# 0.1 mole of N_2O_4(g) was sealed in a tube under one-atmosphere conditions at 25^oC. Calculate the number of moles of NO_2(g) present, if the equilibrium N_2O_4(g) -> 2NO_2(g) (K_p = 0.16) is reached after some time?

## A) $180$ B) $280$ C) $0.034$ D) $0.028$

Jul 24, 2018

$n \approx 0.036 \textcolor{w h i t e}{l} \text{mol}$

#### Explanation:

Calculate the volume of the container given initial conditions of the ${\text{N"_2 "O}}_{4}$ gas.

$\text{V} = \frac{n \cdot R \cdot T}{P}$
$\textcolor{w h i t e}{\text{V") = (0.1 color(white)(l) "mol" * 0.0205 color(white)(l) L * atm * "mol"^(-1) * "K"^(-1) * (273.15 + 25) color(white)(l) "K")/(1.00 color(white)(l) "atm}}$
color(white)("V") = 0.611 color(white)(l) "L"

Let the increase in the partial pressure of ${\text{NO}}_{2}$ be $x \textcolor{w h i t e}{l} \text{atm}$ and construct a partial pressure RICE table:

• R color(white)(I)"N"_2 "O"_4 (g) rightleftharpoons 2 color(white)(l) "NO"_2 (g)
• I color(white)(R) 1 color(white)(l) "atm" color(white)( (g) rightleftharpoons 2 color(white)(l) ) 0 color(white)(l) "atm"
• Ccolor(white)(I) -x/2 color(white)(._4 (g) rightleftharpoons l) +x
• E$\textcolor{w h i t e}{I} 1 \textcolor{w h i t e}{l} \text{atm} - \frac{x}{2} \textcolor{w h i t e}{- 2 l l} x$

$\frac{{x}^{2}}{1 - x / 2} = {K}_{p} = 0.16$

$x \approx 0.36 \textcolor{w h i t e}{l} \text{atm}$

Again, by the ideal gas equation:

$n \left({\text{NO}}_{2}\right) = \frac{P \cdot V}{R \cdot T}$
color(white)(n("NO"_2)) = (0.36 color(white)(l) "atm" * 0.611 color(white)(l)"L")/(0.0205 color(white)(l) L * atm * "mol"^(-1) * "K"^(-1) * (273.15 + 25) color(white)(l) "K")
color(white)(n("NO"_2)) = 0.036 color(white)(l) "mol"