Question

For the given reaction, $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ the number of moles at equilibrium was found to be 2, 6 & 4 respectively at 10 atm pressure. Find the new equilibrium moles of $PCl_5$, if it is restablished at an equilibrium pressure 50 atm.

Answer 1

4

Answer 2

1. Calculate Equilibrium Constant $K_p$: At 10 atm: $n_{PCl_5} = 2$, $n_{PCl_3} = 6$, $n_{Cl_2} = 4$. Total moles $n_{total} = 2 + 6 + 4 = 12$. Partial pressures: $P_{PCl_5} = \frac{2}{12} \times 10 = \frac{10}{6}$ atm $P_{PCl_3} = \frac{6}{12} \times 10 = 5$ atm $P_{Cl_2} = \frac{4}{12} \times 10 = \frac{10}{3}$ atm $K_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{5 \times \frac{10}{3}}{\frac{10}{6}} = 10$ atm.

2. Apply Le Chatelier's Principle: Increasing pressure from 10 atm to 50 atm shifts equilibrium to the side with fewer moles of gas. Here, $PCl_5$ (1 mole) is favored over $PCl_3 + Cl_2$ (2 moles).

3. Calculate New Equilibrium Moles: Let 'y' moles of $PCl_3$ and $Cl_2$ react to form $PCl_5$. New moles: $n'_{PCl_5} = 2+y$, $n'_{PCl_3} = 6-y$, $n'_{Cl_2} = 4-y$. New total moles $n'_{total} = (2+y) + (6-y) + (4-y) = 12-y$. Partial pressures at 50 atm: $P'_{PCl_5} = \frac{2+y}{12-y} \times 50$ $P'_{PCl_3} = \frac{6-y}{12-y} \times 50$ $P'_{Cl_2} = \frac{4-y}{12-y} \times 50$

4. Solve for y: Using $K_p = 10$ atm: $10 = \frac{\left(\frac{6-y}{12-y} \times 50\right) \times \left(\frac{4-y}{12-y} \times 50\right)}{\left(\frac{2+y}{12-y} \times 50\right)}$ $10 = \frac{(6-y)(4-y)}{(12-y)(2+y)} \times 50$ $\frac{1}{5} = \frac{y^2 - 10y + 24}{24 + 10y - y^2}$ $24 + 10y - y^2 = 5(y^2 - 10y + 24)$ $6y^2 - 60y + 96 = 0$ $y^2 - 10y + 16 = 0$ $(y-2)(y-8) = 0$. Since $y$ cannot be greater than the initial moles of reactants (6 and 4), $y=2$ is the valid solution.

5. Find New Equilibrium Moles of $PCl_5$: $n'_{PCl_5} = 2 + y = 2 + 2 = 4$ moles.