Question

Weight of \[13.8g\] of \[{N_2}{O_4}\] was placed in a 1L reaction vessel at \[400K\] and allowed attain equilibrium
\[{N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right)\]
The total pressure at equilibrium was found to be 9.15 bar. Calculate \[{K_C}\] ​, \[{K_P}\]​ and partial pressure at equilibrium.

Answer 1

Hint: \[{K_C}\] is an equilibrium constant in terms of molar concentrations and \[{K_P}\] is an equilibrium constant in terms of partial pressures. We will use the ideal gas equation \[PV = nRT\]. For a gas phase reaction, \[aA + bB \rightleftharpoons cC + dD\] , the expression for \[{K_P}\] is:
\[{K_P} = \dfrac{{{{(PC)}^c}{{(PD)}^d}}}{{{{(PA)}^a}{{(PB)}^b}}}\] .

Complete step by step answer:
Given in the question are,
Weight of \[{N_2}{O_4} = 13.8g\]
Temperature \[ = 400K\]
Total pressure \[ = 9.15\;bar\]
We have to find \[{K_C}\] ​, ​\[{K_P}\] and partial pressure at equilibrium.
Pressure of \[{N_2}{O_4}\] at \[400K\]
\[PV = nRT\]
$P = \dfrac{{nRT}}{V}$
$ \Rightarrow P = \dfrac{{0.15 \times 0.0821 \times 400}}{1}$
$ \Rightarrow P = 4.9bar$
$ \Rightarrow P \approx 5bar$
\[{N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right)\]

${N_2}{O_4}$	$N{O_2}$
$5-x $	$2x$ (at equilibrium)

Total pressure
\[5 + x = 9.15bar\] or
\[x = 4.15{\text{ }}bar\]
So,
\[{K_P} = \dfrac{{{{\left( {{P_{N{O_2}}}} \right)}^2}}}{{{P_{{N_2}{O_4}}}}}\]
\[ \Rightarrow {K_P} = \dfrac{{{{\left( {2 \times 4.15} \right)}^2}}}{{{{(0.85)}^2}}}\]
\[ \Rightarrow {K_P} = 95.34\]
Now, \[{K_C} = {\text{ }}{K_P}R{T^{\Delta ng}}\]
\[ \Rightarrow {K_C} = 95.34 \times 0.0821 \times {400^1}\]
\[ \Rightarrow {K_C} = 3130.96\]

Note: \[{K_C}\] and \[{K_P}\] are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that \[{K_C}\] is defined by molar concentrations, whereas \[{K_P}\] is defined by the partial pressures of the gasses inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solid and they may have units depending on the nature of the reaction. In a mixture of gases, it is the pressure an individual gas exerts. The partial pressure is independent of other gases that may be present in a mixture. According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature.