Question

A 2.4156-g sample of PCl₅ was placed in an empty 2.000-L flask and allowed to decompose to PCl₃ and Cl₂ at 250.0^oC:

$\mathit{P}\mathit{C}{\mathit{I}}_{\mathbf{5}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{P}\mathit{C}{\mathit{I}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At equilibrium the total pressure inside the flask was observed to be 358.7 torr.

a. Calculate the partial pressure of each gas at equilibrium and the value of K_pat 250.0^oC.

b. What are the new equilibrium pressures if 0.250 mole of Cl₂ gas is added to the flask?

Short answer

a. The equilibrium partial pressures of PCI₅(g) , PCI₃(g) and CI₂(g) are 0.0259atm , 0.2230atm and 0.2230atm respectively. The value of K_p is 1.92.

b. The new equilibrium partial pressures of PCI₅(g) , PCI₃(g) and CI₂(g) are 0.1839atm , 0.0650atm and 5.44atm respectively.

Step-by-step solution

Calculate the initial pressure of PCl5.

The initial pressure of PCl₅ can be calculated by the ideal gas law as follows:

$P=\frac{nRT}{V}$ …(1)

Here,

n is the number of moles.

P is the pressure.

V is the volume.

T is the temperature.

R is the gas constant, 0.08206L atm/molK .

$\begin{array}{rcl}{P}_{PC{I}_3}& =& \frac{{\displaystyle \frac{2.4156g}{208.22g/mol}}\times 0.08206Latm/mol\times 523.2K}{2.000L}\\ & =& 0.2490atm\\ & =& 189.2torr\end{array}$

The initial pressure of PCl₅ is 189.2torr.