Question

The vapor density of \[PC{l_3}\] at \[{200^ \circ }C\] and \[{252^ \circ }C\] are 70.2 and 57.2 respectively at one atmosphere. Calculate the value of dissociation constant at these given temperatures.

Answer 1

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.

Complete step by step answer:
The molar mass is given by= 2 × vapour density
The vapour density of \[PC{l_5}\] at \[{200^ \circ }C\] is given as 70.2.
Thus molar mass will be =.

\[PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}\]

The molar mass of the undissociated \[PC{l_5}\]​ 208.2 g/mol, the molar mass of \[PC{l_3}\]​ is 137.3 g/mol and the molar mass of \[C{l_2}\]is​ 71 g/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:
$AMM = \dfrac{{{n_1}{M_1} + {n_2}{M_2} + {n_3}{M_3}}}{{{n_1} + {n_2} + {n_3}}}$
Where, ${n_1} = (1 - x),{n_2} = x,{n_3} = x$
${M_1} = 208.2g/mol$
${M_2} = 137.3g/mol$
${M_3} = 71g/mol$
Thus, substituting the values in the above equation, we have:
\[AMM = \dfrac{{208.2(1 - x) + 137.3x + 71x}}{{1 - x + x + x}} = 140.4g/mol\]​
Thus, solving for the value $x$ now, we get:
\[208.2 + 0.1x = 140.4x + 140.4\]
\[x = 0.483\]
Now we can easily calculate the number of moles of the gases easily,
The number of moles of \[PC{l_3}\]​= the number of moles of \[C{l_2}\]​\[ = x = 0.483\]
The number of moles of \[PC{l_5}\]​\[ = 1 - 0.483 = 0.517\]

The partial pressure of \[PC{l_3}\]​= the partial pressure of \[C{l_2}\]​\[ = 0.483 \times 1atm = 0.483atm\]
The partial pressure of \[PC{l_5}\]​\[ = 0.517 \times 1atm = 0.517atm\]
The equilibrium constant (${K_p}$ ) for the reaction can be written as:
\[{K_p} = \dfrac{{{P_{PC{l_3}}} \times {P_{C{l_2}}}}}{{{P_{PC{l_5}}}}} = \dfrac{{0.483 \times 0.483}}{{0.517}}\]
\[{K_p} = 0.452atm\]
The molar mass is given as 2×vapour density
The vapour density of \[PC{l_5}\]​ at \[{252^ \circ }C\]is \[57.2\]
The molar mass \[ = 2 \times 57.2 = 114.4\;g/mol\] .
\[PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}\]
The molar mass of undissociated \[PC{l_5}\]​ \[ = 208.2{\text{ }}g/mol\] .
The molar mass of \[PC{l_3}\]​ =\[137.3{\text{ }}g/mol\] .
The molar mass of \[C{l_2}\]=\[71{\text{ }}g/mol\]
Now consider \[x\] to be the degree of dissociation at \[{252^ \circ }C\], then
The average molar mass of the mixture will be given as
$AMM = \dfrac{{{n_1}{M_1} + {n_2}{M_2} + {n_3}{M_3}}}{{{n_1} + {n_2} + {n_3}}}$
=\[\dfrac{{208.2(1 - x) + 137.3x + 71x}}{{1 - x + x + x}}\]​=114.4
\[208.2 - 208.2x + 137.3x = 114.4x + 114.4\]
Thus we get \[x\] as,
\[x = 0.821\] Now we can calculate the number of moles of the gases
The number of moles of\[PC{l_3}\]​= the number of moles of \[C{l_2}\]​=\[x = 0.821\]
The number of moles of \[PC{l_5}\]​=\[1 - 0.821 = 0.179\]
The partial pressure of \[PC{l_3}\]​= the partial pressure of \[C{l_2}\]​=\[0.821 \times 1atm = 0.821atm\]
The partial pressure of \[PC{l_5}\]​=0.179×1atm=\[0.179atm\]
Thus, the equilibrium constant (${K_p}$ ) for the reaction will be:
\[{K_p} = \dfrac{{{P_{PC{l_3}}} \times {P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}\]​
${K_p} = \dfrac{{0.821 \times 0.821}}{{0.179}}$
${K_P} = 3.75\,atm$


Note:
In a mixture of gases, we see that each constituent gas has a partial pressure which is the pressure of that constituent gas if it was alone present in the entire volume of the original mixture at the same temperature. The total pressure of an ideal gas mixture is given by the sum of the partial pressures of the gases in the mixture.