Questions: Phosphorus pentachloride decomposes according to the chemical equation PCl5(g) ⇌ PCl3(g) + Cl2(g) Kc=1.80 at 250 C A 0.1915 mol sample of PCl5(g) is injected into an empty 3.15 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium. [PCl5]= [PCl3]=

Solution Steps

First, we need to calculate the initial concentration of \(\mathrm{PCl}_5\) in the reaction vessel. The initial moles of \(\mathrm{PCl}_5\) is 0.1915 mol, and the volume of the vessel is 3.15 L.

\[ [\mathrm{PCl}_5]_{\text{initial}} = \frac{0.1915 \, \text{mol}}{3.15 \, \text{L}} = 0.0608 \, \text{M} \]

The equilibrium expression for the reaction is given by:

\[ K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} \]

Given \(K_c = 1.80\).

Let \(x\) be the change in concentration of \(\mathrm{PCl}_5\) that decomposes at equilibrium. Then, the changes in concentration are:

• \([\mathrm{PCl}_5] = 0.0608 - x\)
• \([\mathrm{PCl}_3] = x\)
• \([\mathrm{Cl}_2] = x\)

Substitute the equilibrium concentrations into the equilibrium expression:

\[ 1.80 = \frac{x \cdot x}{0.0608 - x} = \frac{x^2}{0.0608 - x} \]

Rearrange the equation to form a quadratic equation:

\[ 1.80(0.0608 - x) = x^2 \]

\[ 0.1094 - 1.80x = x^2 \]

\[ x^2 + 1.80x - 0.1094 = 0 \]

Use the quadratic formula to solve for \(x\):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 1\), \(b = 1.80\), and \(c = -0.1094\).

\[ x = \frac{-1.80 \pm \sqrt{(1.80)^2 - 4 \cdot 1 \cdot (-0.1094)}}{2 \cdot 1} \]

\[ x = \frac{-1.80 \pm \sqrt{3.24 + 0.4376}}{2} \]

\[ x = \frac{-1.80 \pm \sqrt{3.6776}}{2} \]

\[ x = \frac{-1.80 \pm 1.9172}{2} \]

The positive root is:

\[ x = \frac{0.1172}{2} = 0.0586 \]

Now, calculate the equilibrium concentrations:

• \([\mathrm{PCl}_5] = 0.0608 - 0.0586 = 0.0022 \, \text{M}\)
• \([\mathrm{PCl}_3] = 0.0586 \, \text{M}\)

Final Answer