Question

Decomposition of non-volatile solute (A) into another non-volatile solute B and C. When dissolved in water follows first order kinetics as: $A(s)\xrightarrow{{{H_2}O}}2B(s) + C(s)$ If initially two moles of A is dissolved in 360g of ${H_2}O$ and left for decomposition at constant temperature $(25^\circ C)$. Then ${P_1}$ in the given table is: (assuming A, B and C are miscible in water)

S.No.	Time	The vapour pressure of solution
1	1 hr	20 mm Hg
2	80 hr	${P_1}$

Vapour pressure of ${H_2}O$ at $(25^\circ C)$ is 24mm Hg. (log 2 $ = $ 0.30)
A) 15mm Hg
B) 18.75mm Hg
C) 24mm Hg
D) 19.2mm Hg

Answer 1

Hint: The solute undergoing decomposition is non-volatile so, the vapour pressure for non-volatile solute is defined in terms of vapour pressure of solvent at the given temperature with respect to its mole fraction. The formula of vapour pressure of non-volatile solute can be represented as follows: ${P_{sol}} = P^\circ (1 - {\chi _{solute}})$, here ${P_{sol}}$ is the vapour pressure of the solution, $P^\circ $ is the given value of pressure and ${\chi _{solute}}$represents the mole fraction of solute.

Complete step by step answer:
As per the given question Initial concentration (in moles) of A is 2 and ${H_2}O$ is 20
It is given in question that $P^\circ = $ 24 mm Hg and ${P_{sol}} = $ 20 mm Hg after 12 h (where $P$ stands for vapour pressure).
We know that,
${P_{sol}} = P^\circ (1 - {\chi _{solute}})$
Where ${\chi _{solute}} = $ mole fraction of solute
So after 12 hr of reaction let us assume $x$ mole of A reacted with 20 mole of water to produce $2x$ mole of B and $x$ mole of C.
Therefore mole fraction of different solute present in the reaction is given by:
${\chi _{solute}} = {\chi _A} + {\chi _B} + {\chi _C}$
Where ${\chi _A} = \dfrac{{{n_A}}}{n}$ (where, ${n_A} = $ number of moles of A and $n = $ total number of moles of solute)
Let us calculate the mole fraction of different solute present in the reaction
${\chi _A} = \dfrac{{2 - x}}{{22 + 2x}}$
${\chi _B} = \dfrac{{2x}}{{22 + 2x}}$
${\chi _C} = \dfrac{x}{{22 + 2x}}$
Therefore,
${\chi _{solute}} = \dfrac{{2 - x}}{{22 + 2x}} + \dfrac{{2x}}{{22 + 2x}} + \dfrac{x}{{22 + 2x}} = \dfrac{{2 + 2x}}{{22 + 2x}}$
Substituting value of ${\chi _{solute}}$ we get
$\dfrac{4}{{24}} = \dfrac{{2 + 2x}}{{22 + 2x}}$
$ \Rightarrow x = 1$
Now to find out the mole fraction after 80 hours let us head towards the reaction kinetics of the given reaction:
We know that for first order reaction:
$\ln \left( {\dfrac{{[{A_0}]}}{{[{A_t}]}}} \right) = kt$ (Where ${A_0},{A_t}$ are the concentration and $k$ is the rate constant)
After 12 hour the concentration are as follows:
$[{A_0}] = \dfrac{{(2) \times 1000}}{{360}}$
$[{A_{12}}] = \dfrac{{(2 - x) \times 1000}}{{360}} = \dfrac{{(1) \times 1000}}{{360}}$
On substituting the above concentration in first order reaction we get
$\ln \left( {\dfrac{2}{1}} \right) = k \times 12$
$ \Rightarrow k = \dfrac{1}{{40}}$
Let the number of moles of A, B, C after 80 h be:
${n_A} = 2 - y,{n_B} = 2y,{n_C} = y$
Using the first order rate law we get:
$\ln \left( {\dfrac{2}{{2 - y}}} \right) = 80 \times k$
$\ln \left( {\dfrac{2}{{2 - y}}} \right) = 80 \times \dfrac{1}{{40}}$ or $\ln (22 - y) = 80 \times 140 = 2$
$ \Rightarrow 2 - y = {e^{ - 17}}$
$ \Rightarrow y = 1.82$
Therefore mole fraction of solute after 80 h is:
${\chi _{solute}} = \dfrac{{2 + 2(1.82)}}{{22 + 2(1.82)}}$
$ \Rightarrow {\chi _{solute}} = 0.22$
 Now as we know that
${P_{sol}} = P^\circ (1 - {\chi _{solute}})$
Therefore,
${P_{sol}} = 24 \times (1 - 0.22)$
$ \Rightarrow {P_{sol}} = 18.72mm\;of\;Hg$

Therefore, option B is correct.

Note: Vapour pressure is defined as the measure of the ability of a material to change its state either into the vapour or gaseous state. Vapour pressure is directly proportional to temperature, with increase in temperature it increases.
In this question, vapour pressure for non-volatile is used whereas for volatile solute it can be defined as summation of vapour pressure of the solvents present in the solution and vapour pressure of the solutes present in the solution.