Question

For the following reaction in equilibrium

$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

Vapour density is found to be 100 when 1 mole of $PCl_5$ is taken in a 10 litre flask at $27^\circ C$. Calculate the equilibrium pressure.

• A. 1.57 atm
• B. 2.57 atm
• C. 3.57 atm
• D. 4.57 atm

Answer 1

Answer: (B)

2.57 atm

Answer 2

The reaction is $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. 1 mole of $PCl_5$ is taken. At equilibrium, let $\alpha$ be the degree of dissociation. The moles at equilibrium are: $PCl_5: (1-\alpha)$ $PCl_3: \alpha$ $Cl_2: \alpha$ Total moles at equilibrium, $n_{total} = (1-\alpha) + \alpha + \alpha = 1+\alpha$.

The molar mass of $PCl_5$ ($M_0$) is $31 + 5 \times 35.5 = 208.5$ g/mol. The vapour density (VD) of the equilibrium mixture is given as 100. The molar mass of the equilibrium mixture ($M_{mix}$) is $2 \times VD = 2 \times 100 = 200$ g/mol.

For a reaction $A(g) \rightleftharpoons nB(g)$, the relationship between molar masses and degree of dissociation is $M_{mix} = \frac{M_0}{1+(n-1)\alpha}$. In this case, $PCl_5 \rightleftharpoons PCl_3 + Cl_2$, so $n=2$. Thus, $M_{mix} = \frac{M_0}{1+(2-1)\alpha} = \frac{M_0}{1+\alpha}$.

Substituting the values: $200 = \frac{208.5}{1+\alpha}$ $1+\alpha = \frac{208.5}{200} = 1.0425$ So, the total moles at equilibrium are $n_{total} = 1+\alpha = 1.0425$ moles.

Now, we use the Ideal Gas Law, $PV = nRT$, to calculate the equilibrium pressure. Given: $n_{total} = 1.0425$ moles $V = 10$ litres $T = 27^\circ C = 27 + 273 = 300$ K $R = 0.0821$ L atm mol$^{-1}$ K$^{-1}$

$P = \frac{n_{total}RT}{V}$ $P = \frac{1.0425 \text{ mol} \times 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \times 300 \text{ K}}{10 \text{ L}}$ $P = 1.0425 \times 0.0821 \times 30$ $P = 2.56754625$ atm.

Rounding to two decimal places, the equilibrium pressure is approximately 2.57 atm.