Chapter 14, Problem 14.128QP

Interpretation Introduction

Interpretation:

The amount of $\text{NO}$ produced from the given amount of dry air has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$: For gaseous substances the concentration will be proportional to its partial pressure.  The equilibrium constant for gaseous  reactions can be expressed in terms of the partial pressures of the gaseous products and reactants, and it is called equilibrium constant ${\text{K}}_{\text{p}}$.  The value of ${\text{K}}_{\text{p}}$ is different from ${\text{K}}_c$.  The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_c$ is given below.

The partial pressure of an ideal gas in a gaseous mixture is related to the mole fraction of the gas by the given equation.

${\text{P}}_{\text{i}}\text{=}{\text{x}}_{\text{i}}{\text{×P}}_{\text{total}}$

Where,

${\text{P}}_{\text{i}}$ is the individual partial pressure of the gas

${\text{x}}_{\text{i}}$ is the individual mole fraction of the gas

${\text{P}}_{\text{total}}$ total pressure of the gaseous mixture

Answer to Problem 14.128QP

The amount of $\text{NO}$ produced from the $\text{100 g}$ of the dry air is was found to be $\text{2}\text{.1g}$

Explanation of Solution

Given,

Volume of the flask $\text{=2}\text{.00 L}$

$\text{Pressure}\text{= 1}\text{.00atm}$

$\text{Partial pressure of}{\text{N}}_2\text{=0}\text{.781 atm}$

$\text{Partial pressure of}{\text{O}}_{\text{2}}\text{=0}\text{.209 atm}$

$\text{Temperature,}{\text{T=2127}}^{°}\text{C}$

${\text{K}}_{\text{p}}=0.0025$

$\text{Amount}\text{of}\text{dry}\text{air=100}\text{.0g}$

${\text{Molar mass of O}}_2{\text{=16gmol}}^{-1}$

${\text{Molar mass of N}}_2{\text{=14gmol}}^{-1}$

To find the mole fractions of ${\text{N}}_{\text{2}}$and ${\text{O}}_{\text{2}}$

Partial pressure is related to mole fraction by the equation given below.

${\text{P}}_{\text{partial}}{\text{=X}}_{}{\text{P}}_{\text{total}}$

Hence,

Mole fractions of ${\text{N}}_{\text{2}}$

$\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}}{\text{=X}}_{{\text{N}}_{\text{2}}}{\text{P}}_{\text{total}}\\ \text{0}{\text{.781atm=X}}_{{\text{N}}_{\text{2}}}\text{(1}\text{.000atm)}\\ {\text{X}}_{{\text{N}}_{\text{2}}}=0.781\end{array}$

Mole fractions of ${\text{O}}_{\text{2}}$

$\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}{\text{=X}}_{{\text{O}}_{\text{2}}}{\text{P}}_{\text{total}}\\ \text{0}{\text{.209atm=X}}_{{\text{N}}_{\text{2}}}\text{(1}\text{.000atm)}\\ {\text{X}}_{{\text{O}}_{\text{2}}}\text{=0}\text{.209}\end{array}$

To find the molar mass of dry air

The molar mass of the dry air will be the sum of the products of the molar mass of the constituent gas with its molecular weight.

$\begin{array}{l}{\text{M}}_{\text{air}}{\text{=X}}_{{\text{N}}_{\text{2}}}{\text{M}}_{{\text{N}}_{\text{2}}}{\text{+X}}_{{\text{O}}_{\text{2}}}{\text{M}}_{{\text{O}}_{\text{2}}}\\ \text{=(0}\text{.781×28}{\text{.02gmol}}^{\text{-1}}\text{)+(0}\text{.209×32}{\text{.00gmol}}^{\text{-1}}\text{)}\\ \text{=}\text{28}{\text{.57gmol}}^{\text{-1}}\end{array}$

To find the number of moles of ${\text{N}}_{\text{2}}$and ${\text{O}}_{\text{2}}$

$100g$ of dry air has 3.500 moles of dry air.  The number of moles of each constituent gas can be found by multiplying the mole fraction of the gas with the molar mass of dry air.

Hence,

$\begin{array}{l}\text{Number of moles of}{\text{N}}_{\text{2}}={\text{X}}_{{\text{N}}_{\text{2}}}{\text{mol}}_{\text{dry}\text{air}}\\ \text{=0}\text{.781×3}\text{.500mol}\\ \text{=2}\text{.7335}\text{mol}\end{array}$

$\begin{array}{l}\text{Number of moles of}{\text{O}}_{\text{2}}={\text{X}}_{{\text{O}}_{\text{2}}}{\text{mol}}_{\text{dry}\text{air}}\\ \text{=0}\text{.209×3}\text{.500mol}\\ \text{=0}\text{.7315}\text{mol}\end{array}$

To find the equilibrium composition

The equilibrium composition was expressed by the table format.

$\begin{array}{l}\text{Pressure }\text{(atm)}{\text{N}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{O}}_{\text{2(g)}}\rightleftharpoons {\text{2NO}}_{\text{(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}2.7350.7315\rightleftharpoons \text{0}\\ \text{Change}\text{-x}-x\rightleftharpoons +2x\\ \text{Equillibrium}\left(2.735-\text{x}\right)(0.7315-x)\rightleftharpoons 2x\\ \end{array}$

Equilibrium constant ${\text{K}}_{\text{p}}$ is proportional to the partial pressures.

Hence,

$\begin{array}{l}{\text{K}}_{\text{p}}=\frac{{\text{P}}^2{}_{NO}}{{\text{P}}_{N_2}{P}_{O_2}}=\frac{{\text{n}}^{\text{2}}{}_{\text{NO}}}{{\text{n}}_{{\text{N}}_{\text{2}}}{\text{n}}_{{\text{O}}_{\text{2}}}}\\ \end{array}$

$\frac{{\text{n}}^{\text{2}}{}_{\text{NO}}}{{\text{n}}_{{\text{N}}_{\text{2}}}{\text{n}}_{{\text{O}}_{\text{2}}}}=\frac{{(2x)}^2}{(2.7335-x)(0.7315-x)}=0.0025$

On rearranging we get a quadratic equation.

$3.0095x^2-(8.6626\times {10}^{-3})x-4.999\times {10}^{-3}=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-(\text{8}\text{.6625}\times {\text{10}}^{-3})\pm \sqrt{{(\text{8}\text{.6625}\times {\text{10}}^{-3})}^2-4(3.997)(4.999\times {\text{10}}^{-3})}}{2(3.997)}$

$\text{x}=-3.65\times {10}^{-2}or3.43\times {10}^{-2}$

On solving we get two values for x, the logical value for x is taken.

$\text{x}=3.43\times {10}^{-2}$

$\text{Number of moles of NO}\text{=2x}\text{=}\text{2×3}{\text{.43×10}}^{\text{-2}}\text{=}\text{6}{\text{.86×10}}^{\text{-2}}\text{mol}$

To find the mass of $\text{NO}$

The product of number of moles and molecular weight will the amount of substance.

$\text{Mass}\text{of}\text{NO}\text{formed}\text{=}\text{6}{\text{.86×10}}^{\text{-2}}\text{mol ×30}{\text{.01gmol}}^{\text{-1}}=\text{2}\text{.1g}\text{NO}$

Conclusion

The equilibrium composition of the dry air was found. The partial pressure of a gas is proportional to number of moles of the gas.  By substituting number of moles of each constituent in the expression for ${\text{K}}_{\text{p}}$, the number of moles of $\text{NO}$ was found.  From the number of moles and the molecular mass the amount of $\text{NO}$ was found.