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Characteristics of Equilibrium Constant

At 21.5oC a...

Question

At ${21.5}^{o} C$ and a total pressure of $0.0787 a t m$ , $N_{2} O_{4}$ is $48.3$ % dissociated into ${N O}_{2}$ . At what total pressure will the percent dissociation be $10.0$ %?

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Solution

$N_{2} O_{4} \to 2 N O_{2}$
Let P atm be the initial pressure of $N_{2} O_{4}$
48.3% of it or $P \times \frac{48.3}{100} = 0.483 P$ atm of it will dissociate to form $2 \times 0.483 P = 0.966 P$ atm of $N O_{2}$
The equilibrium pressure of $N_{2} O_{4} = P - 0.483 P = 0.517 P$ atm.
Total pressure $= 0.517 P + 0.966 P = 1.483 P = 0.0787$ atm
$P = \frac{0.0787}{1.483} = 0.0531$ atm
The equilibrium constant $K_{P} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}}$
$K_{p} = \frac{(0.966 P)^{2}}{0.517 P}$
$K_{P} = 1.8 P$
$K_{P} = 1.8 \times 0.0531$
$K_{p} = 0.0958$
10% of $N_{2} O_{4}$ or $P \times \frac{10}{100} = 0.1 P$ atm of it will dissociate to form $2 \times 0.1 P = 0.2 P$ atm of $N O_{2}$
The equilibrium pressure of $N_{2} O_{4} = P - 0.1 P = 0.9 P$ atm.
Total pressure $= 0.9 P + 0.2 P = 1.1 P$ atm
The equilibrium constant $K_{P} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}}$
$0.0958 = \frac{(0.2 P)^{2}}{0.9 P}$
$0.0958 = 0.0444 P$
$P = 2.16$
Total pressure $= 1.1 P = 1.1 \times 2.16 = 2.37$ atm.

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