Paragraph:The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc.
H2SO4 are used to absorb the selective gases like
O2,O3,CO2 and
H2O (v) respectively. A 2.0 mole mixture of
H2( g),O2( g) and
He(g) are placed together in a closed container at pressure equal to 50 atm . An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm , keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions.
Question:
The mole fraction of
H2( g) present in the initial mixture is:
Step-by-step solution
Let the no. of moles of
H2,O2andHeare\mathrm{x}, \mathrm{y}
andzrespectively\therefore x+y+z=2.0
Pressureexertedby2molesofgaseousmixtureis50atm.Afterthefirstelectricsparkdecreaseinpressure=50-12.5=37.5 \mathrm{~atm}\therefore
decreaseinno.ofmolesofgaseousmixtureis1.5Butfromthegiveninformationlimitingreagentinfirststepis\mathrm{O}_2\therefore
from\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\ell)\therefore \mathrm{H}_2
and\mathrm{O}_2
reactedareintheratioof2: 1\therefore 0.5
moleof\mathrm{O}_2
shouldbereactedwith1.0moleof\mathrm{H}_2\therefore
No.ofmoleof\mathrm{O}_2
inthemixture(\mathrm{y})=0.5
Whenagain\mathrm{O}_2
ispassed,pressureincreasedfrom12.5to25.0atm\therefore
Changeinpressure=12.5 \mathrm{~atm}\therefore
No.ofmolesof\mathrm{O}_2
added=\frac{12.5 \times 2}{50}=0.5 \mathrm{~mole}
Now\mathrm{H}_2
willbecompletelyreactedNo.ofmoles\mathrm{H}_2
actuallyleftafterfirstelectricspark=x-1
H2+21O2⟶H2O(x−1)(2x−1)Nowchangeinpressurein25-10=15 \mathrm{~atm}\therefore
changeinno.ofmolesis\frac{15 \times 2}{50}=0.6
∴(x−1)+(2x−1)=0.6⇒x=1.4\therefore
no.ofmolesof\mathrm{H}_2=1.4\therefore
no.ofmolesof\mathrm{He}=2.0-(1.4+0.5)=0.1\therefore
molefractionof\mathrm{H}_2=\frac{1.4}{2.0}=0.7
