Question

Equilibrium Exercise #1 A flask is charged with 2.00 atm of nitrogen dioxide and 1.00 atm of dinitrogen tetroxide at 25 oC and allowed to reach equilibrium. When equilibrium is established, the partial pressure of NO2 has decreased by 1.24 atm. (a) What are the partial pressures of NO2 and N2O4 at equilibrium? (b) Calculate Kp and Kc for following reaction at 25 oC.

2 NO2(g) ⇄ N2O4(g)

(Answer: Kp = 2.80; Kc = 68.6)

Answer 1

The reaction is $2\,NO_{2(g)} ⇄ N_2O_{4(g)}$ and since the pressure is directly linked to the concentration of substances we can define the equilibrium stare for both gases

$\begin{matrix} & 2\,NO_{2(g)} & ⇄ & N_2O_{4(g)} \\ start & 2\,atm=2P_o & & 1\,atm=P_o \\ reaction & -2xP_o & & +xP_o \\ equilibrium & 2P_o(1-x) & & P_o(1+x) \end{matrix}$

Then we find that for $NO_{2(g)}$ we have $\Delta P=-2P_ox$ and from there

$x=-\frac{\Delta P}{2P_o}=\frac{-(-1.24\,atm)}{(2)(1.00\,atm)}=0.62$

This defines $P_{NO_{2(g)}(equilibrium)}=2P_o(1-x)=(2*1\,atm)(1-0.62)=0.76\,atm$

and as well $P_{N_2O_{4(g)}(equilibrium)}=P_o(1+x)=(1\,atm)(1+0.62)=1.62\,atm$

With that information, we can calculate $K_c$ and $K_p$ as

$K_p=\large{ \frac{P_{N_2O_4}}{P_{NO_2}^2} }=\large{ \frac{1.62}{(0.76)^2} }=2.805$

$K_c= \frac{[N_2O_4]}{ [{NO_2}]^2 }=\large{ \frac{ \frac{ P_{N_2O_4}}{RT} }{ ( \frac{P_{NO_2}}{RT} )^2} }= \frac{ P_{N_2O_4} }{ ( P_{NO_2} )^2} * \frac{ (RT)^2 }{ RT } = {K_p}{RT}$

Then we use R=0.0821 atmL/molK and T=298.15 K to find:

$\implies K_c = {K_p}{RT}=(2.805)(0.0821)(298.15)=68.66$

In conclusion, (a) the partial pressures of NO₂ and N₂O₄ at equilibrium are 0.76 atm and 1.62 atm, respectively ; (b) the values calculated for K_c and K_p are 68.66 and 2.805 at 25 °C, respectively.

Reference:

• Atkins, P., & De Paula, J. (2011). Physical chemistry for the life sciences. Oxford University Press, USA.