Chapter 17, Problem 17.103P

(a)

Interpretation Introduction

Interpretation:

Equilibrium pressure of $N_2$, $O_2$ and $NO$ for the given reaction has to be calculated.

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 17.103P

• Equilibrium pressure of $N_2$ is $0.780atm.$
• Equilibrium pressure of $O_2$ is $0.210atm.$
• Equilibrium pressure of $NO$ is $2.669\times 10^{-16}atm.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2NO}\left(g\right)\\ \\ Atmospheric{\text{ partial pressure of N}}_{\text{2}}=\text{ 0}\text{.780 atm}\\ \\ Atmospheric{\text{ partial pressure of O}}_{\text{2}}=\text{ 0}\text{.210 atm}\\ \\ {\text{K}}_P\text{ = 4}\text{.35}\times {10}^{-31}\end{array}$

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

  $K_p\text{ = }\frac{{\left(P_{NO}\right)}^2}{\left(P_{N_2}\right)\left(P_{O_2}\right)}$

Equilibrium pressure of $N_2$, $O_2$ and $NO$ can be determined by constructing ICE table.

  $\begin{array}{l}{\text{ N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2NO}\left(g\right)\\ \\ \begin{array}{cccc}Initial& \text{0}\text{.780}& \text{0}\text{.210}& 0\\ Change& -x& -x& +2x\\ Equilibrium& \left(\text{0}\text{.780}-x\right)& \left(\text{0}\text{.210}-x\right)& 2x\end{array}\end{array}$

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

  $\begin{array}{c}{K}_p\text{ = }\frac{{\left(P_{NO}\right)}^2}{\left(P_{N_2}\right)\left(P_{O_2}\right)}\\ \\ \text{ 4}\text{.35}\times {10}^{-31}\text{ = }\frac{{\left(2x\right)}^2}{\left(\text{0}\text{.780}-x\right)\left(\text{0}\text{.210}-x\right)}\end{array}$

Here, assume that $x<<K$. Hence,

  $\begin{array}{c}\text{4}\text{.35}\times {10}^{-31}\text{ = }\frac{{\left(2x\right)}^2}{\left(\text{0}\text{.780}\right)\left(\text{0}\text{.210}\right)}\\ \\ x\text{ = 1}\text{.3345}\times {\text{10}}^{-16}\end{array}$

Therefore,

Equilibrium pressure of $N_2$, $O_2$ and $NO$ is calculated as follows,

  $\begin{array}{c}{P}_{N_2}\left(Equlibrium\right)\text{ = }\left(0.780-\text{1}\text{.3345}\times {\text{10}}^{-16}\right)\\ \\ =\text{ 0}{\text{.780 atm N}}_2\\ \\ P_{O_2}\left(Equlibrium\right)\text{ = }\left(0.210-\text{1}\text{.3345}\times {\text{10}}^{-16}\right)\\ \\ =\text{ 0}{\text{.210 atm O}}_2\\ \\ P_{NO}\left(Equlibrium\right)\text{ = 2}\left(\text{1}\text{.3345}\times {\text{10}}^{-16}\right)\\ \\ =\text{ 2}\text{.669}\times {\text{10}}^{-16}\text{atm N}O\end{array}$

Equilibrium pressure of $N_2$ is $0.780atm.$

Equilibrium pressure of $O_2$ is $0.210atm.$

Equilibrium pressure of $NO$ is $2.669\times 10^{-16}atm.$

(b)

Interpretation Introduction

Interpretation:

$P_{container}$ in the container ${\text{N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2NO}\left(g\right)$ has to be calculated.

Concept Introduction:

According to Dalton’s law of partial pressures, the total pressure is the sum partial pressure of all the gaseous component present.

Answer to Problem 17.103P

${\text{K}}_{\text{p}}$ for the given reaction is $4.9\times 10^5.$

Explanation of Solution

Given data is shown below:

Total pressure is the sum of partial pressure of $N_2$, $O_2$ and $NO$

Therefore,

$P_{container}$ is determined as follows,

  $\begin{array}{c}{P}_{container}\text{ = }{P}_{N_2}\left(Equlibrium\right)+P_{O_2}\left(Equlibrium\right)+P_{NO}\left(Equlibrium\right)\\ \\ =\text{ 0}{\text{.780 atm N}}_2+\text{ 0}{\text{.210 atm O}}_2\text{ + 2}\text{.669}\times {\text{10}}^{-16}\text{atm N}O\\ \\ =\text{ 0}\text{.990 atm}\end{array}$

$P_{container}$ in the container is $0.990atm.$

(c)

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{C}}$ for the given reaction ${\text{N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2NO}\left(g\right)$ has to be calculated.

Concept Introduction:

The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ is given by the following equation.

${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\boxed{\begin{array}{l}{\text{K}}_{\text{p}}\text{ = }Equilibriumcons\tan tintermsofpartialpressure\\ {\text{K}}_{\text{c}}\text{ = }Equilibriumcons\tan tintermsofconcentration\\ \Delta {\text{n}}_{\text{gas}}\text{ = moles of gaseous product }-\text{ moles of gaseous reactant}\end{array}}$

Only moles of gaseous products and reactants are used for calculating $\Delta {\text{n}}_{\text{gas}}.$

Answer to Problem 17.103P

${\text{K}}_{\text{C}}$ for the given reaction is $4.9\times 10^5.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{K}}_P\text{ = 4}\text{.35}\times {10}^{-31}\\ \\ T\text{ = 298 K}\end{array}$

• Determine $\Delta n_{gas}$:

The total number of moles of gaseous reactants is 3 and the moles of gaseous product is 2.  $\Delta n_{gas}$ is determined as follows,

  $\begin{array}{l}\Delta {\text{n}}_{\text{gas}}\text{= moles of gaseous product - moles of gaseous reactant}\\ =2-2\\ =0\end{array}$

• Determine ${\text{K}}_{\text{C}}$:

${\text{K}}_{\text{C}}$ of the reaction can be determined from given ${\text{K}}_{\text{p}}$ as below,

  $\begin{array}{c}{\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\\ \\ \text{4}\text{.35}\times {10}^{-31}={\text{ K}}_{\text{c}}{\left[\left(0.0821\text{ atm}\text{.L/mol}\text{.K}\right)\left(298\text{ K}\right)\right]}^0\\ \\ {\text{K}}_{\text{c}}=\text{ 4}\text{.35}\times {10}^{-31}\end{array}$

Therefore,

${\text{K}}_{\text{C}}$ for the given reaction is $4.35\times 10^{-31}.$