Chapter 5, Problem 5.105PAE

105 The decomposition of mercury(II) thiocyanate produces an odd brown snake-like mass that is so unusual the process was once used in fireworks displays. There are actually several reactions that take place when the solid Hg(SCN)₂ is ignited:

   $\begin{array}{l}\text{2Hg}{\left(\text{SCN}\right)}_{\text{2}}\left(\text{s}\right)\to \text{2 HgS}\left(\text{s}\right)+{\text{CS}}_{\text{2}}\left(\text{s}\right)+{\text{C}}_{\text{3}}{\text{N}}_{\text{4}}\left(\text{s}\right)\\ {\text{CS}}_{\text{2}}\left(\text{s}\right)+{\text{3O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right)+{\text{2 SO}}_{\text{2}}\left(\text{g}\right)\\ {\text{2C}}_{\text{3}}{\text{N}}_{\text{4}}\left(\text{s}\right)\to \text{3}{\left(\text{CN}\right)}_{\text{2}}\left(\text{g}\right)+{\text{N}}_{\text{2}}\left(\text{g}\right)\\ \text{HgS}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{Hg}\left(l\right)+{\text{SO}}_{\text{2}}\left(\text{g}\right)\end{array}$

A 42.4-g sample of Hg(SCN)₂ is placed into a 2.4-L vessel at 21°C. The vessel also contains air at a pressure of 758 torr. The container is sealed and the mixture is ignited, causing the reaction sequence above to occur. Once the reaction is complete, the container is cooled back to the original temperature of 21°C. (a) Without doing numerical calculations, predict whether the final pressure in the vessel will be greater than, less than, or equal to the initial pressure. Explain your answer. (b) Calculate the final pressure and compare your result with your prediction. (Assume that the mole fraction of O₂ in air is 0.21.)

Interpretation Introduction

Interpretation:

The gas moles will contribute to the pressure change of the system.

Concept introduction:

• The ideal gas equation relates the pressure, volume, and temperature of a system.

$\text{Pressure }\times \text{ Volume = moles }\times \text{ Gas Constant }\times \text{ Temperature}$

• The gases will contribute to the pressure change of the system.
• If gases are produced then the pressure of the system is expected to change.

Answer to Problem 5.105PAE

Solution:

The pressure will increase.

Explanation of Solution

2 moles of mercury (II) thiocyanate will produce 8 gas moles and consumes 4 gas moles since there will be more gas moles after the reaction then the pressure of the system is expected to increase.

Interpretation Introduction

Interpretation:

It is required to calculate the number of initial moles of the reactants, identify the limiting and excess reactant, calculate the moles of the products and use the ideal gas equation to find the final pressure of the system.

Concept introduction:

• The ideal gas equation relates the pressure, volume, and temperature of a system.
• $\text{Pressure }\times \text{ Volume = moles }\times \text{ Gas Constant }\times \text{ Temperature}$
• The limiting reactant is the chemical species that will be consumed totally, stoichiometric calculations must be done based on the limiting reactant.
• The excess reactant is the chemical species that will not be consumed totally.
• To change units of a measured quantity a proper conversion factor is required.
• The gases will contribute to the pressure change of the system.
• If gases are produced then the pressure of the system is expected to change.

Answer to Problem 5.105PAE

Solution:

The final pressure will be 903.58 torrs.

Explanation of Solution

Step1: Calculate the amount of oxygen

It is possible to calculate the amount of oxygen in the system by assuming that the oxygen is behaving like an ideal gas.

The ideal gas equation is:

$\text{Pressure }\times \text{ Volume = moles }\times \text{ Gas Constant }\times \text{ Temperature}$

Rearrange the equation to find the number of moles:

$\frac{\text{Pressure }\times \text{Volume}}{\text{Gas Constant }\times \text{ Temperature}}=\text{moles}$

Pressure must be in atmospheres

Gas constant is $\text{0}\text{.0821 }\frac{\text{atm L}}{\text{K mole}}$

The temperature must be in Kelvin

The volume must be in liters

From the statement:

Pressure is 758 torr, change torr to atm:

$\text{1 atm = 760 torr}$ ;

$\text{758 torr }\times \frac{\text{1 atm}}{\text{760 torr}}=0.997\text{ atm}$

Temperature is $\text{21 }\text{C}$, change it to Kelvin:

$\text{Temperature = 21 + 273}\text{.15 = 294}\text{.15 K}$

Volume is 2.4 Liters

Calculate the number of moles:

$\frac{\text{0}\text{.997 atm }\times \text{ 2}\text{.4 L}}{\text{0}\text{.082 }\times \text{ 294}\text{.15 K}}=0.0992\text{ moles}$

Air is composed by:

21 % Oxygen and 79 % Nitrogen

So, there are:

$0.21\times \text{ 0}\text{.0992 = 0}\text{.02 moles of Oxygen}$

and

$0.79\times \text{ 0}\text{.0992 = 0}\text{.0783 moles of Nitrogen}$

Step 2: Calculate the number of moles of mercury (II) thiocyanate

The number of moles is the ratio of the mass of the compound to its molar mass:

$\text{moles = }\frac{\text{mass}}{\text{molar mass}}$

There are 42.4 grams of mercury (II) thiocyanate, molar mass is $\text{316}\text{.75 }\frac{\text{grams}}{\text{mole}}$

$\text{moles = }\frac{\text{42}\text{.4 grams}}{\text{316}\text{.75 }\frac{\text{grams}}{\text{mole}}}\text{=0}\text{.13386 moles}$

According to the reaction, 2 moles of mercury (II) thiocyanate need 4 moles of oxygen, 0.13386 moles of mercury (II) thiocyanate will require:

$\text{0}\text{.13386 moles of mercury (II) thiocyanate }\times \frac{4\text{ moles of Oxygen}}{2\text{ moles of mercury (II) thiocyanate}}\text{ = 0}\text{.26772 moles of Oxygen}$

Since there are only 0.02 moles of oxygen then it is correct to say that the oxygen is the limiting reactant.

Some mercury (II) thiocyanate will remain unreacted; all the oxygen will react to produce the new compounds.

Step 3: Calculate the number of gas moles produced

According to the set of chemical reactions 4 moles of oxygen are consumed to produce 8 gas moles, we are interested in the number of gas moles that will be produced so calculate the gas moles expected:

$0.02\text{ moles of oxygen }\times \text{ 2 = 0}\text{.04 gas moles produced}$

The total number of gas moles will be:

$\begin{array}{l}0.04\text{ moles of oxygen}\\ \underset{\_}{\text{+0}\text{.0783 moles of nitrogen}}\\ 0.1183\text{ moles of gas}\end{array}$

Step 4: Calculate the final pressure of the system

The ideal gas equation is:

$\text{Pressure x Volume = moles x Gas Constant x Temperature}$

Rearrange the equation to calculate the pressure:

$\frac{\text{moles }\times \text{ Gas Constant}\times \text{Temperature}}{\text{Volume}}\text{=Pressure}$

Substitute the values:

$\frac{\text{0}\text{.1183 moles }\times \text{ 0}\text{.082 }\times \text{294}\text{.15 K}}{2.4\text{ L}}\text{=1}\text{.1889 atm}$

Change the units from atm to torr:

Pressure is 758 torr, change torr to atm:

$\text{1 atm = 760 torr}$ ;

$\text{1}\text{.1889 atm }\times \frac{760\text{ torr}}{\text{1 atm}}=903.586\text{ torr}$

Interpretation Introduction

Interpretation:

It is required to calculate the number of initial moles of the reactants, identify the limiting and excess reactant, calculate the moles of the products and use the ideal gas equation to find the final pressure of the system.

Concept introduction:

• The ideal gas equation relates the pressure, volume, and temperature of a system.
• $\text{Pressure }\times \text{ Volume = moles }\times \text{ Gas Constant }\times \text{ Temperature}$
• The limiting reactant is the chemical species that will be consumed totally, stoichiometric calculations must be done based on the limiting reactant.
• The excess reactant is the chemical species that will not be consumed totally.
• To change units of a measured quantity a proper conversion factor is required.
• The gases will contribute to the pressure change of the system.
• If gases are produced then the pressure of the system is expected to change.

Answer to Problem 5.105PAE

Solution:

The final pressure will be 903.58 torrs.

Explanation of Solution

Step1: Calculate the amount of oxygen

It is possible to calculate the amount of oxygen in the system by assuming that the oxygen is behaving like an ideal gas.

The ideal gas equation is:

$\text{Pressure }\times \text{ Volume = moles }\times \text{ Gas Constant }\times \text{ Temperature}$

Rearrange the equation to find the number of moles:

$\frac{\text{Pressure }\times \text{Volume}}{\text{Gas Constant }\times \text{ Temperature}}=\text{moles}$

Pressure must be in atmospheres

Gas constant is $\text{0}\text{.0821 }\frac{\text{atm L}}{\text{K mole}}$

The temperature must be in Kelvin

The volume must be in liters

From the statement:

Pressure is 758 torr, change torr to atm:

$\text{1 atm = 760 torr}$ ;

$\text{758 torr }\times \frac{\text{1 atm}}{\text{760 torr}}=0.997\text{ atm}$

Temperature is $\text{21 }\text{C}$, change it to Kelvin:

$\text{Temperature = 21 + 273}\text{.15 = 294}\text{.15 K}$

Volume is 2.4 Liters

Calculate the number of moles:

$\frac{\text{0}\text{.997 atm }\times \text{ 2}\text{.4 L}}{\text{0}\text{.082 }\times \text{ 294}\text{.15 K}}=0.0992\text{ moles}$

Air is composed by:

21 % Oxygen and 79 % Nitrogen

So, there are:

$0.21\times \text{ 0}\text{.0992 = 0}\text{.02 moles of Oxygen}$

and

$0.79\times \text{ 0}\text{.0992 = 0}\text{.0783 moles of Nitrogen}$

Step 2: Calculate the number of moles of mercury (II) thiocyanate

The number of moles is the ratio of the mass of the compound to its molar mass:

$\text{moles = }\frac{\text{mass}}{\text{molar mass}}$

There are 42.4 grams of mercury (II) thiocyanate, molar mass is $\text{316}\text{.75 }\frac{\text{grams}}{\text{mole}}$

$\text{moles = }\frac{\text{42}\text{.4 grams}}{\text{316}\text{.75 }\frac{\text{grams}}{\text{mole}}}\text{=0}\text{.13386 moles}$

According to the reaction, 2 moles of mercury (II) thiocyanate need 4 moles of oxygen, 0.13386 moles of mercury (II) thiocyanate will require:

$\text{0}\text{.13386 moles of mercury (II) thiocyanate }\times \frac{4\text{ moles of Oxygen}}{2\text{ moles of mercury (II) thiocyanate}}\text{ = 0}\text{.26772 moles of Oxygen}$

Since there are only 0.02 moles of oxygen then it is correct to say that the oxygen is the limiting reactant.

Some mercury (II) thiocyanate will remain unreacted; all the oxygen will react to produce the new compounds.

Step 3: Calculate the number of gas moles produced

According to the set of chemical reactions 4 moles of oxygen are consumed to produce 8 gas moles, we are interested in the number of gas moles that will be produced so calculate the gas moles expected:

$0.02\text{ moles of oxygen }\times \text{ 2 = 0}\text{.04 gas moles produced}$

The total number of gas moles will be:

$\begin{array}{l}0.04\text{ moles of oxygen}\\ \underset{\_}{\text{+0}\text{.0783 moles of nitrogen}}\\ 0.1183\text{ moles of gas}\end{array}$

Step 4: Calculate the final pressure of the system

The ideal gas equation is:

$\text{Pressure x Volume = moles x Gas Constant x Temperature}$

Rearrange the equation to calculate the pressure:

$\frac{\text{moles }\times \text{ Gas Constant}\times \text{Temperature}}{\text{Volume}}\text{=Pressure}$

Substitute the values:

$\frac{\text{0}\text{.1183 moles }\times \text{ 0}\text{.082 }\times \text{294}\text{.15 K}}{2.4\text{ L}}\text{=1}\text{.1889 atm}$

Change the units from atm to torr:

Pressure is 758 torr, change torr to atm:

$\text{1 atm = 760 torr}$ ;

$\text{1}\text{.1889 atm }\times \frac{760\text{ torr}}{\text{1 atm}}=903.586\text{ torr}$