Chemistry - Chemical Equilibrium Question with Solution | TestHub

The dissociation of PCl₅ is given by the reaction:

PCl₅(g) $\rightleftharpoons$ PCl₃(g) + Cl₂(g)

Initial moles:

PCl₅ = 5 mol

N₂ = 2 mol (inert gas)

Total initial moles = $n_{total,initial}=5+2=7$ mol

Let $\alpha$ be the degree of dissociation of PCl₅.

At equilibrium:

Moles of PCl₅ = $5(1-\alpha )$

Moles of PCl₃ = $5\alpha$

Moles of Cl₂ = $5\alpha$

Moles of N₂ = 2 mol

Total moles at equilibrium = $n_{total,eq}=5(1-\alpha )+5\alpha +5\alpha +2=7+5\alpha$

Given:

Volume, $V=200$ L

Temperature, $T=600$ K

Equilibrium pressure, $P_{total}=2.46$ atm

Gas constant, $R=0.082$ L atm K⁻¹ mol⁻¹

Using the ideal gas law, $P_{total}V=n_{total,eq}RT$:

$2.46\times 200=(7+5\alpha )\times 0.082\times 600$

$492=(7+5\alpha )\times 49.2$

$10=7+5\alpha$

$3=5\alpha$

$\alpha =3/5=0.6$

Now, calculate the partial pressures at equilibrium.

Partial pressure of a gas = (mole fraction of gas) $\times$$P_{total}$

Mole fraction of PCl₅ = $\frac{5(1-\alpha )}{7+5\alpha }=\frac{5(1-0.6)}{7+5(0.6)}=\frac{5(0.4)}{7+3}=\frac{2}{10}=0.2$

Partial pressure of PCl₅ = $0.2\times 2.46=0.492$ atm

Mole fraction of PCl₃ = $\frac{5\alpha }{7+5\alpha }=\frac{5(0.6)}{7+3}=\frac{3}{10}=0.3$

Partial pressure of PCl₃ = $0.3\times 2.46=0.738$ atm

Mole fraction of Cl₂ = $\frac{5\alpha }{7+5\alpha }=\frac{5(0.6)}{7+3}=\frac{3}{10}=0.3$

Partial pressure of Cl₂ = $0.3\times 2.46=0.738$ atm

The equilibrium constant $K_p$ is given by:

$$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}$$

$$K_p=\frac{0.738\times 0.738}{0.492}=\frac{0.544644}{0.492}\approx 1.107$$

In the required format, $K_p=1.107\times 10^0=1107\times 10^{-3}$.

The nearest integer is 1107.