Chapter 20, Problem 20.22E

Assume that thermal decomposition of mercuric oxide, $\text{HgO}$, follows first-order kinetics. It can be followed by the production of oxygen gas as a product:

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

At a particular temperature, $k=6.02\times {10}^{-4}{\text{s}}^{-1}$. If $1.00\text{gram}$ of $\text{HgO}$ were present initially, how long would it take to produce (a) $1.00\text{mL}$ of ${\text{O}}_2\left(\text{g}\right)$ at STP; (b) $10.0\text{mL}$ of ${\text{O}}_2\left(\text{g}\right)$ at STP? (STP = standard temperature and pressure for a gas.)

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.

Concept introduction:

The rate law for first order reaction is represented as,

$-\frac{d[\text{A}]}{dt}=k\cdot {[\text{A}]}^1$

In the first order kinetics rate of the reaction depends linearly on the concentration of one of the reactant. The integrated rate law for first order reaction is represented as,

$\ln \frac{{[\text{A}]}_0}{{[\text{A}]}_t}=kt$

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 20.22E

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $32.4\text{s}$.

Explanation of Solution

It is given that the thermal decomposition of mercuric oxide follows first 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 $1.00\text{mL}$ 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.

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

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,

$\begin{array}{l}\text{Amount}\text{of}\text{HgO}\text{left}=\text{Initial}\text{amount}\text{of}\text{HgO}-\text{Amount}\text{of}\text{HgO}\text{reacted}\\ \text{Amount}\text{of}\text{HgO}\text{left}=\text{1}\text{.00}\text{g}-0.0193\\ \text{Amount}\text{of}\text{HgO}\text{left}=0.9807\text{g}\end{array}$

Thus, the amount of $\text{HgO}$ is $0.9807\text{g}$.

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

$\ln \frac{{\left(\text{HgO}\right)}_0}{{\left(\text{HgO}\right)}_t}=kt$

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.

$\begin{array}{rll}\ln \frac{1.00\text{g}}{0.9807\text{g}}& =& 6.02\times {10}^{-4}{\text{s}}^{-1}t\\ t& =& \frac{1}{6.02\times {10}^{-4}{\text{s}}^{-1}}\ln \frac{1.00\text{g}}{0.9807\text{g}}\\ t& =& 32.4\text{s}\end{array}$

Thus, the time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $32.4\text{s}$.

Conclusion

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $1.00\text{mL}$ of ${\text{O}}_2$ is $32\text{s}$.

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.

Concept introduction:

The rate law for first order reaction is represented as,

$-\frac{d[\text{A}]}{dt}=k\cdot {[\text{A}]}^1$

In the first order kinetics, rate of the reaction depends linearly on the concentration of one of the reactant. The integrated rate law for first order reaction is represented as,

$\ln \frac{{[\text{A}]}_0}{{[\text{A}]}_t}=kt$

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 20.22E

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $356\text{s}$.

Explanation of Solution

It is given that the thermal decomposition of mercuric oxide follows first 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.

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

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 first order reaction is given by,

$\ln \frac{{\left(\text{HgO}\right)}_0}{{\left(\text{HgO}\right)}_t}=kt$

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.

$\begin{array}{rll}\ln \frac{1.00\text{g}}{0.807\text{g}}& =& 6.02\times {10}^{-4}{\text{s}}^{-1}t\\ t& =& \frac{1}{6.02\times {10}^{-4}{\text{s}}^{-1}}\ln \frac{1.00\text{g}}{0.807\text{g}}\\ t& =& 356\text{s}\end{array}$

Thus, the time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $356\text{s}$.

Conclusion

The time taken by the given thermal decomposition reaction of mercuric oxide to produce $10.0\text{mL}$ of ${\text{O}}_2$ is $356\text{s}$.