Question

The dissociation of ${N_2}{O_4}$ takes place as per the equation ${N_2}{O_4}(g) \Leftrightarrow 2N{O_2}(g)$. ${N_2}{O_4}$ is $20\% $ dissociated while the equilibrium pressure of the mixture is $600$ mm of Hg. Calculate ${K_p}$ assuming the volume to be constant.
A.50
B.100
C.166.8
D.600

Answer 1

Hint: $Kc$ and ${K_p}$ are the equilibrium constant of the gaseous mixture. The difference between the two constants is that $Kc$ is defined by the molar concentrations and ${K_p}$ is defined by the partial pressure of the gases inside a closed system.

Complete answer:
The standard example of writing gas equilibrium constants are:
$aA + bB \rightleftarrows cC + dD$
${K_p} = \dfrac{{{{(C)}^c}{{(D)}^d}}}{{{{(A)}^a}{{(B)}^b}}}$
It is given that volume is constant. Let the concentration of ${N_2}{O_4}$ be $x$.

	$N{O_2}$${N_2}{O_4}$	$N{O_2}$
Initial	$x$	$0$
Equilibrium	$1 - x$	$2x$

If $20\% $of ${N_2}{O_4}$ is dissociated, then $0.2$ mol will be consumed.
Thus, the total concentration at equilibrium will be: $1 - x + 2x = 1 + x$
The total concentration will be= $1 + 0.2 = 1.2$
Mole fraction of each gas will be:
${X_{N{O_2}}} = \dfrac{{{n_{N{O_2}}}}}{{{n_{total}}}} = \dfrac{{2 \times 0.2}}{{1.2}} = 0.33$ (as $x = 0.2$)
${X_{{N_2}{O_2}}} = \dfrac{{{n_{{N_2}{O_4}}}}}{{{n_{total}}}} = \dfrac{{1 - 0.2}}{{1.2}} = 0.67$
Since the pressure is constant, ${P_{total}} = 600$ mm of Hg
Pressure at $N{O_2}$ will be:
${P_{N{O_2}}} = {X_{N{O_2}}} \times {P_{total}} = 0.33 \times 600 = 200$
Pressure at ${N_2}{O_4}$ will be:
${P_{{N_2}{O_4}}} = {X_{{N_2}{O_4}}} \times {P_{total}} = 0.67 \times 600 = 400$
The equilibrium constant ${K_p}$ can be written as:
${K_P} = \dfrac{{{{({P_{N{O_2}}})}^2}}}{{{P_{{N_2}{O_4}}}}}$
${K_P} = \dfrac{{{{(200)}^2}}}{{400}} = 100$

Note:
The difference between $Kc$ and ${K_p}$ is important to know, ${K_p}$ is an equilibrium constant in terms of partial pressure and uses parenthesis (), whereas $Kc$ is an equilibrium constant in terms of molar concentration and uses brackets [].