Chapter 8, Problem 94QRT

(a)

Interpretation Introduction

Interpretation:

Partial pressure of $H_2$ and ${\text{Cl}}_2$ before reaction has to be determined.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Molar mass can be determined using the given equation,

  $\text{n}\text{= }\frac{\text{m}}{\text{M}}$

Here, n is the number of moles.

  M is the Molar mass.

    m is the Mass.

Answer to Problem 94QRT

Partial pressure of $H_2$ is $3.7\text{atm}$.

Partial pressure of ${\text{Cl}}_2$ is $4.9\text{ atm}$.

Explanation of Solution

Reaction is shown below,

  $H_2\left(g\right){\text{+ Cl}}_2\left(g\right)\underset{}{\to }\text{ 2 HCl}\left(g\right)$

Number of moles of $H_2$ and ${\text{Cl}}_2$ can be determined using the given formula,

  $No.\text{ of moles = }\frac{Mass}{Molar\text{ mass}}$

Substitute the values to obtain the number of moles of $H_2$ and ${\text{Cl}}_2$ as follows,

  $\begin{array}{l}{n}_{H_2}\text{ = }\frac{3.0\text{ g}}{2.02\text{ g/mol}}\text{ = 1}{\text{.5 mol H}}_2\\ \\ n_{C{l}_2}\text{ = }\frac{140.0\text{ g}}{\text{70}\text{.91 g/mol}}\text{ = 1}{\text{.97 mol Cl}}_2\end{array}$

Partial pressure of $H_2$ and ${\text{Cl}}_2$ is determined using Ideal Gas equation as follows,

  $\text{PV = nRT}$

Partial pressure of $H_2$ can be obtained by substituting the values as follows,

  $\begin{array}{c}T\text{ = 28}°\text{C = }\left(28+273\right)K=301\text{ K}\\ \\ {\text{P}}_{H_2}\text{ = }\frac{n_{H_2}RT}{V}\\ \\ =\frac{\left(1.5\text{ mol}\right)\left(0.0821\text{ L}\text{.atm}.{\text{K}}^{-1}{\text{.mol}}^{-1}\right)\left(301\text{ K}\right)}{10\text{L}}\\ \\ =3.7\text{atm}\end{array}$

Partial pressure of ${\text{Cl}}_2$ can be obtained by substituting the values as follows,

  $\begin{array}{c}{\text{P}}_{{\text{Cl}}_2}=\frac{\text{nRT}}{\text{V}}\\ \begin{array}{l}\begin{array}{l}\\ =\frac{\left(1.97\text{ mol}\right)\left(0.0821\text{ L}\text{.atm}.{\text{K}}^{-1}{\text{.mol}}^{-1}\right)\left(301\text{ K}\right)}{10\text{ L}}\\ \end{array}\hfill \\ =4.9\text{ atm}\hfill \end{array}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Total pressure before the reaction has to be determined.

Concept Introduction:

Dalton’s law of partial pressure:

According to this law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases.

  $\begin{array}{c}{P}_{total}\text{ = }{\displaystyle \sum p_i}\\ \\ P_{total}\text{ = }{p}_1{\text{ + p}}_2{\text{ + p}}_3\text{ + }\mathrm{....}\end{array}$

Answer to Problem 94QRT

Total pressure before the reaction is $\text{8}\text{.6 atm}$

Explanation of Solution

Partial pressure of $H_2$ is $3.7\text{atm}$.

Partial pressure of ${\text{Cl}}_2$ is $4.9\text{ atm}$.

According to Dalton’s law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases at constant temperature and volume.

Therefore, total pressure before the reaction is determined as follows,

  $\begin{array}{c}{P}_{total}{\text{ = P}}_{H_2}+P_{C{l}_2}\\ \\ =\left(3.7\text{atm}\right)+\left(4.9\text{ atm}\right)\\ \\ =\text{ 8}\text{.6 atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Total pressure after the reaction has to be determined.

Answer to Problem 94QRT

Total pressure after the reaction is $\text{8}\text{.6 atm}$

Explanation of Solution

Total pressure before the reaction is $\text{8}\text{.6 atm}$

Number of moles of gas reactants and the number of moles of gaseous products are equal. Hence, the total pressure after and before the reaction will be same.

Total pressure after the reaction is $\text{8}\text{.6 atm}$

(d)

Interpretation Introduction

Interpretation:

The reactant that remains in the flask after the reaction has to be determined. Also the amount of remaining reactant has to be calculated.

Concept Introduction:

Answer to Problem 94QRT

$\text{0}{\text{.5 mol Cl}}_2$ remains in the flask after the reaction.

Explanation of Solution

Reaction is shown below,

  $H_2\left(g\right){\text{+ Cl}}_2\left(g\right)\underset{}{\to }\text{ 2 HCl}\left(g\right)$

Number of moles of $H_2$ and ${\text{Cl}}_2$ can be determined using the given formula,

  $No.\text{ of moles = }\frac{Mass}{Molar\text{ mass}}$

Substitute the values to obtain the number of moles of $H_2$ and ${\text{Cl}}_2$ as follows,

  $\begin{array}{l}{n}_{H_2}\text{ = }\frac{3.0\text{ g}}{2.02\text{ g/mol}}\text{ = 1}{\text{.5 mol H}}_2\\ \\ n_{C{l}_2}\text{ = }\frac{140.0\text{ g}}{\text{70}\text{.91 g/mol}}\text{ = 1}{\text{.97 mol Cl}}_2\end{array}$

From the balanced equation, it is clear that $1\text{ mol }{H}_2$ and $1{\text{ mol Cl}}_2$ reacts to give $2\text{ mol HCl}$. Hence, $H_2$ the limiting reactant and the reactant that is in excess will be ${\text{Cl}}_2$.

Here, $\text{1}{\text{.5 mol H}}_2$ and $\text{1}{\text{.5 mol Cl}}_2$ is reacted. Therefore, number of moles of excess ${\text{Cl}}_2$ is determined as follows,

    $\begin{array}{c}{\text{Remaining n}}_{C{l}_2}{\text{ = Initial n}}_{C{l}_2}-{\text{Reacted n}}_{C{l}_2}\\ \\ =\left(\text{2}\text{.0 mol}\right)-\left(1.5\text{ mol}\right)\\ \\ =\text{}\text{ 0}{\text{.5 mol Cl}}_2\end{array}$

(e)

Interpretation Introduction

Interpretation:

Partial pressure of each gas after the reaction has to be determined.

Answer to Problem 94QRT

Partial pressure of $\text{HCl}$ is $7.4\text{ atm}$.

Partial pressure of ${\text{Cl}}_2$ is $\text{1}\text{.2 atm}$.

Explanation of Solution

Reaction is shown below,

  $H_2\left(g\right){\text{+ Cl}}_2\left(g\right)\underset{}{\to }\text{ 2 HCl}\left(g\right)$

From the balanced equation, it is clear that $1\text{ mol }{H}_2$ and $1{\text{ mol Cl}}_2$ reacts to give $2\text{ mol HCl}$. Hence, $1.5\text{ mol }{H}_2$ and $1.5{\text{ mol Cl}}_2$ reacts to give $3\text{ mol HCl}$.

Therefore, partial pressure of $\text{HCl}$ can be calculated as follows,

  $\begin{array}{c}\text{PV}=\text{nRT}\\ \\ \text{P}\left(10\text{ L}\right)=\left(3.0\text{ mol}\right)\left(0.0821\text{ L}\text{.atm}.{\text{K}}^{-1}{\text{.mol}}^{-1}\right)\left(301\text{ K}\right)\\ \begin{array}{l}\begin{array}{l}\\ =\frac{\left(3.0\text{ mol}\right)\left(0.0821\text{ L}\text{.atm}.{\text{K}}^{-1}{\text{.mol}}^{-1}\right)\left(301\text{ K}\right)}{10\text{ L}}\\ \end{array}\hfill \\ =7.4\text{ atm}\hfill \end{array}\end{array}$

In the given reaction, $H_2$ the limiting reactant and the reactant that is in excess will be ${\text{Cl}}_2$.

Total pressure after the reaction is $\text{8}\text{.6 atm}$

Therefore, partial pressure of ${\text{Cl}}_2$ is determined as follows,

  $\begin{array}{c}{P}_{C{l}_2}{\text{ = P}}_{tot}-P_{HCl}\\ \\ =\left(\text{8}\text{.6 atm}\right)-\left(\text{7}\text{.4 atm}\right)\\ \\ =\text{1}{\text{.2 atm Cl}}_{\text{2}}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Pressure inside the flask if the temperature is raised to $\text{40}°\text{C}$ has to be determined.

Answer to Problem 94QRT

Pressure inside the flask is $\text{8}\text{.9 atm}$.

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{T}_1\text{ = 28}°\text{C = }\left(28+273\right)K=301\text{ K}\\ \\ T_2\text{ = 40}°\text{C = }\left(40+273\right)K=313\text{ K}\\ \\ {\text{P}}_1\text{ = 8}\text{.6 atm}\end{array}$

Pressure inside the flask can be calculated using combined gas law that is proposed by combining the Boyle’s law and Charles’ law,

  $\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

Here, the volume of the tire is constant and equation becomes,

  $\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

Therefore, pressure inside the flask can be determined as shown below,

  $\begin{array}{c}\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}}\\ \\ \frac{\text{8}\text{.6 atm}}{\text{303 K}}\text{=}\frac{{\text{P}}_{\text{2}}}{\text{313 K}}\\ \\ {\text{P}}_{\text{2}}\text{ = }\frac{\left(\text{8}\text{.6 atm}\right)\left(\text{313 K}\right)}{\left(\text{303 K}\right)}\\ \\ =\text{ 8}\text{.9 atm}\end{array}$