Question

At $353K$, the vapour pressure of pure ethylene bromide and propylene bromide are $22.93$ and $16.93kN{m^{ - 2}}$ respectively, and these compounds form a nearly ideal solution. $3mol$ of ethylene bromide and $2mol$ of propylene bromide are equilibrated at $553K$ and a total pressure of $20.4kN{m^{ - 2}}$
(a) what is the composition of the liquid phase?
(b) what amount of each compound is present in the vapour phase?
A. (a) $0.578,0.422$ (b) $0.9967mol,0.5374mol$
B. (a) $0.345,0.422$ (b) $0.9967mol,0.434mol$
C. (a) $0.578,0.212$ (b) $0.9967mol,0.5374mol$
D. (a) $0.578,0.212$ (b) $0.497mol,0.5374mol$

Answer 1

We can use Raoult's Law to find the composition of the liquid phase. With this, we can then find the partial pressure of each component and use that to find the composition in the vapour phase. As we are given the number of moles initially present, if we can find the total number of moles present in the vapour phase at equilibrium, we can also find the amount of each component.
Formula used: ${p_A} = {x_A}P_A^S,{p_B} = {x_B}P_B^S$
Where ${p_A}$ and ${p_B}$ are the partial pressures of component A (ethylene bromide) and component B (propylene bromide) respectively, ${x_A}$ and ${x_B}$ are the mole fractions of each component in the liquid phase and $P_A^S,P_B^S$ are the vapour pressures of components A and B.
${p_A} = {y_A}P,{p_B} = {y_B}P$
Where ${y_A}$ and ${y_B}$ are the mole fractions of respective components in the vapour phase and $P$ is the total pressure.
${y_A} = \dfrac{{n_A^V}}{{n_A^V + n_B^V}}$ and ${x_A} = \dfrac{{n_A^L}}{{n_A^L + n_B^L}}$
Where $n_A^V$ and $n_B^V$ are the moles of A and B that are present in the vapour phase, and $n_A^L$ and $n_B^L$ are the moles of A and B present in the liquid phase.

Complete answer:
Raoult’s Law states that partial pressure of a component is the product of its mole fraction in liquid phase and vapour pressure at a particular temperature. Thus,
${p_A} = {x_A}P_A^S,{p_B} = {x_B}P_B^S$
Where ${p_A}$ and ${p_B}$ are the partial pressures of component A (ethylene bromide) and component B (propylene bromide) respectively, ${x_A}$ and ${x_B}$ are the mole fractions of each component in the liquid phase and $P_A^S,P_B^S$ are the vapour pressures of components A and B.
The sum of both partial pressures is the total pressure. Thus, by adding, we get:
${p_A} + {p_B} = {x_A}P_A^S + {x_B}P_B^S = P$
As we know, sum of mole fractions is 1. Therefore,
${x_A} + {x_B} = 1 \Rightarrow {x_B} = 1 - {x_A}$
Substituting this in our previous equation, we get:
$P = {x_A}P_A^S + (1 - {x_A})P_B^S = {x_A}P_A^S + P_B^S - {x_A}P_B^S$
Rearranging this, we get:
$P - P_B^S = {x_A}P_A^S - {x_A}P_B^S \Rightarrow P - P_B^S = {x_A}(P_A^S - P_B^S)$
Therefore, mole fraction of component A in liquid phase,
${x_A} = \dfrac{{P - P_B^S}}{{(P_A^S - P_B^S)}}$
From the data given, $P = 20.4kN{m^{ - 2}},P_A^S = 22.93kN{m^{ - 2}},P_B^S = 16.93kN{m^{ - 2}}$
Substituting our values into this equation, we get:
${x_A} = \dfrac{{20.4 - 16.93}}{{22.93 - 16.93}}$
On solving, we get: ${x_A} = 0.578$
$\Rightarrow {x_B} = 1 - {x_A} = 1 - 0.578 = 0.422$
Thus, we have now obtained the liquid phase compositions
The Dalton’s Law states that the partial pressure of a component in a gaseous mixture is equal to the product of its mole fraction in vapour phase and the total pressure. That is,
${p_A} = {y_A}P,{p_B} = {y_B}P$
Therefore, mole fraction in vapour phase,
${y_A} = \dfrac{{{p_A}}}{P},{y_B} = \dfrac{{{p_B}}}{P}$
Where ${y_A}$ and ${y_B}$ are the mole fractions of respective components in the vapour phase and $P$ is the total pressure
But we know ${p_A} = {x_A}P_A^S,{p_B} = {x_B}P_B^S$
Substituting this in our previous equation, we get:
${y_A} = \dfrac{{{x_A}P_A^S}}{P}$
As we have obtained earlier, ${x_A} = 0.578,P_A^S = 22.93kN{m^{ - 2}},P = 20.4kN{m^{ - 2}}$
Substituting our values, we get:
${y_A} = \dfrac{{0.578 \times 22.93}}{{20.4}} = 0.6497$
Now, let us take $n_A^V$ and $n_B^V$ as the moles of A and B that are present in the vapour phase, and $n_A^L$ and $n_B^L$ as the moles of A and B present in the liquid phase.
Hence, by the definition of mole fraction, ${y_A} = \dfrac{{n_A^V}}{{n_A^V + n_B^V}} = 0.6497$
On rearrangement, we get:
$\dfrac{{n_A^V}}{{0.6497}} = n_A^V + n_B^V$
As initially $3mol$ of ethylene bromide and $2mol$ of propylene bromide are present, sum of moles in liquid and vapour phase for A and B will be equal to $3mol$ and $2mol$ respectively. Hence,
$3mol = n_A^L + n_A^V,2mol = n_B^L + n_B^V$
Therefore, since ${x_A} = \dfrac{{n_A^L}}{{n_A^L + n_B^L}}$
$\Rightarrow {x_A} = \dfrac{{3 - n_A^V}}{{(3 - n_A^V) + (2 - n_B^V)}} = 0.578$
Rearranging this, we get:
$\dfrac{{3 - n_A^V}}{{(3 + 2) - (n_A^V + n_B^V)}} = 0.578 \Rightarrow 3 - n_A^V = 0.578\left[ {5 - (n_A^V + n_B^V)} \right]$
Substituting the value we got earlier, $n_A^V + n_B^V = \dfrac{{n_A^V}}{{0.6497}}$
$\Rightarrow 3 - n_A^V = 0.578\left[ {5 - \dfrac{{n_A^V}}{{0.6497}}} \right]$
On solving this, we get:
$3 - n_A^V = 2.89 - 0.8896n_A^V$
Hence, $n_A^V = 0.9967mol$
Substituting this into our equation for mole fraction in vapour phase, $\dfrac{{n_A^V}}{{0.6497}} = n_A^V + n_B^V$, we get:
$\dfrac{{0.9967}}{{0.6497}} = 0.9967 + n_B^V \Rightarrow n_B^V = 1.53409 - 0.9967$
On solving this, we get:
$n_B^V = 0.5374mol$
Therefore, the values we have obtained are: liquid phase composition $= 0.578,0.422$ and amount of each compound present in vapour phase $n_A^V = 0.9967mol,n_B^V = 0.5374mol$
Hence, the correct set of options is A.

Note:
While the Raoult’s Law can be applied only when the binary mixture of components forms an ideal solution, the Dalton’s Law of partial pressures is applicable to all gaseous mixtures. Note that the vapour pressure of a compound is a direct function of the temperature. That is, these experiments are thought to be carried out in isothermal conditions, so that the vapour pressures don’t change. Also note that the sum of individual mole fractions of all the components in any mixture will always be equal to 1.