Chapter 11, Problem 11.108QP

Interpretation Introduction

Interpretation:

The partial pressure of the given ${\text{O}}_{\text{2}}{\text{ and NO}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

• The pressure exerted by an individual gas in a mixture is known as its partial pressure.
• Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
• Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:
• ${\text{P}}_{\text{Total}}\text{}{\text{=P}}_{\text{gas 1}}\text{}{\text{+P}}_{\text{gas 2}}\text{}{\text{+P}}_{\text{gas 3}}\text{}\mathrm{....}$
• ${\text{χ}}_{\text{i}}\text{=}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{total}}}$

Answer to Problem 11.108QP

The partial pressure of ${\text{O}}_{\text{2}}{\text{ and NO}}_{\text{2}}$ was found to be $0.166\text{atm}$ $\text{and 0}\text{.333}\text{atm}$ respectively.

Explanation of Solution

To calculate the initial number of moles of ${\text{O}}_{\text{2}}\text{ and NO}$

$\begin{array}{l}{\text{n}}_{\text{NO}}\text{=}\frac{{\text{P}}_{\text{NO}}\text{V}}{\text{RT}}\text{=}\frac{\text{(0}\text{.500}\text{atm)}\text{(4}\text{.00}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(298}\text{K)}}\text{=}\text{0}\text{.0817}\text{mol}\text{NO}\\ {\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{P}}_{{\text{O}}_{\text{2}}}\text{V}}{\text{RT}}\text{=}\frac{\text{(1}\text{.00}\text{atm)}\text{(2}\text{.00}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(298}\text{K)}}\text{=}\text{0}\text{.0817}\text{mol}{\text{O}}_{\text{2}}\end{array}$

To determine the molar amounts of ${\text{NO, O}}_{\text{2}}{\text{, and NO}}_{\text{2}}$after complete reaction.

$\text{mol NO = 0 mol}$ (All NO is consumed during reaction)

$\begin{array}{l}\text{mol}{\text{NO}}_{\text{2}}\text{=}\text{0}\text{.0817}\text{mol}\text{NO}\text{×}\frac{\text{2}\text{mol}{\text{NO}}_{\text{2}}}{\text{2}\text{mol}\text{NO}}\text{=}\text{0}\text{.0817}\text{mol}{\text{NO}}_{\text{2}}\\ \\ \text{mol}{\text{O}}_{\text{2}}\text{consumed}\text{=}\text{0}\text{.0817}\text{mol}\text{NO}\text{×}\frac{\text{1}\text{mol}{\text{O}}_{\text{2}}}{\text{2}\text{mol}\text{NO}}\text{=}\text{0}\text{.0409}\text{mol}{\text{O}}_{\text{2}}\text{consumed}\end{array}$

$\text{mol}{\text{O}}_{\text{2}}\text{remaining}\text{=}\text{0}\text{.0817}\text{mol}{\text{O}}_{\text{2}}\text{initial}\text{-}\text{0}\text{.0409}\text{mol}{\text{O}}_{\text{2}}\text{consumed}\text{=0}\text{.0408}\text{mol}{\text{O}}_{\text{2}}$

To calculate the partial pressure of the ${\text{O}}_{\text{2}}{\text{ and NO}}_{\text{2}}$

$\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{RT}}{\text{V}}\text{=}\frac{\text{(0}\text{.0408}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(298}\text{K)}}{\text{(6}\text{.00}\text{L)}}\text{=}\text{0}\text{.166}\text{atm}\\ \\ {\text{P}}_{{\text{NO}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{NO}}_{\text{2}}}\text{RT}}{\text{V}}\text{=}\frac{\text{(0}\text{.0817}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(298}\text{K)}}{\text{(6}\text{.00}\text{L)}}\text{=}\text{0}\text{.333}\text{atm}\end{array}$

Conclusion

The partial pressure of the given ${\text{O}}_{\text{2}}{\text{ and NO}}_{\text{2}}$ was calculated.