Question

Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation: pcl5(g)→pcl3(g)+cl2(g) at 250° 0.125 m pcl5 is added to the flask. if kc = 1.80, what are the equilibrium concentrations of each gas?

Answer 1

Answer:

$\boxed{\text{[PCl _{5} ] = 0.0077 mol/L; [PCl _{3} ] = [Cl _{2} ] = 0.117 mol/L}}$

Explanation:

The balanced equation is

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

You don't give the volume of the flask, so I assume it is 1 L.

We can set up an ICE table to organize our calculations.

$\begin{array}{lccccc} & \text{PCl}_{5} & \rightleftharpoons & \text{PCl}_{3} & + & \text{Cl}_{2} \\\text{I/mol}\cdot\text{L}^{-1}: & 0.125 & & 0 & & 0 &\\\text{C/mol}\cdot\text{L}^{-1}: & -x & & +x & & +x &\\\text{E/mol}\cdot\text{L}^{-1}:& 0.125-x & & x & & x &\\\end{array}$

$K_{\text{c}} = \dfrac{\text{[PCl _3 ][Cl _2 ]}}{\text{[PCl _5 ]}} = \dfrac{x^{2}}{0.125-x} = 1.80$

Check for negligibility

$\dfrac{0.125}{1.80} = 0.0674 < 400.$

x is not negligible, so we must solve a quadratic.

$x^{2} = 1.80(0.125 - x)\\x^{2} = 0.225 - 1.80x\\x^{2} + 1.80x - 0.225 = 0$

Solve for x.

x = 0.1173

[PCl₅] = 0.125 - x = 0.125 – 0.1173 = 0.0077 mol·L⁻¹

[PCl₃] = x = 0.117 mol·L⁻¹

 [Cl₂] = x = 0.117 mol·L⁻¹

$\boxed{\textbf{[PCl _{5} ] = 0.0077 mol/L; [PCl _{3} ] = [Cl _{2} ] = 0.117 mol/L}}$

Check:

$\dfrac{0.117^{2}}{0.0077} = 1.80\\\\\dfrac{0.0138}{0.0077} = 1.79$

Close enough. It checks.