Question

1. phosphorous pentachloride is a fine grained powder, ranging in colour from colourless to pale yellow. it has a pungent odour. in the gaseous state it shows the equilibrium shown in the equation below. initially, a sample of phosphorus pentachloride is introduced to an empty vessel. if the equilibrium constant for the above reaction is 4.0 × 10⁻⁴ and the equilibrium concentration of pcl₅(g) is 0.090 mol/l, the equilibrium concentration of pcl₃(g) is ______ mol/l. show all your work.

pcl₅(g) ⇌ pcl₃(g) + cl₂(g)

Explanation:

Step1: Recall Equilibrium Constant Formula

For the reaction $\ce{PCl_{5}(g)
ightleftharpoons PCl_{3}(g) + Cl_{2}(g)}$, the equilibrium constant expression $K_c$ is given by $K_c = \frac{[\ce{PCl_{3}}][\ce{Cl_{2}}]}{[\ce{PCl_{5}}]}$. At equilibrium, $[\ce{PCl_{3}}] = [\ce{Cl_{2}}]$ (since they are produced in a 1:1 ratio from $\ce{PCl_{5}}$ decomposition, and initially only $\ce{PCl_{5}}$ is present). Let $x = [\ce{PCl_{3}}] = [\ce{Cl_{2}}]$.

Step2: Substitute Known Values into $K_c$ Expression

We know $K_c = 4.0 \times 10^{-4}$, $[\ce{PCl_{5}}] = 0.090\ \text{mol/L}$. Substituting into $K_c = \frac{[\ce{PCl_{3}}][\ce{Cl_{2}}]}{[\ce{PCl_{5}}]}$ gives $4.0 \times 10^{-4} = \frac{x \cdot x}{0.090}$.

Step3: Solve for $x$

First, multiply both sides by $0.090$: $4.0 \times 10^{-4} \times 0.090 = x^2$. Calculate the left - hand side: $4.0 \times 10^{-4} \times 0.090 = 3.6\times 10^{-5}$. Then take the square root of both sides: $x=\sqrt{3.6\times 10^{-5}}$. $\sqrt{3.6\times 10^{-5}}=\sqrt{36\times 10^{-6}} = 6.0\times 10^{-3}\ \text{mol/L}$.

Answer:

$6.0\times 10^{-3}$