Question

the vapour density of pcl5 at 200 c and 252 c are 70.2 and 57.3 at one hemisphere.Calculate its value of dissociatoon constant ta these temperatures.

Answer 1

Answer:

0.235 at 200 degree C & 0.671 at 252 degree C

here's the answer and hope it helps you.

Answer 2

Answer:

Value of dissociation constant ta these temperatures is $K_{P}=3.75 \mathrm{~atm}$

Explanation:

Given: Vapour density

To find: Value of dissociation constant of these temperatures

Solution:

Hint: Vapour density is the density of a vapour in relation to that of hydrogen. It may be defined as the mass of a certain volume of a substance divided by mass of the same volume of hydrogen. vapour density = mass of n molecules of gas / mass of n molecules of hydrogen.

The molar mass is given by $=2 \times$ vapour density

The vapour density of $P C l_{5} at 200^{\circ} \mathrm{C}$is given as $70.2$.

Thus molar mass will be $=$.

$P C l_{5} \rightleftharpoons P C l_{3}+C l_{2}$

The molar mass of the undissociated $P C l_{5} 208.2 \mathrm{~g} / \mathrm{mol}$, the molar mass of $P C l_{3} is 137.3 \mathrm{~g} / \mathrm{mol}$ and the molar mass of $C l_{2} is 71 \mathrm{~g} / \mathrm{mol}$.Now consider $x$ to be the degree of dissociation at $200^{\circ} C$, then the average molar mass of the mixture is given by:

$AMM=\frac{n_{1} M_{1}+n_{2} M_{2}+n_{3} M_{3}}{n_{1}+n_{2}+n_{3}}$

Where, $n_{1}=(1-x), n_{2}=x, n_{3}=x$

$M_{1}=208.2 \mathrm{~g} / \mathrm{mol} \\ M_{2}=137.3 \mathrm{~g} / \mathrm{mol} \\ M_{3}=71 \mathrm{~g} / \mathrm{mol}$

Thus, substituting the values in the above equation, we have:

$AMM=\frac{208.2(1-x)+137.3 x+71 x}{1-x+x+x}=140.4 g / m o l$

Thus, solving for the value $x$ now, we get:

$208.2+0.1 x=140.4 x+140.4$

$x=0.483$

Now we can easily calculate the number of moles of the gases easily,

The number of moles of $P C l_{3}= the number of moles of C l_{2}=x=0.483$

The number of moles of $P C l_{5}=1-0.483=0.517$

The partial pressure of $P C l_{3}= the partial pressure of C l_{2}=0.483 \times 1 a t m=0.483 a t m$

The partial pressure of $P C l_{5}=0.517 \times 1 a t m=0.517 a t m$

The equilibrium constant $\left(K_{p}\right)$ for the reaction can be written as:

$\begin{aligned}K_{p} &=\frac{P_{P C l_{3}} \times P_{C l_{2}}}{P_{P C l_{5}}}=\frac{0.483 \times 0.483}{0.517} \\K_{p} &=0.452 a t m\end{aligned}$

The molar mass is given as $2 \times$ vapour density

The vapour density of $P C l_{5} at 252^{\circ} \mathrm{C} is 57.2$

The molar mass $=2 \times 57.2=114.4 \mathrm{~g} / \mathrm{mol} .$

The number of moles of $P C l_{3}= the number of moles of C l_{2}=x=0.821$

The number of moles of $P C l_{5}=1-0.821=0.179$

The partial pressure of $P C l_{3}= the partial pressure of C l_{2}=0.821 \times 1 \mathrm{~atm}=0.821 \mathrm{~atm}$

The partial pressure of $P C l_{5}=0.179 \times 1 \mathrm{~atm}=0.179 \mathrm{~atm}$

Thus, the equilibrium constant $\left(K_{p}\right)$ for the reaction will be:

$\begin{aligned}&K_{p}=\frac{P_{P C l_{3}} \times P_{C l_{2}}}{P_{P C l_{5}}} \\&K_{p}=\frac{0.821 \times 0.821}{0.179} \\&K_{P}=3.75 \mathrm{~atm}\end{aligned}$