Chapter 17, Problem 17.100P

(a)

Interpretation Introduction

Interpretation:

Equilibrium pressure of $N$ in $N_2$ for the given reaction has to be calculated at $1000\text{ K}$ and $200\text{ atm}\text{.}$

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

$\text{aA}\text{+}\text{bB}\underset{}{\rightleftharpoons }\text{cC}\text{+}\text{dD}$

The equilibrium constant in terms of  partial pressure is, $K=\frac{{\left(P_{\text{C}}\right)}^{\text{c}}{\left(P_D\right)}^{\text{d}}}{{\left(P_A\right)}^{\text{a}}{\left(P_{\text{B}}\right)}^{\text{b}}}$

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

Answer to Problem 17.100P

Equilibrium pressure of $N$ is $4.0\times 10^{-21}atm.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{N}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2N}\left(g\right)\log {\text{K}}_P\text{ =}-\text{43}\text{.10}\\ \\ Initial{\text{ partial pressure of N}}_{\text{2}}=\text{ 200 atm}\\ \\ Temperature\text{ = 1000 K}\end{array}$

Equilibrium constant for the given reaction is given in terms of partial,

  $K_p\text{ = }\frac{{\left(P_N\right)}^2}{\left(P_{N_2}\right)}$

Calculate $K_p$ as follows,

  $\begin{array}{c}\log {\text{ K}}_p\text{ = }-43.10\\ \\ K_p\text{ = 7}\text{.943}\times {\text{10}}^{-44}\end{array}$

$K_p$ for the given reaction is $\text{7}\text{.943}\times {\text{10}}^{-44}.$

Equilibrium pressure of $N$ in $N_2$ can be determined by constructing ICE table.

  $\begin{array}{l}{\text{ N}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2N}\left(g\right)\\ \\ \begin{array}{ccc}Initial& 200& 0\\ Change& -x& +2x\\ Equilibrium& 200-x& 2x\end{array}\end{array}$

Substitute these values for the equation of equilibrium constant as given,

  $\begin{array}{c}{K}_p\text{ = }\frac{{\left(P_N\right)}^2}{\left(P_{N_2}\right)}\\ \\ \text{7}\text{.943}\times {\text{10}}^{-44}\text{ = }\frac{{\left(2x\right)}^2}{\left(200-x\right)}\end{array}$

Here, assume that $\left(200-x\right)\approx 200$. Hence,

  $\begin{array}{c}\text{7}\text{.943}\times {\text{10}}^{-44}\text{ = }\frac{{\left(2x\right)}^2}{\left(200\right)}\\ \\ x\text{ = 1}\text{.99}\times {\text{10}}^{-21}\end{array}$

Therefore,

Equilibrium pressure of $N$ is calculated as follows,

  $\begin{array}{c}{P}_N\left(Equlibrium\right)\text{ = 2}\left(\text{1}\text{.99}\times {\text{10}}^{-21}\right)\\ \\ =\text{ 4}\text{.0}\times {\text{10}}^{-21}\text{ atm N}\end{array}$

Equilibrium pressure of $N$ is $4.0\times 10^{-21}atm.$

(b)

Interpretation Introduction

Interpretation:

Equilibrium pressure of $H$ in $H_2$ for the given reaction has to be calculated at $1000\text{ K}$ and $600\text{ atm}\text{.}$

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

$\text{aA}\text{+}\text{bB}\underset{}{\rightleftharpoons }\text{cC}\text{+}\text{dD}$

The equilibrium constant in terms of partial pressure is, $K=\frac{{\left(P_{\text{C}}\right)}^{\text{c}}{\left(P_D\right)}^{\text{d}}}{{\left(P_A\right)}^{\text{a}}{\left(P_{\text{B}}\right)}^{\text{b}}}$

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

Answer to Problem 17.100P

Equilibrium pressure of $H$ is $5.5\times 10^{-8}atm.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{H}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2H}\left(g\right)\log {\text{K}}_P\text{ =}-\text{17}\text{.30}\\ \\ Initial{\text{ partial pressure of H}}_{\text{2}}=\text{ 600 atm}\\ \\ Temperature\text{ = 1000 K}\end{array}$

Equilibrium constant for the given reaction is given in terms of partial,

  $K_p\text{ = }\frac{{\left(P_H\right)}^2}{\left(P_{H_2}\right)}$

Calculate $K_p$ as follows,

  $\begin{array}{c}\log {\text{ K}}_p\text{ = }-\text{17}\text{.30}\\ \\ K_p\text{ = 5}\text{.0119}\times {\text{10}}^{-18}\end{array}$

$K_p$ for the given reaction is $\text{5}\text{.0119}\times {\text{10}}^{-18}.$

Equilibrium pressure of $H$ in $H_2$ can be determined by constructing ICE table.

  $\begin{array}{l}{\text{ H}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2H}\left(g\right)\\ \\ \begin{array}{ccc}Initial& 600& 0\\ Change& -x& +2x\\ Equilibrium& 600-x& 2x\end{array}\end{array}$

Substitute these values for the equation of equilibrium constant as given,

  $\begin{array}{c}{K}_p\text{ = }\frac{{\left(P_H\right)}^2}{\left(P_{H_2}\right)}\\ \\ \text{5}\text{.0119}\times {\text{10}}^{-18}\text{ = }\frac{{\left(2x\right)}^2}{\left(600-x\right)}\end{array}$

Here, assume that $\left(600-x\right)\approx 600$. Hence,

  $\begin{array}{c}\text{5}\text{.0119}\times {\text{10}}^{-18}\text{ = }\frac{{\left(2x\right)}^2}{\left(600\right)}\\ \\ x\text{ = 2}\text{.74}\times {\text{10}}^{-8}\end{array}$

Therefore,

Equilibrium pressure of $H$ is calculated as follows,

  $\begin{array}{c}{P}_H\left(Equlibrium\right)\text{ = 2}\left(\text{2}\text{.74}\times {\text{10}}^{-8}\right)\\ \\ =\text{ 5}\text{.5}\times {\text{10}}^{-8}\text{ atm H}\end{array}$

Equilibrium pressure of $H$ is $5.5\times 10^{-8}atm.$

(c)

Interpretation Introduction

Interpretation:

Number of $N$ atoms and $H$ atoms for the given reaction has to be calculated.

Concept Introduction:

According to Ideal gas equation,

  $\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Number of atoms can be determined using the given formula,

  $No.\text{ of atoms = No}\text{. of moles }\times \text{ Avogadro's number}$

Answer to Problem 17.100P

• Number of $N$ atoms is $29Natoms/L.$
• Number of $H$ atoms is $4.0\times 10^{14}Hatoms/L.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{N}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2N}\left(g\right)\log {\text{K}}_P\text{ =}-\text{43}\text{.10}\\ \\ {\text{H}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2H}\left(g\right)\log {\text{K}}_P\text{ =}-\text{17}\text{.30}\\ \\ Equilibriumpressureof\text{ N = 4}{\text{.0×10}}^{\text{-21}}\text{ atm}\\ \\ Equilibriumpressureof\text{ H = 5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\\ \\ \text{Volume of both reactions = 1 L}\\ \\ \text{Temperature of both reactions = 1000 K}\end{array}$

• Calculate number of moles of $N$:

Number of moles of $N$ can be calculated using Ideal gas equation as follows,

$\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{4}{\text{.0×10}}^{-\text{21}}\text{ atm}\right)\left(1L\right)=\text{ n}\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)\\ \\ n\text{ = }\frac{\left(\text{4}{\text{.0×10}}^{-\text{21}}\text{ atm}\right)\left(1L\right)}{\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)}\\ \\ =\text{ 4}\text{.87}\times {\text{10}}^{-23}\text{ mol N}\end{array}$

Number of moles of $N$ is $\text{4}\text{.87}\times {\text{10}}^{-23}\text{ mol}\text{.}$

• Calculate number of $N$ atoms:

Number of $N$ atoms can be obtained by multiplying number of moles of $N$ with Avogadro’s number.  Number of $N$ atoms is determined as shown below,

  $\begin{array}{c}No.\text{ of N atoms = }\left(\text{4}\text{.87}\times {\text{10}}^{-23}\text{ N mol}\right)\left(\frac{6.022\times {10}^{23}\text{ N atoms}}{1\text{ mol N}}\right)\\ \\ =\text{ 29 N atoms/L }\end{array}$

Number of $N$ atoms is $29Natoms/L.$

• Calculate number of moles of $H$:

Number of moles of $H$ can be calculated using Ideal gas equation as follows,

$\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\right)\left(1L\right)=\text{ n}\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)\\ \\ n\text{ = }\frac{\left(\text{5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\right)\left(1L\right)}{\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)}\\ \\ =\text{ 6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\end{array}$

Number of moles of $H$ is $\text{6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\text{.}$

• Calculate number of $H$ atoms:

Number of $H$ atoms can be obtained by multiplying number of moles of $H$ with Avogadro’s number.  Number of $H$ atoms is determined as shown below,

  $\begin{array}{c}No.\text{ of N atoms = }\left(\text{6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\right)\left(\frac{6.022\times {10}^{23}\text{ N atoms}}{1\text{ mol N}}\right)\\ \\ =\text{ 4}\text{.0}\times {\text{10}}^{14}\text{ H atoms/L }\end{array}$

Number of $H$ atoms is $4.0\times 10^{14}Hatoms/L.$

(c)

Interpretation Introduction

Interpretation:

Number of $N$ atoms and $H$ atoms for the given reaction has to be calculated.

Concept Introduction:

According to Ideal gas equation,

  $\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Number of atoms can be determined using the given formula,

  $No.\text{ of atoms = No}\text{. of moles }\times \text{ Avogadro's number}$

Answer to Problem 17.100P

• Number of $N$ atoms is $29Natoms/L.$
• Number of $H$ atoms is $4.0\times 10^{14}Hatoms/L.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{N}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2N}\left(g\right)\log {\text{K}}_P\text{ =}-\text{43}\text{.10}\\ \\ {\text{H}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2H}\left(g\right)\log {\text{K}}_P\text{ =}-\text{17}\text{.30}\\ \\ Equilibriumpressureof\text{ N = 4}{\text{.0×10}}^{\text{-21}}\text{ atm}\\ \\ Equilibriumpressureof\text{ H = 5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\\ \\ \text{Volume of both reactions = 1 L}\\ \\ \text{Temperature of both reactions = 1000 K}\end{array}$

• Calculate number of moles of $N$:

Number of moles of $N$ can be calculated using Ideal gas equation as follows,

$\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{4}{\text{.0×10}}^{-\text{21}}\text{ atm}\right)\left(1L\right)=\text{ n}\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)\\ \\ n\text{ = }\frac{\left(\text{4}{\text{.0×10}}^{-\text{21}}\text{ atm}\right)\left(1L\right)}{\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)}\\ \\ =\text{ 4}\text{.87}\times {\text{10}}^{-23}\text{ mol N}\end{array}$

Number of moles of $N$ is $\text{4}\text{.87}\times {\text{10}}^{-23}\text{ mol}\text{.}$

• Calculate number of $N$ atoms:

Number of $N$ atoms can be obtained by multiplying number of moles of $N$ with Avogadro’s number.  Number of $N$ atoms is determined as shown below,

  $\begin{array}{c}No.\text{ of N atoms = }\left(\text{4}\text{.87}\times {\text{10}}^{-23}\text{ N mol}\right)\left(\frac{6.022\times {10}^{23}\text{ N atoms}}{1\text{ mol N}}\right)\\ \\ =\text{ 29 N atoms/L }\end{array}$

Number of $N$ atoms is $29Natoms/L.$

• Calculate number of moles of $H$:

Number of moles of $H$ can be calculated using Ideal gas equation as follows,

$\begin{array}{c}PV\text{ = nRT}\\ \\ \left(\text{5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\right)\left(1L\right)=\text{ n}\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)\\ \\ n\text{ = }\frac{\left(\text{5}\text{.5}\times {\text{10}}^{-8}\text{ atm}\right)\left(1L\right)}{\left(0.0821\text{ L}\text{.atm/mol}\text{.K}\right)\left(1000\text{ K}\right)}\\ \\ =\text{ 6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\end{array}$

Number of moles of $H$ is $\text{6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\text{.}$

• Calculate number of $H$ atoms:

Number of $H$ atoms can be obtained by multiplying number of moles of $H$ with Avogadro’s number.  Number of $H$ atoms is determined as shown below,

  $\begin{array}{c}No.\text{ of N atoms = }\left(\text{6}\text{.7}\times {\text{10}}^{-10}\text{ mol H}\right)\left(\frac{6.022\times {10}^{23}\text{ N atoms}}{1\text{ mol N}}\right)\\ \\ =\text{ 4}\text{.0}\times {\text{10}}^{14}\text{ H atoms/L }\end{array}$

Number of $H$ atoms is $4.0\times 10^{14}Hatoms/L.$

(d)

Interpretation Introduction

Interpretation:

More reasonable step to continue the mechanism after the catalytic dissociation has to be explained.

Explanation of Solution

Number of $N$ atoms present in $1.0\text{ L}$ is only $29$ and it indicates that basically no $NH$ molecules are produced during the first reaction.

There are more $N_2$ molecules than $N$ atoms in order of magnitude and hence the more reasonable step is the second reaction.

  ${\text{N}}_{\text{2}}\left(g\right)\text{ + H}\left(g\right)\underset{}{\to }\text{ NH}\left(g\right)\text{ + N}\left(g\right)$