Chapter 10, Problem 89GQ

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide:

2 NO(g) + O₂(g) → 2 NO₂(g)

1. (a) Place the three gases in order of increasing rms speed at 298 K.
2. (b) If you mix NO and O₂ in the correct stoichiometric ratio and NO has a partial pressure of 150 mm Hg, what is the partial pressure of O₂?
3. (c) After reaction between NO and O₂ is complete; what is the pressure of NO₂ if the NO originally had a pressure of 150 mm Hg and O₂ was added in the correct stoichiometric amount?

Interpretation Introduction

Interpretation:

The given gases reaction should be arranged in the order of increasing rms speed at given temperature and $\left({\text{O}}_{\text{2}}\right)\text{ and }\left({\text{NO}}_{\text{2}}\right)$ partial pressure should calculate and identified.

Concept introduction:    

The root mean square velocity $\mu$ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

  ${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ (1)

  $\begin{array}{l}\left(\text{gas constant}\right)R=\frac{8.314J}{Kmol}\\ M=Molarmass\end{array}$

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The mixture of gases each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the entire volume of the original mixture at the same temperature.

Answer to Problem 89GQ

The increasing order of $\text{rms}$ speed for given set of gases at given temperature is as follows, ${\text{NO}}_{\text{2}}{\text{ < O}}_{\text{2}}\text{ < NO}$    

Explanation of Solution

The expression used to calculate the $\text{rms}$ speed is as follows,

  ${\text{μ}}_{\text{rms}}\text{=}{\left(\frac{\text{3}\text{R}\text{T}}{\text{M}}\right)}^{\text{1/2}}$

The $\text{rms}$ speed for the ${\text{NO}}_{\text{2}}$ molecule is

    $\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}{\left(\frac{\text{3}\text{R}\text{T}}{\text{M}}\right)}^{\text{1/2}}\\ \text{Molar }\text{mass}\text{of }{\text{NO}}_{\text{2}}=46.0055\text{g/mol}\\ ={\left(\frac{\text{3}\times \text{8}\text{.314}\times \text{298}}{46.0055}\right)}^{\text{1/2}}\\ =12.577\text{m/s}\end{array}$

The $\text{rms}$ speed for the ${\text{O}}_{\text{2}}$ molecule is

  $\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}{\left(\frac{\text{3}\text{R}\text{T}}{\text{M}}\right)}^{\text{1/2}}\\ ={\left(\frac{\text{3}\times \text{8}\text{.314}\times \text{298}}{32.00}\right)}^{\text{1/2}}\\ =15.24\text{m/s}\end{array}$

The $\text{rms}$ speed for the $\text{NO}$ molecule is

  $\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}{\left(\frac{\text{3}\text{R}\text{T}}{\text{M}}\right)}^{\text{1/2}}\\ ={\left(\frac{\text{3}\times \text{8}\text{.314}\times \text{298}}{30.01}\right)}^{\text{1/2}}\\ =15.73\text{m/s}\end{array}$

According to the calculated data the increasing order for the $\text{rms}$ speed value for given gases is ${\text{NO}}_{\text{2}}{\text{< O}}_{\text{2}}\text{ < NO}$.

Interpretation Introduction

Interpretation:

The given gases reaction should be arranged in the order of increasing rms speed at given temperature and $\left({\text{O}}_{\text{2}}\right)\text{ and }\left({\text{NO}}_{\text{2}}\right)$ partial pressure should calculate and identified.

Concept introduction:   

The root mean square velocity $\mu$ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

  ${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ (1)

  $\begin{array}{l}\left(\text{gas constant}\right)R=\frac{8.314J}{Kmol}\\ M=Molarmass\end{array}$

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The mixture of gases each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the entire volume of the original mixture at the same temperature.

Answer to Problem 89GQ

The partial pressure of oxygen $\left(O_2\right)hasP\left(O_2\right)=75mmHg$    

Explanation of Solution

Analyzing for partial pressure of oxygen molecule (O₂)

Let us consider the given the equation

  $\begin{array}{l}2NO(g)+O_2(g)\to 2N{O}_2(g)\\ ThepartialpressureofNO=150mmHg\\ Here\\ partialpressure=partialtemprature\times molefraction\\ partialpressure=150mmHg\times \frac{1}{2}\\ partialpressure=75mmHg\end{array}$

The partial pressure of oxygen molecule has $P(O_2)=75mmHg$

Interpretation Introduction

Interpretation:

The given gases reaction should be arranged in the order of increasing rms speed at given temperature and $\left({\text{O}}_{\text{2}}\right)\text{ and }\left({\text{NO}}_{\text{2}}\right)$ partial pressure should calculate and identified.

Concept introduction:   

The root mean square velocity $\mu$ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

  ${\mu }_{rms}={\left(\frac{3RT}{M}\right)}^{1/2}$ (1)

  $\begin{array}{l}\left(\text{gas constant}\right)R=\frac{8.314J}{Kmol}\\ M=Molarmass\end{array}$

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The mixture of gases each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the entire volume of the original mixture at the same temperature.

Answer to Problem 89GQ

The partial pressure of product molecule $\left(N{O}_2\right)wasP\left(N{O}_2\right)=150mmHg$

Explanation of Solution

Analyzing for partial pressure of (NO₂) molecule.

  $\begin{array}{l}2NO(g)+O_2(g)\to 2N{O}_2(g)\\ ThepartialpressureofN{O}_2=150mmHg\\ Here\\ partialpressure=partialtemprature\times molefraction\\ partialpressure=150mmHg\times \frac{1}{1}\\ partialpressure=150mmHg\end{array}$

The partial pressure of (NO₂) molecule has $P(N{O}_2)=150mmHg$