# Question 26e77

Jan 29, 2017

$\textsf{\left(a\right)}$ 1.00 Atm

$\textsf{\left(b\right)}$ 0.0745 Atm

$\textsf{\left(c\right)}$ 1.92 Atm

$\textsf{\left(d\right)}$ $\textsf{\alpha = 0.925}$

#### Explanation:

$\textsf{\left(a\right)}$

$\textsf{P C {l}_{5 \left(g\right)} r i g h t \le f t h a r p \infty n s P C {l}_{3 \left(g\right)} + C {l}_{2 \left(g\right)}}$

For which:

$\textsf{{K}_{p} = \frac{{p}_{P C {l}_{3}} \times {p}_{C {l}_{2}}}{{p}_{P C {l}_{5}}} = 11.5 \text{ ""Atm}}$

No dimensions have been given for $\textsf{{K}_{p}}$ or any value for the gas constant R. Subsequent calculations suggest it is in atmospheres so I have gone with that.

To find the pressure if no dissociation has taken place we use the ideal gas equation:

$\textsf{P V = n R T}$

$\therefore$$\textsf{P = \frac{n R T}{V}}$

To find $\textsf{n}$ we use:

$\textsf{n = \frac{m}{M} _ \left(r\right) = \frac{2.010}{208.24} = 0.009652}$

$\therefore$$\textsf{P = \frac{0.009652 \times 0.082 \times 600}{0.475} \text{ ""Atm}}$

$\textsf{P = 1.00 \textcolor{w h i t e}{x} \text{Atm}}$

$\textsf{\left(b\right)}$

We set up an ICE table based on atmospheres:

$\textsf{\text{ } P C {l}_{5 \left(g\right)} r i g h t \le f t h a r p \infty n s P C {l}_{3 \left(g\right)} + C {l}_{2 \left(g\right)}}$

$\textsf{I \text{ "1.00" "0" } 0}$

$\textsf{C \text{ "-x" "+x" } + x}$

$\textsf{E \text{ "(1.00-x)" "x" } x}$

$\therefore$$\textsf{{K}_{p} = {x}^{2} / \left(1.00 - x\right) = 11.5}$

$\therefore$$\textsf{{x}^{2} = 11.5 \left(1.00 - x\right)}$

$\Rightarrow$$\textsf{{x}^{2} + 11.5 x - 11.5 = 0}$

Applying the quadratic formula and ignoring the -ve root we get:

$\textsf{x = 0.9255 \textcolor{w h i t e}{x} \text{Atm}}$

$\therefore$$\textsf{{p}_{P C {l}_{5}} = 1.00 - 0.9255 = 0.0745 \textcolor{w h i t e}{x} \text{Atm}}$

$\textsf{\left(c\right)}$

The total pressure is given by:

sf(P_("tot")=(1.00-x)+x+x=(1.00+x)

$\therefore$sf(P_("tot")=1.00+0.9255=1.9255color(white)(x)"Atm")#

$\textsf{\left(d\right)}$

The degree of dissociation $\textsf{\alpha}$ is the fraction of moles of $\textsf{P C {l}_{5}}$ which have dissociated.

Since partial pressure is proportional to the number of moles we get:

$\textsf{\alpha = \frac{0.9255}{1.00} = 0.9255}$

$\textsf{\alpha = 0.925}$

This tells us that the $\textsf{P C {l}_{5}}$ is 92.5% dissociated.