Question

The process of eudiometry is used to calculate the volume of the various reacting gases present in a closed vessel. A spark is made and the change in volume or pressure is noted depending on the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. $H_{2}S{O}_4$ are used to absorb the selective gases like $O_2,O_3,C{O}_2$ and $H_{2}O$ (v) respectively.
A 4.0 mole mixture of $H_2\left(g\right),O_2\left(g\right)$ and He(g) are placed in a closed container at a pressure of 50 atm. An electric spark is passed, the pressure is noted to be 12.5 atm after cooling the contents to the room temperature. Oxygen gas is introduced to increase the pressure to 25 atm, keeping volume and the temperature constant. Again, an electrical spark is passed, and the pressure drops to 10 atm at room temperature conditions. List-I contains questions and List-II contains their answers. $\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ \text{(I)}& \text{What is the mole fraction of}{H}_2\left(g\right)\text{present}& \text{(P)}& 0.6\\ & \text{in the initial mixture?}& & \\ \text{(II)}& \text{How many moles of He(g) are present in the sample?}& \text{(Q)}& 1.0\\ & & & \\ \text{(III)}& \text{How many moles of oxygen remain in the final gaseous}& \text{(R)}& 0.8\\ & \text{mixture that is left?}& & \\ \text{(IV)}& \text{How many moles of}{O}_2\text{(g) are introduced in the vessel}& \text{(S)}& 0.2\\ & \text{after the first electrical spark?}& & \\ & & \text{(T)}& 0.7\\ & & \text{(U)}& 0.4\end{array}$
Which of the following options has the correct combination considering List-I and List-II

• A. (I), (R) and (II), (S)
• B. (II), (R) and (IV), (Q)
• C. (I), (T) and (III), (P)
• D. (III), (S) and (II), (T)

Answer 1

The correct option is C (I), (T) and (III), (P)

Let partial pressure of $H_2$ be x and that of $O_2$ be y then that of He will be 50-x-y all in atm.
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$
$\begin{array}{ccc}x& y& \text{before reaction}\\ x-2y& 0& \text{after reaction}\end{array}$
As there is decrease in pressure is 50-12.5 atm
So, 2y+y=50-12.5 $\Rightarrow$ y=12.5 atm
After first spark, 12.5 atm $O_2$ is introduced as pressure becomes 25 atm
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$
$\begin{array}{ccc}x-2y& 12.5& \text{before reaction in second stage reaction}\\ 0& 12.5-\frac{1}{2}(x-2y)& \text{after reaction}\end{array}$
Now $H_2$ will be limiting and decrease in pressure will be $\frac{3}{2}(x-2y)=15$ atm
$\Rightarrow \frac{3}{2}(x-2\times 12.5)=15\Rightarrow x=35atm$
So, partial pressure of He = 50-35-12.5 = 2.5 atm.
As 50 atm = 4 mol, So, 2.5 atm = 0.2 mol of He.
II) Thus moles of He present in the mixture sample is 0.2 mol.
III) Similarly since 12.5 atm of $O_2$ is added after the first spark, it means 1 mol of $O_2$ is introduced after the first spark.
Again,
Mole fraction of $H_2=\frac{35}{50}=0.7$
I) So mole fraction of $H_2$ present in the initial mixture is 0.7
Partial pressure of $O_2$ left in the final gaseous mixture =10-2.5 = 7.5 atm.
IV) So moles of oxygen remained in the final gaseous mixture = 0.6 mol.