. From the stoichiometry of the given reaction the number of moles of $\text{HgO}$ reacted is twice as the number of moles of ${\text{O}}_2$ produced. Thus, the number of moles of $\text{HgO}$ is $2\times 4.46\times {10}^{-5}\text{moles}$.

The amount of $\text{HgO}$ reacted is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Amount}\text{of}\text{HgO}}{\text{Molar}\text{mass}\text{of}\text{HgO}}$

The molar mass of $\text{HgO}$ is $\text{216}\text{.59}\text{g}/\text{mol}$.

Substitute the number of moles and molar mass of $\text{HgO}$ in the given formula,

$\begin{array}{l}2\times 4.46\times {10}^{-5}\text{moles}=\frac{\text{Amount}\text{of}\text{HgO}}{\text{216}\text{.59}\text{g}/\text{mol}}\\ \text{Amount}\text{of}\text{HgO}=2\times 4.46\times {10}^{-5}\text{moles}\times \text{216}\text{.59}\text{g}/\text{mol}\\ \text{Amount}\text{of}\text{HgO}=0.0193\text{g}\end{array}$

The amount of $\text{HgO}$ reacted is $0.0193\text{g}$.

The amount of $\text{HgO}$ left at time $t$ is given by,

$\begin{array}{l}\text{Amount}\text{of}\text{HgO}\text{left}=\text{Initial}\text{amount}\text{of}\text{HgO}-\text{Amount}\text{of}\text{HgO}\text{reacted}\\ \text{Amount}\text{of}\text{HgO}\text{left}=\text{1}\text{.00}\text{g}-0.0193\\ \text{Amount}\text{of}\text{HgO}\text{left}=0.981\text{g}\end{array}$

Thus, the amount of $\text{HgO}$ is $0.981\text{g}$.

The rate law for the given second order reaction is given by,

$\frac{1}{{\left(\text{HgO}\right)}_t}-\frac{1}{{\left(\text{HgO}\right)}_0}=k\cdot t$

Where,

• ${\left(\text{HgO}\right)}_0$ is the initial amount of $\text{HgO}$.

• ${\left(\text{HgO}\right)}_t$ is the amount of $\text{HgO}$ at time $t$.

• $k$ is the rate constant.

Substitute the values of initial amount, amount at time $t$ and rate constant in the given formula.

$\begin{array}{l}\frac{1}{0.981\text{g}}-\frac{1}{1.00\text{g}}=6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}\cdot t\\ t=\frac{1.02{\text{g}}^{-1}-1.00{\text{g}}^{-1}}{6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}}\\ t=\frac{0.02{\text{g}}^{-1}}{6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}}\end{array}$

Under the assumption of standard temperature and pressure, units cancel out such that the equation becomes,

$\begin{array}{l}t=\frac{0.02}{6.02\times {10}^{-4}{\text{s}}^{-1}}\\ t=33.2\text{s}\end{array}$

Thus, the time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $33.2\text{s}$.

On comparison of the time taken by first order kinetics and second order kinetics, it is observed that the amount of time taken in second order kinetics to produce the same amount of gas is higher than the first order kinetics.

Conclusion

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $33.2\text{s}$.

The amount of time taken in second order kinetics to produce the same amount of gas is higher than the first order kinetics.

Interpretation Introduction

(b)

Interpretation:

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is to be calculated. The answer is to be compared with that of first order kinetics.

Concept introduction:

The rate law for second order reaction is represented as,

$-\frac{d\text{[A]}}{dt}=k{\text{[A]}}^2$

This represents the case in which identical reactants are presents. In the second order kinetics rate of the reaction is proportional to the square of concentration of the reactant. The integrated rate law for second order reaction is represented as,

$\frac{1}{{\text{[A]}}_t}-\frac{1}{{\text{[A]}}_0}=k\cdot t$

Where,

• ${[\text{A}]}_0$ is the initial concentration.

• ${[\text{A}]}_t$ is the concentration at time $t$.

• $k$ is the rate constant.

Answer to Problem 20.23E

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $398.67\text{s}$.

The amount of time taken in second order kinetics to produce the same amount of gas is higher than that of the first order kinetics.

Explanation of Solution

It is given that the thermal decomposition of mercuric oxide follows second order kinetics and the rate constant is $6.02\times {10}^{-4}{\text{s}}^{-1}$. The given reaction is,

$2\text{HgO}\left(\text{s}\right)\to 2\text{Hg}\left(l\right)+{\text{O}}_2\left(\text{g}\right)$

The initial amount of $\text{HgO}$ is $1.00\text{gram}$.

The number of moles of $10.0\text{mL}$ of ${\text{O}}_2$ at STP ($1\text{atm}$, $273\text{K}$) is calculated by the formula,

$PV=nRT$

Where,

• $P$ is the pressure.

• $V$ is the volume.

• $R$ is the gas constant and its value is $0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T$ is the temperature.

• $n$ is the number of moles.

Substitute the values of pressure, volume, gas constant and temperature in the above formula.

$\begin{array}{rll}n& =& \frac{PV}{RT}\\ n& =& \frac{1\text{atm}\left(\frac{10.0\text{mL}}{1000\text{mL}}\right)}{0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\cdot 273\text{K}}\\ n& =& 4.46\times {10}^{-4}\text{moles}\end{array}$

Thus, the number of moles of ${\text{O}}_2$ is $4.46\times {10}^{-4}\text{moles}$. From the stoichiometry of the given reaction the number of moles of $\text{HgO}$ reacted is twice as the number of moles of ${\text{O}}_2$ produced. Thus, the number of moles of $\text{HgO}$ is $2\times 4.46\times {10}^{-4}\text{moles}$.

The amount of $\text{HgO}$ reacted is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Amount}\text{of}\text{HgO}}{\text{Molar}\text{mass}\text{of}\text{HgO}}$

The molar mass of $\text{HgO}$ is $\text{216}\text{.59}\text{g}/\text{mol}$.

Substitute the number of moles and molar mass of $\text{HgO}$ in the given formula,

$\begin{array}{l}2\times 4.46\times {10}^{-4}\text{moles}=\frac{\text{Amount}\text{of}\text{HgO}}{\text{216}\text{.59}\text{g}/\text{mol}}\\ \text{Amount}\text{of}\text{HgO}=2\times 4.46\times {10}^{-4}\text{moles}\times \text{216}\text{.59}\text{g}/\text{mol}\\ \text{Amount}\text{of}\text{HgO}=0.193\text{g}\end{array}$

The amount of $\text{HgO}$ reacted is $0.193\text{g}$.

The amount of $\text{HgO}$ left at time $t$ is given by,

$\begin{array}{l}\text{Amount}\text{of}\text{HgO}\text{left}=\text{Initial}\text{amount}\text{of}\text{HgO}-\text{Amount}\text{of}\text{HgO}\text{reacted}\\ \text{Amount}\text{of}\text{HgO}\text{left}=\text{1}\text{.00}\text{g}-0.193\text{g}\\ \text{Amount}\text{of}\text{HgO}\text{left}=0.807\text{g}\end{array}$

Thus, the amount of $\text{HgO}$ is $0.807\text{g}$.

The rate law for the given second order reaction is given by,

$\frac{1}{{\left(\text{HgO}\right)}_t}-\frac{1}{{\left(\text{HgO}\right)}_0}=k\cdot t$

Where,

• ${\left(\text{HgO}\right)}_0$ is the initial amount of $\text{HgO}$.

• ${\left(\text{HgO}\right)}_t$ is the amount of $\text{HgO}$ at time $t$.

• $k$ is the rate constant.

Substitute the values of initial amount, amount at time $t$ and rate constant in the given formula.

$\begin{array}{l}\frac{1}{0.807\text{g}}-\frac{1}{1.00\text{g}}=6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}\cdot t\\ t=\frac{1.24{\text{g}}^{-1}-1.00{\text{g}}^{-1}}{6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}}\\ t=\frac{0.24{\text{g}}^{-1}}{6.02\times {10}^{-4}{\text{M}}^{-1}\cdot {\text{s}}^{-1}}\end{array}$

Under the assumption of standard temperature and pressure, units cancel out such that the equation becomes,

$\begin{array}{l}t=\frac{0.24}{6.02\times {10}^{-4}{\text{s}}^{-1}}\\ t=398.67\text{s}\end{array}$

Thus, the time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $398.67\text{s}$.

On comparison of the time taken by first order kinetics and second order kinetics, it is observed that the amount of time taken in second order kinetics to produce the same amount of gas is higher than that of the first order kinetics.

Conclusion

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $398.67\text{s}$.

The amount of time taken in second order kinetics to produce the same amount of gas is higher than that of the first order kinetics.