Chapter 15, Problem 53GQ

A 15-L flask at 300 K contains 6.44 g of a mixture of NO₂ and N₂O₄ in equilibrium. What is the total pressure in the flask? (K_p for 2 NO₂ (g) ⇄ N₂O₄(g) is 7.1.)

Interpretation Introduction

Interpretation:

The total pressure in the flask that contain $\text{6}\text{.44}\text{g}$ mixture of ${\text{NO}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ has to be calculated

Concept Introduction:

Equilibrium constant in terms of pressure${\text{[K}}_{\text{p}}\text{]}$: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{\text{C}}{}^{\text{c}}\text{×}{\text{P}}_{\text{D}}{}^{\text{d}}}{{\text{P}}_{\text{A}}{}^{\text{a}}\text{×}{\text{P}}_{\text{B}}{}^{\text{b}}}\end{array}$

Ideal gas equation:

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\text{-}\text{Pressure}\\ \text{V}\text{-}\text{Volume}\\ \text{R}\text{-}\text{Universal}\text{gas}\text{constant}\\ \text{T}\text{-}\text{Temperature}\\ \text{n}\text{-}\text{number}\text{of}\text{moles}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}{\text{K}}_{\text{c}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{concentration}\\ {\text{K}}_{\text{p}}\text{-}\text{Equilibrium}\text{constant}\text{in}\text{terms}\text{of}\text{pressure}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Answer to Problem 53GQ

The total pressure in the flask is $\text{0}\text{.13}\text{atm}$.

Explanation of Solution

Given:

$\begin{array}{l}{\text{2NO}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}\\ {\text{K}}_{\text{P}}\text{=}\text{7}\text{.1}\\ \text{V}\text{=}\text{15}\text{L}\\ \text{T}\text{=}\text{300}\text{K}\end{array}$

Using the equation ${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}$, we can find the value of ${\text{K}}_{\text{c}}$

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{p}}}{{\left(\text{RT}\right)}^{\text{Δn}}}\text{=}\frac{\text{7}\text{.1}}{{\left(\text{0}\text{.082×300}\right)}^{\text{-1}}}\\ \text{=}\text{174}\text{.87}\end{array}$

For the reaction ${\text{2NO}}_{\text{2}}{}_{\text{(g)}}\rightleftarrows {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}$

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}{{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}}\\ \text{174}\text{.87}\text{=}\frac{\text{x}}{{\text{y}}^{\text{2}}}\\ \text{x}\text{=}\text{174}\text{.87}{\text{y}}^{\text{2}}\text{Equation}\text{(1)}\\ \end{array}$

$\begin{array}{l}\text{Molarity}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\frac{\text{No}\text{.}\text{of}\text{moles}}{\text{Volume}}\\ \text{x}\text{=}\frac{{\text{n}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{15}}\\ {\text{n}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}\text{15x}\\ \end{array}$

$\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}\frac{{\text{m}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\\ {\text{m}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{n}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{×}\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\\ \text{=}\text{15x×92}\\ \text{=}\text{1380x}\end{array}$

$\begin{array}{l}{\text{m}}_{{\text{NO}}_{\text{2}}}\text{=}{\text{n}}_{{\text{NO}}_{\text{2}}}\text{×}\text{Molar}\text{mass}\text{of}{\text{NO}}_{\text{2}}\\ \text{=}\text{15y×46}\\ \text{=}\text{690y}\end{array}$

Total mass is given as $\text{6}\text{.44}\text{g}$

$\begin{array}{l}\text{1380x}\text{+}\text{690y}\text{=}\text{6}\text{.44}\text{g}\\ \text{690}\left(\text{2x+y}\right)\text{=}\text{6}\text{.44}\\ \left(\text{2x+y}\right)\text{=}\text{0}\text{.0093}\\ \text{y}\text{=}\text{0}\text{.0093}\text{-2x}\text{Equation}\text{(2)}\end{array}$

Substituting (2) into (1),

$\begin{array}{l}\text{x=}\text{174}\text{.87}{\text{×y}}^{\text{2}}\\ \text{=}\text{174}\text{.87×}{\left(\text{0}\text{.0093-2x}\right)}^{\text{2}}\\ \text{Solving}\text{this}\text{we}\text{get}\\ \text{x}\text{=}\text{0}\text{.0038}\end{array}$

$\begin{array}{l}\text{y}\text{=}\text{0}\text{.0093-2×0}\text{.0038}\\ \text{=}\text{0}\text{.0017}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}\text{15x}\\ \text{=}\text{15×0}\text{.0038}\\ \text{=0}\text{.057}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{NO}}_{\text{2}}}\text{=}\text{15y}\\ \text{=}\text{15×0}\text{.0017}\\ \text{=}\text{0}\text{.025}\end{array}$

$\begin{array}{l}\text{Total}\text{number}\text{of}\text{moles}\text{=}\text{0}\text{.057+}\text{0}\text{.025}\\ \text{=}\text{0}\text{.082}\end{array}$

According to ideal gas equation

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\text{=}\frac{\text{nRT}}{\text{V}}\\ \text{=}\frac{\text{0}\text{.082×0}\text{.082×300}}{\text{15}}\\ \text{=}\text{0}\text{.13}\text{atm}\end{array}$

Conclusion

The total pressure in the flask that contain $\text{6}\text{.44}\text{g}$ mixture of ${\text{NO}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ was calculated